11.6 SPECIFIC HEAT CAPACITY average energy of 2 × ½ kBT = kBT. In three dimensions, the average energy is 3 kBT.Suppose an amount of heat ∆Q supplied to a For a mole of a solid, the total energy is substance changes its temperature from T to T + ∆T. We define heat capacity of a substance U = 3 kBT × NA = 3 RT (∵kBT × NA = R) (see Chapter 10) to be Now, at constant pressure, ∆Q = ∆U + P ∆V ≅ ∆ Q ∆U, since for a solid ∆V is negligible. Therefore, S = (11.4) ∆ T ∆ Q ∆U C = = = 3 R (11.7) We expect ∆Q and, therefore, heat capacity S ∆ T ∆ T to be proportional to the mass of the substance. Table 11.1 Specific and molar heat capacities Further, it could also depend on the of some solids at room temperature, i.e., a different amount of heat may temperature and atmospheric be needed for a unit rise in temperature at pressure different temperatures. To define a constant –v Speci"c heat Molar speci"c characteristic of the substance and Substance –1 –1 –1 –1 (J kg K ) heat (J mol K ) independent of its amount, we divide S by the mass of the substance m in kg : S  1  ∆Q s = (11.5) m =  m  ∆T s is known as the specific heat capacity of the substance. It depends on the nature of the As Table 11.1 shows, the experimentally substance and its temperature. The unit of measured values which generally agrees withspecific heat capacity is J kg–1 K–1. Reprint 2025-26 232 PHYSICS predicted value 3R at ordinary temperatures. ideal gas, we have a simple relation. (Carbon is an exception.) The agreement is known to break down at low temperatures. Cp – Cv = R (11.8) Specific heat capacity of water where Cp and Cv are molar specific heat capacities of an ideal gas at constant pressure The old unit of heat was calorie. One calorie and volume respectively and R is the universal was earlier defined to be the amount of heat gas constant. To prove the relation, we begin required to raise the temperature of 1g of water with Eq. (11.3) for 1 mole of the gas : by 1°C. With more precise measurements, it was found that the specific heat of water varies ∆Q = ∆U + P ∆V slightly with temperature. Figure 11.5 shows If ∆Q is absorbed at constant volume, ∆V = 0this variation in the temperature range 0 to 100 °C.  ∆Q   ∆ U   ∆ U  (11.9) C v =  ∆T  v =  ∆T  v =  ∆T  where the subscript v is dropped in the last step, since U of an ideal gas depends only on temperature. (The subscript denotes the quantity kept fixed.) If, on the other hand, ∆Q is absorbed at constant pressure,  ∆ Q   ∆ U   ∆V  (11.10) C p =  ∆T  p =  ∆T  p + P  ∆ T  p The subscript p can be dropped from the Fig. 11.5 Variation of specific heat capacity of first term since U of an ideal gas depends only water with temperature. on T. Now, for a mole of an ideal gas For a precise definition of calorie, it was, PV = RTtherefore, necessary to specify the unit temperature interval. One calorie is defined which gives to be the amount of heat required to raise the temperature of 1g of water from 14.5 °C to  ∆ V  P = R (11.11)15.5 °C. Since heat is just a form of energy,  ∆ T  p it is preferable to use the unit joule, J. In SI units, the specific heat capacity of water Equations (11.9) to (11.11) give the desired is 4186 J kg–1 K–1 i.e. 4.186 J g–1 K–1. The so relation, Eq. (11.8). called mechanical equivalent of heat defined 11.7THERMODYNAMIC STATE VARIABLES as the amount of work needed to produce AND EQUATION OF STATE 1 cal of heat is in fact just a conversion factor between two different units of energy : calorie Every equilibrium state of a thermodynamic to joule. Since in SI units, we use the unit joule system is completely described by specific for heat, work or any other form of energy, the values of some macroscopic variables, also term mechanical equivalent is now called state variables. For example, an superfluous and need not be used. equilibrium state of a gas is completely As already remarked, the specific heat specified by the values of pressure, volume, capacity depends on the process or the temperature, and mass (and composition if conditions under which heat capacity transfer there is a mixture of gases). A thermodynamic takes place. For gases, for example, we can system is not always in equilibrium. For example, define two specific heats : specific heat a gas allowed to expand freely against vacuum capacity at constant volume and specific is not an equilibrium state [Fig. 11.6(a)]. During heat capacity at constant pressure. For an the rapid expansion, pressure of the gas may Reprint 2025-26 THERMODYNAMICS 233 not be uniform throughout. Similarly, a mixture temperature do not. To decide which variable is of gases undergoing an explosive chemical extensive and which intensive, think of a reaction (e.g. a mixture of petrol vapour and relevant system in equilibrium, and imagine that air when ignited by a spark) is not an it is divided into two equal parts. The variables that remain unchanged for each part areequilibrium state; again its temperature and intensive. The variables whose values get halvedpressure are not uniform [Fig. 11.6(b)]. in each part are extensive. It is easily seen, forEventually, the gas attains a uniform example, that internal energy U, volume V, total temperature and pressure and comes to mass M are extensive variables. Pressure P, thermal and mechanical equilibrium with its temperature T, and density ρ are intensive surroundings. variables. It is a good practice to check the consistency of thermodynamic equations using this classification of variables. For example, in the equation ∆Q = ∆U + P ∆V quantities on both sides are extensive*. (The product of an intensive variable like P and an extensive quantity ∆V is extensive.) 11.8 THERMODYNAMIC PROCESSES 11.8.1 Quasi-static process Consider a gas in thermal and mechanical equilibrium with its surroundings. The pressure of the gas in that case equals the external Fig. 11.6 (a) The partition in the box is suddenly pressure and its temperature is the same as removed leading to free expansion of the that of its surroundings. Suppose that the gas. (b) A mixture of gases undergoing an external pressure is suddenly reduced (say by explosive chemical reaction. In both lifting the weight on the movable piston in the situations, the gas is not in equilibrium and container). The piston will accelerate outward. cannot be described by state variables. During the process, the gas passes through In short, thermodynamic state variables states that are not equilibrium states. The non- describe equilibrium states of systems. The equilibrium states do not have well-defined various state variables are not necessarily pressure and temperature. In the same way, if independent. The connection between the state a finite temperature difference exists between variables is called the equation of state. For the gas and its surroundings, there will be a example, for an ideal gas, the equation of state rapid exchange of heat during which the gas is the ideal gas relation will pass through non-equilibrium states. In due course, the gas will settle to an equilibrium P V = µ R T state with well-defined temperature and For a fixed amount of the gas i.e. given µ, there pressure equal to those of the surroundings. The are thus, only two independent variables, say P free expansion of a gas in vacuum and a mixture and V or T and V. The pressure-volume curve of gases undergoing an explosive chemical for a fixed temperature is called an isotherm. reaction, mentioned in section 11.7 are also Real gases may have more complicated examples where the system goes through non- equations of state. equilibrium states. The thermodynamic state variables are of two Non-equilibrium states of a system are difficult kinds: extensive and intensive. Extensive to deal with. It is, therefore, convenient to variables indicate the ‘size’ of the system. imagine an idealised process in which at every Intensive variables such as pressure and stage the system is an equilibrium state. Such a * As emphasised earlier, Q is not a state variable. However, ∆Q is clearly proportional to the total mass of system and hence is extensive. Reprint 2025-26 234 PHYSICS process is, in principle, infinitely slow, hence the A process in which the temperature of the name quasi-static (meaning nearly static). The system is kept fixed throughout is called an system changes its variables (P, T, V ) so slowly isothermal process. The expansion of a gas in that it remains in thermal and mechanical a metallic cylinder placed in a large reservoir of equilibrium with its surroundings throughout. fixed temperature is an example of an isothermal In a quasi-static process, at every stage, the process. (Heat transferred from the reservoir to difference in the pressure of the system and the the system does not materially affect the external pressure is infinitesimally small. The temperature of the reservoir, because of its very large heat capacity.) In isobaric processes thesame is true of the temperature difference pressure is constant while in isochoricbetween the system and its surroundings processes the volume is constant. Finally, if the(Fig.11.7). To take a gas from the state (P, T ) to another state (P ′, T ′ ) via a quasi-static process, system is insulated from the surroundings and no heat flows between the system and thewe change the external pressure by a very small surroundings, the process is adiabatic. The amount, allow the system to equalise its pressure definitions of these special processes are with that of the surroundings and continue the summarised in Table. 11.2 process infinitely slowly until the system achieves the pressure P ′. Similarly, to change Table 11.2 Some special thermodynamic the temperature, we introduce an infinitesimal processes temperature difference between the system and the surrounding reservoirs and by choosing reservoirs of progressively different temperatures T to T ′, the system achieves the temperature T ′. We now consider these processes in some detail : 11.8.2 Isothermal process For an isothermal process (T fixed), the ideal gas equation gives PV = constant i.e., pressure of a given mass of gas varies inversely as its volume. This is nothing but Boyle’s Law. Suppose an ideal gas goes isothermally (at temperature T ) from its initial state (P1, V1) to Fig. 11.7 In a quasi-static process, the temperature the final state (P2, V 2). At any intermediate stage of the surrounding reservoir and the with pressure P and volume change from V to external pressure differ only infinitesimally V + ∆V (∆V small) from the temperature and pressure of the system. ∆W = P ∆ V A quasi-static process is obviously a Taking (∆V → 0) and summing the quantity hypothetical construct. In practice, processes ∆W over the entire process, that are sufficiently slow and do not involve V2 accelerated motion of the piston, large W = ∫ P d V temperature gradient, etc., are reasonably V1 approximation to an ideal quasi-static process. V2 d V V2We shall from now on deal with quasi-static = µ RT = µRT In ∫ (11.12) Vprocesses only, except when stated otherwise. V1 V 1 Reprint 2025-26 THERMODYNAMICS 235 where in the second step we have made use of We can calculate, as before, the work done in the ideal gas equation PV = µ RT and taken the an adiabatic change of an ideal gas from the constants out of the integral. For an ideal gas, state (P1, V1, T1) to the state (P2, V2, T2). internal energy depends only on temperature. V 2Thus, there is no change in the internal energy W = ∫ P d Vof an ideal gas in an isothermal process. The V1First Law of Thermodynamics then implies that heat supplied to the gas equals the work done by the gas : Q = W. Note from Eq. (11.12) that for V2 > V1, W > 0; and for V2 < V1, W < 0. That is, in an isothermal expansion, the gas absorbs heat and does work while in an isothermal (11.15) compression, work is done on the gas by the environment and heat is released. From Eq. (11.14), the constant is P1V1γ or P2V2γ 11.8.3 Adiabatic process γ  γ  P2 V 2 P1V 1In an adiabatic process, the system is insulated W = 1 − 1   −1 γ γ − from the surroundings and heat absorbed or 1 − γ  V2 V1  released is zero. From Eq. (11.1), we see that work done by the gas results in decrease in its 1 µR(T1 − T2 ) = [ P2 V2 − P1V1 ] = (11.16)internal energy (and hence its temperature for 1 − γ γ − 1 an ideal gas). We quote without proof (the result that you will learn in higher courses) that for As expected, if work is done by the gas in an an adiabatic process of an ideal gas. adiabatic process (W > 0), from Eq. (11.16), P V γ = const (11.13) T2 < T1. On the other hand, if work is done on the gas (W < 0), we get T2 > T1 i.e., thewhere γ is the ratio of specific heats (ordinary temperature of the gas rises. or molar) at constant pressure and at constant volume. 11.8.4 Isochoric process Cp In an isochoric process, V is constant. No work γ = Cv is done on or by the gas. From Eq. (11.1), the heat absorbed by the gas goes entirely to change Thus if an ideal gas undergoes a change in its internal energy and its temperature. The its state adiabatically from (P1, V1) to (P2, V2) : change in temperature for a given amount of γ = P2 V2γ (11.14) heat is determined by the specific heat of the P1 V1 gas at constant volume. Figure11.8 shows the P-V curves of an ideal gas for two adiabatic processes connecting two 11.8.5 Isobaric process isotherms. In an isobaric process, P is fixed. Work done by the gas is W = P (V2 – V1) = µ R (T2 – T1) (11.17) Since temperature changes, so does internal energy. The heat absorbed goes partly to increase internal energy and partly to do work. The change in temperature for a given amount of heat is determined by the specific heat of the gas at constant pressure. 11.8.6 Cyclic process In a cyclic process, the system returns to its initial state. Since internal energy is a stateFig. 11.8 P-V curves for isothermal and adiabatic variable, ∆U = 0 for a cyclic process. From processes of an ideal gas. Reprint 2025-26 236 PHYSICS Eq. (11.1), the total heat absorbed equals the 11.10 REVERSIBLE AND IRREVERSIBLE work done by the system. PROCESSES Imagine some process in which a thermodynamic11.9 SECOND LAW OF THERMODYNAMICS system goes from an initial state i to a final state The First Law of Thermodynamics is the f. During the process the system absorbs heat Q principle of conservation of energy. Common from the surroundings and performs work W on experience shows that there are many it. Can we reverse this process and bring both conceivable processes that are perfectly the system and surroundings to their initial allowed by the First Law and yet are never states with no other effect anywhere ? Experience observed. For example, nobody has ever seen suggests that for most processes in nature this a book lying on a table jumping to a height by is not possible. The spontaneous processes ofitself. But such a thing would be possible if nature are irreversible. Several examples can bethe principle of conservation of energy were cited. The base of a vessel on an oven is hotterthe only restriction. The table could cool than its other parts. When the vessel is removed,spontaneously, converting some of its internal heat is transferred from the base to the otherenergy into an equal amount of mechanical parts, bringing the vessel to a uniformenergy of the book, which would then hop to a height with potential energy equal to the temperature (which in due course cools to the mechanical energy it acquired. But this never temperature of the surroundings). The process happens. Clearly, some additional basic cannot be reversed; a part of the vessel will not principle of nature forbids the above, even get cooler spontaneously and warm up the base. though it satisfies the energy conservation It will violate the Second Law of Thermodynamics, principle. This principle, which disallows if it did. The free expansion of a gas is irreversible. many phenomena consistent with the First The combustion reaction of a mixture of petrol Law of Thermodynamics is known as the and air ignited by a spark cannot be reversed. Second Law of Thermodynamics. Cooking gas leaking from a gas cylinder in the The Second Law of Thermodynamics gives kitchen diffuses to the entire room. The diffusion a fundamental limitation to the efficiency of a process will not spontaneously reverse and bring heat engine and the co-efficient of the gas back to the cylinder. The stirring of a performance of a refrigerator. In simple terms, liquid in thermal contact with a reservoir will it says that efficiency of a heat engine can convert the work done into heat, increasing the never be unity. For a refrigerator, the Second internal energy of the reservoir. The process Law says that the co-efficient of performance cannot be reversed exactly; otherwise it would can never be infinite. The following two amount to conversion of heat entirely into work, statements, one due to Kelvin and Planck violating the Second Law of Thermodynamics. denying the possibility of a perfect heat engine, Irreversibility is a rule rather an exception and another due to Clausius denying the in nature. possibility of a perfect refrigerator or heat Irreversibility arises mainly from two causes: pump, are a concise summary of these one, many processes (like a free expansion, or observations. an explosive chemical reaction) take the system Kelvin-Planck statement to non-equilibrium states; two, most processes No process is possible whose sole result is the involve friction, viscosity and other dissipative absorption of heat from a reservoir and the effects (e.g., a moving body coming to a stop and complete conversion of the heat into work. losing its mechanical energy as heat to the floor and the body; a rotating blade in a liquid coming Clausius statement to a stop due to viscosity and losing its No process is possible whose sole result is the mechanical energy with corresponding gain in transfer of heat from a colder object to a the internal energy of the liquid). Since hotter object. dissipative effects are present everywhere and It can be proved that the two statements can be minimised but not fully eliminated, most above are completely equivalent. processes that we deal with are irreversible. Reprint 2025-26 THERMODYNAMICS 237 A thermodynamic process (state i → state f ) in a reversible heat engine operating between is reversible if the process can be turned back two temperatures, heat should be absorbed such that both the system and the surroundings (from the hot reservoir) isothermally and return to their original states, with no other released (to the cold reservoir) isothermally. We change anywhere else in the universe. From the thus have identified two steps of the reversible preceding discussion, a reversible process is an heat engine : isothermal process at temperature idealised notion. A process is reversible only if T1 absorbing heat Q1 from the hot reservoir, and another isothermal process at temperature T2it is quasi-static (system in equilibrium with the releasing heat Q2 to the cold reservoir. Tosurroundings at every stage) and there are no complete a cycle, we need to take the systemdissipative effects. For example, a quasi-static from temperature T1 to T2 and then back fromisothermal expansion of an ideal gas in a temperature T2 to T1. Which processes shouldcylinder fitted with a frictionless movable piston we employ for this purpose that are reversible?is a reversible process. A little reflection shows that we can only adopt Why is reversibility such a basic concept in reversible adiabatic processes for these thermodynamics ? As we have seen, one of the purposes, which involve no heat flow from any concerns of thermodynamics is the efficiency reservoir. If we employ any other process that is with which heat can be converted into work. not adiabatic, say an isochoric process, to take The Second Law of Thermodynamics rules out the system from one temperature to another, we the possibility of a perfect heat engine with 100% shall need a series of reservoirs in the efficiency. But what is the highest efficiency temperature range T2 to T1 to ensure that at each possible for a heat engine working between two stage the process is quasi-static. (Remember reservoirs at temperatures T1 and T2? It turns again that for a process to be quasi-static and out that a heat engine based on idealised reversible, there should be no finite temperature reversible processes achieves the highest difference between the system and the reservoir.) efficiency possible. All other engines involving But we are considering a reversible engine that irreversibility in any way (as would be the case operates between only two temperatures. Thus for practical engines) have lower than this adiabatic processes must bring about the limiting efficiency. temperature change in the system from T1 to T2 and T2 to T1 in this engine.11.11 CARNOT ENGINE Suppose we have a hot reservoir at temperature T1 and a cold reservoir at temperature T2. What is the maximum efficiency possible for a heat engine operating between the two reservoirs and what cycle of processes should be adopted to achieve the maximum efficiency ? Sadi Carnot, a French engineer, first considered this question in 1824. Interestingly, Carnot arrived at the correct answer, even though the basic concepts of heat and thermodynamics had yet to be firmly established. We expect the ideal engine operating between two temperatures to be a reversible engine. Irreversibility is associated with dissipative Fig. 11.9 Carnot cycle for a heat engine with an effects, as remarked in the preceding section, ideal gas as the working substance. and lowers efficiency. A process is reversible if it is quasi-static and non-dissipative. We have A reversible heat engine operating between seen that a process is not quasi-static if it two temperatures is called a Carnot engine. We involves finite temperature difference between have just argued that such an engine must have the system and the reservoir. This implies that the following sequence of steps constituting one Reprint 2025-26 238 PHYSICS cycle, called the Carnot cycle, shown in Fig. 11.9. We have taken the working substance In  V 3  of the Carnot engine to be an ideal gas. T2   V4  = 1 −  T1   V2  (11.23)(a) Step 1 → 2 Isothermal expansion of the gas In  V1  taking its state from (P1, V1, T1) to (P2, V2, T1). Now since step 2 → 3 is an adiabatic process, The heat absorbed by the gas (Q1) from the γ −1 γ −1 reservoir at temperature T1 is given by T1 V 2 = T2 V3 Eq. (11.12). This is also the work done (W1 → 2) by the gas on the environment. 1 /( γ −1 ) V2  T2   V2  i.e. = (11.24) 3  T1  W1 → 2 = Q1 = µ R T1 ln  V1  (11.18) V (b) Step 2 → 3 Adiabatic expansion of the gas Similarly, since step 4 → 1 is an adiabatic from (P2, V2, T1) to (P3, V3, T2) process Work done by the gas, using γ −1 γ −1 Eq. (11.16), is T2 V 4 = T1 V1 µR ( T1 − T2 ) 1 /γ −1 W2 →=3 (11.19) V1  T2  γ − 1 i.e. = (11.25) V4  T1  (c) Step 3 → 4 Isothermal compression of the gas from (P3, V3, T2) to (P4, V4, T2). From Eqs. (11.24) and (11.25), 2 Heat released (Q2) by the gas to the reservoir V3 = V (11.26) at temperature T2 is given by Eq. (11.12). This V4 V1 is also the work done (W3 → 4) on the gas by the environment. Using Eq. (11.26) in Eq. (11.23), we get  V 3  2 W 3 → 4 = Q2 = µRT2 ln η = 1 −T (Carnot engine) (11.27)  V 4  (11.20) T1 (d) Step 4 → 1 Adiabatic compression of the gas from (P4, V4, T2) to (P1,V1, T1). We have already seen that a Carnot engine is a reversible engine. Indeed it is the only Work done on the gas, [using Eq.(11.16), is reversible engine possible that works between two reservoirs at different temperatures. Each  T1 − T2  W4 → 1 = µR (11.21) step of the Carnot cycle given in Fig. 11.9 can  γ -1  be reversed. This will amount to taking heat Q2 from the cold reservoir at T2, doing work W on From Eqs. (11.18) to (11.21) total work done by the gas in one complete cycle is the system, and transferring heat Q1 to the hot reservoir. This will be a reversible refrigerator. W = W1 → 2 + W2 → 3 – W3 → 4 – W4 → 1 We next establish the important result  V2   V3  (sometimes called Carnot’s theorem) that = µ RT1 ln  V1  – µ RT2 ln  V4  (11.22) (a) working between two given temperatures T1 and T2 of the hot and cold reservoirs respectively, The efficiency η of the Carnot engine is no engine can have efficiency more than that of the Carnot engine and (b) the efficiency of the W Q 2 η = = 1 − Carnot engine is independent of the nature of Q1 Q1 the working substance. Reprint 2025-26 THERMODYNAMICS 239 To prove the result (a), imagine a reversible reservoir and delivers the same amount of work (Carnot) engine R and an irreversible engine I in one cycle, without any change in the source working between the same source (hot reservoir) or anywhere else. This is clearly against the and sink (cold reservoir). Let us couple the Kelvin-Planck statement of the Second Law of engines, I and R, in such a way so that I acts Thermodynamics. Hence the assertion ηI > ηR like a heat engine and R acts as a refrigerator. is wrong. No engine can have efficiency greater Let I absorb heat Q1 from the source, deliver than that of the Carnot engine. A similar work W ′ and release the heat Q1- W′ to the sink. argument can be constructed to show that a We arrange so that R returns the same heat Q1 reversible engine with one particular substance to the source, taking heat Q2 from the sink and cannot be more efficient than the one using requiring work W = Q1 – Q2 to be done on it. another substance. The maximum efficiency of Now suppose ηR < ηI i.e. if R were to act a Carnot engine given by Eq. (11.27) is as an engine it would give less work output independent of the nature of the system performing the Carnot cycle of operations. Thus we are justified in using an ideal gas as a system in the calculation of efficiency η of a Carnot engine. The ideal gas has a simple equation of I state, which allows us to readily calculate η, but the final result for η, [Eq. (11.27)], is true for any Carnot engine. R This final remark shows that in a Carnot cycle, Q1 T1 = W (11.28) T2 Q 2 is a universal relation independent of the natureFig. 11.10 An irreversible engine (I) coupled to a reversible refrigerator (R). If W ′ > W, this of the system. Here Q1 and Q2 are respectively, would amount to extraction of heat the heat absorbed and released isothermally W′ – W from the sink and its full (from the hot and to the cold reservoirs) in a conversion to work, in contradiction with Carnot engine. Equation (11.28), can, therefore, the Second Law of Thermodynamics. be used as a relation to define a truly universal thermodynamic temperature scale that is than that of I i.e. W < W ′ for a given Q1. With R independent of any particular properties of the acting like a refrigerator, this would mean system used in the Carnot cycle. Of course, for Q2 = Q1 – W > Q1 – W ′. Thus, on the whole, an ideal gas as a working substance, this the coupled I-R system extracts heat universal temperature is the same as the ideal (Q1 – W) – (Q1 – W ′) = (W ′ – W ) from the cold gas temperature introduced in section 11.9. SUMMARY 1. The zeroth law of thermodynamics states that ‘two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other’. The Zeroth Law leads to the concept of temperature. 2. Internal energy of a system is the sum of kinetic energies and potential energies of the molecular constituents of the system. It does not include the over-all kinetic energy of the system. Heat and work are two modes of energy transfer to the system. Heat is the energy transfer arising due to temperature difference between the system and the surroundings. Work is energy transfer brought about by other means, such as moving the piston of a cylinder containing the gas, by raising or lowering some weight connected to it. Reprint 2025-26 240 PHYSICS 3. The first law of thermodynamics is the general law of conservation of energy applied to any system in which energy transfer from or to the surroundings (through heat and work) is taken into account. It states that ∆Q = ∆U + ∆W where ∆Q is the heat supplied to the system, ∆W is the work done by the system and ∆U is the change in internal energy of the system. 4. The specific heat capacity of a substance is defined by 1 ∆Q s = m ∆T where m is the mass of the substance and ∆Q is the heat required to change its temperature by ∆T. The molar specific heat capacity of a substance is defined by 1 ∆ Q C = µ ∆T where µ is the number of moles of the substance. For a solid, the law of equipartition of energy gives C = 3 R which generally agrees with experiment at ordinary temperatures. Calorie is the old unit of heat. 1 calorie is the amount of heat required to raise the temperature of 1 g of water from 14.5 °C to 15.5 °C. 1 cal = 4.186 J. 5. For an ideal gas, the molar specific heat capacities at constant pressure and volume satisfy the relation Cp – Cv = R where R is the universal gas constant. 6. Equilibrium states of a thermodynamic system are described by state variables. The value of a state variable depends only on the particular state, not on the path used to arrive at that state. Examples of state variables are pressure (P ), volume (V ), temperature (T ), and mass (m ). Heat and work are not state variables. An Equation of State (like the ideal gas equation PV = µ RT ) is a relation connecting different state variables. 7. A quasi-static process is an infinitely slow process such that the system remains in thermal and mechanical equilibrium with the surroundings throughout. In a quasi-static process, the pressure and temperature of the environment can differ from those of the system only infinitesimally. 8. In an isothermal expansion of an ideal gas from volume V1 to V2 at temperature T the heat absorbed (Q) equals the work done (W ) by the gas, each given by  V 2   Q = W = µ R T ln   V1  9. In an adiabatic process of an ideal gas γ PV = constant C p where γ = C v Work done by an ideal gas in an adiabatic change of state from (P1, V1, T1) to (P2, V2, T2) is µR ( T1 − T2 ) W = γ – 1 Reprint 2025-26 THERMODYNAMICS 241 10. The second law of thermodynamics disallows some processes consistent with the First Law of Thermodynamics. It states Kelvin-Planck statement No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work. Clausius statement No process is possible whose sole result is the transfer of heat from a colder object to a hotter object. Put simply, the Second Law implies that no heat engine can have efficiency η equal to 1 or no refrigerator can have co-efficient of performance α equal to infinity. 11. A process is reversible if it can be reversed such that both the system and the surroundings return to their original states, with no other change anywhere else in the universe. Spontaneous processes of nature are irreversible. The idealised reversible process is a quasi-static process with no dissipative factors such as friction, viscosity, etc. 12. Carnot engine is a reversible engine operating between two temperatures T1 (source) and T2 (sink). The Carnot cycle consists of two isothermal processes connected by two adiabatic processes. The efficiency of a Carnot engine is given by T 2 η= 1 − (Carnot engine) T1 No engine operating between two temperatures can have efficiency greater than that of the Carnot engine. 13. If Q > 0, heat is added to the system If Q < 0, heat is removed to the system If W > 0, Work is done by the system If W < 0, Work is done on the system Quantity Symbol Dimensions Unit Remark Co-efficienty of volume αv [K–1] K–1 αv = 3 α1 expansion Heat supplied to a system ∆Q [ML2 T–2] J Q is not a state variable Specific heat capacity s [L2 T–2 K–1] J kg–1 K–1 dt Thermal Conductivity K [MLT–3 K–1] J s–1 K–1 H = – KA dx POINTS TO PONDER 1. Temperature of a body is related to its average internal energy, not to the kinetic energy of motion of its centre of mass. A bullet fired from a gun is not at a higher temperature because of its high speed. 2. Equilibrium in thermodynamics refers to the situation when macroscopic variables describing the thermodynamic state of a system do not depend on time. Equilibrium of a system in mechanics means the net external force and torque on the system are zero. Reprint 2025-26 242 PHYSICS 3. In a state of thermodynamic equilibrium, the microscopic constituents of a system are not in equilibrium (in the sense of mechanics). 4. Heat capacity, in general, depends on the process the system goes through when heat is supplied. 5. In isothermal quasi-static processes, heat is absorbed or given out by the system even though at every stage the gas has the same temperature as that of the surrounding reservoir. This is possible because of the infinitesimal difference in temperature between the system and the reservoir. EXERCISES 11.1 A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 104 J/g ? 11.2 What amount of heat must be supplied to 2.0 × 10–2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure ? (Molecular mass of N2 = 28; R = 8.3 J mol–1 K–1.) 11.3 Explain why (a) Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1 + T2 )/2. (b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat. (c) Air pressure in a car tyre increases during driving. (d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude. 11.4 A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume ? 11.5 In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case ? (Take 1 cal = 4.19 J) 11.6 Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following : (a) What is the final pressure of the gas in A and B ? (b) What is the change in internal energy of the gas ? (c) What is the change in the temperature of the gas ? (d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface ? 11.7 An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing? Reprint 2025-26 THERMODYNAMICS 243 11.8 A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (11.13) Fig. 11.11 Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F Reprint 2025-26 CHAPTER TWELVE KINETIC THEORY 12.1 INTRODUCTION Boyle discovered the law named after him in 1661. Boyle, Newton and several others tried to explain the behaviour of 12.1 Introduction gases by considering that gases are made up of tiny atomic 12.2 Molecular nature of matter particles. The actual atomic theory got established more than 12.3 Behaviour of gases 150 years later. Kinetic theory explains the behaviour of gases 12.4 Kinetic theory of an ideal gas based on the idea that the gas consists of rapidly moving 12.5 Law of equipartition of energy atoms or molecules. This is possible as the inter-atomic forces, 12.6 Specific heat capacity which are short range forces that are important for solids and liquids, can be neglected for gases. The kinetic theory12.7 Mean free path was developed in the nineteenth century by Maxwell, Summary Boltzmann and others. It has been remarkably successful. It Points to ponder gives a molecular interpretation of pressure and temperature Exercises of a gas, and is consistent with gas laws and Avogadro’s hypothesis. It correctly explains specific heat capacities of many gases. It also relates measurable properties of gases such as viscosity, conduction and diffusion with molecular parameters, yielding estimates of molecular sizes and masses. This chapter gives an introduction to kinetic theory. 12.2 MOLECULAR NATURE OF MATTER Richard Feynman, one of the great physicists of 20th century considers the discovery that “Matter is made up of atoms” to be a very significant one. Humanity may suffer annihilation (due to nuclear catastrophe) or extinction (due to environmental disasters) if we do not act wisely. If that happens, and all of scientific knowledge were to be destroyed then Feynman would like the ‘Atomic Hypothesis’ to be communicated to the next generation of creatures in the universe. Atomic Hypothesis: All things are made of atoms - little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repelling upon being squeezed into one another. Speculation that matter may not be continuous, existed in many places and cultures. Kanada in India and Democritus Reprint 2025-26 KINETIC THEORY 245 Atomic Hypothesis in Ancient India and Greece Though John Dalton is credited with the introduction of atomic viewpoint in modern science, scholars in ancient India and Greece conjectured long before the existence of atoms and molecules. In the Vaiseshika school of thought in India founded by Kanada (Sixth century B.C.) the atomic picture was developed in considerable detail. Atoms were thought to be eternal, indivisible, infinitesimal and ultimate parts of matter. It was argued that if matter could be subdivided without an end, there would be no difference between a mustard seed and the Meru mountain. The four kinds of atoms (Paramanu — Sanskrit word for the smallest particle) postulated were Bhoomi (Earth), Ap (water), Tejas (fire) and Vayu (air) that have characteristic mass and other attributes, were propounded. Akasa (space) was thought to have no atomic structure and was continuous and inert. Atoms combine to form different molecules (e.g. two atoms combine to form a diatomic molecule dvyanuka, three atoms form a tryanuka or a triatomic molecule), their properties depending upon the nature and ratio of the constituent atoms. The size of the atoms was also estimated, by conjecture or by methods that are not known to us. The estimates vary. In Lalitavistara, a famous biography of the Buddha written mainly in the second century B.C., the estimate is close to the modern estimate of atomic size, of the order of 10 –10 m. In ancient Greece, Democritus (Fourth century B.C.) is best known for his atomic hypothesis. The word ‘atom’ means ‘indivisible’ in Greek. According to him, atoms differ from each other physically, in shape, size and other properties and this resulted in the different properties of the substances formed by their combination. The atoms of water were smooth and round and unable to ‘hook’ on to each other, which is why liquid /water flows easily. The atoms of earth were rough and jagged, so they held together to form hard substances. The atoms of fire were thorny which is why it caused painful burns. These fascinating ideas, despite their ingenuity, could not evolve much further, perhaps because they were intuitive conjectures and speculations not tested and modified by quantitative experiments - the hallmark of modern science. in Greece had suggested that matter may consist of matter. The theory is now well accepted by of indivisible constituents. The scientific ‘Atomic scientists. However even at the end of the Theory’ is usually credited to John Dalton. He nineteenth century there were famous scientists proposed the atomic theory to explain the laws who did not believe in atomic theory ! of definite and multiple proportions obeyed by From many observations, in recent times we elements when they combine into compounds. now know that molecules (made up of one or The first law says that any given compound has, more atoms) constitute matter. Electron a fixed proportion by mass of its constituents. microscopes and scanning tunnelling The second law says that when two elements microscopes enable us to even see them. The form more than one compound, for a fixed mass size of an atom is about an angstrom (10 -10 m). of one element, the masses of the other elements In solids, which are tightly packed, atoms are are in ratio of small integers. spaced about a few angstroms (2 Å) apart. In To explain the laws Dalton suggested, about liquids the separation between atoms is also 200 years ago, that the smallest constituents about the same. In liquids the atoms are not of an element are atoms. Atoms of one element as rigidly fixed as in solids, and can move are identical but differ from those of other around. This enables a liquid to flow. In gases elements. A small number of atoms of each the interatomic distances are in tens of element combine to form a molecule of the angstroms. The average distance a molecule compound. Gay Lussac’s law, also given in early can travel without colliding is called the mean 19th century, states: When gases combine free path. The mean free path, in gases, is of chemically to yield another gas, their volumes the order of thousands of angstroms. The atoms are in the ratios of small integers. Avogadro’s are much freer in gases and can travel long law (or hypothesis) says: Equal volumes of all distances without colliding. If they are not gases at equal temperature and pressure have enclosed, gases disperse away. In solids and the same number of molecules. Avogadro’s law, liquids the closeness makes the interatomic force when combined with Dalton’s theory explains important. The force has a long range attraction Gay Lussac’s law. Since the elements are often and a short range repulsion. The atoms attract in the form of molecules, Dalton’s atomic theory when they are at a few angstroms but repel when can also be referred to as the molecular theory they come closer. The static appearance of a gas Reprint 2025-26 246 PHYSICS is misleading. The gas is full of activity and the is 6.02 × 1023. This is known as Avogadro number equilibrium is a dynamic one. In dynamic and is denoted by NA. The mass of 22.4 litres of equilibrium, molecules collide and change their any gas is equal to its molecular weight in grams speeds during the collision. Only the average at S.T.P (standard temperature 273 K and properties are constant. pressure 1 atm). This amount of substance is Atomic theory is not the end of our quest, but called a mole (see Chapter 1 for a more precise the beginning. We now know that atoms are not definition). Avogadro had guessed the equality of indivisible or elementary. They consist of a numbers in equal volumes of gas at a fixed nucleus and electrons. The nucleus itself is made temperature and pressure from chemical up of protons and neutrons. The protons and reactions. Kinetic theory justifies this hypothesis. neutrons are again made up of quarks. Even The perfect gas equation can be written as quarks may not be the end of the story. There PV = µ RT (12.3)may be string like elementary entities. Nature always has surprises for us, but the search for where µ is the number of moles and R = NA truth is often enjoyable and the discoveries kB is a universal constant. The temperature T is absolute temperature. Choosing kelvin scale forbeautiful. In this chapter, we shall limit ourselves absolute temperature, R = 8.314 J mol–1K–1.to understanding the behaviour of gases (and a Herelittle bit of solids), as a collection of moving molecules in incessant motion. M N µ = = (12.4) M 0 N A

5.1 Thermodynamic terms the system from the surroundings is called We are interested in chemical reactions and boundary. This is designed to allow us to control and keep track of all movements ofthe energy changes accompanying them. For matter and energy in or out of the system.this we need to know certain thermodynamic terms. These are discussed below. 5.1.2 Types of the System 5.1.1 The System and the Surroundings We, further classify the systems according A system in thermodynamics refers to that to the movements of matter and energy in or part of universe in which observations are out of the system. made and remaining universe constitutes 1. Open Systemthe surroundings. The surroundings include everything other than the system. System In an open system, there is exchange of energy and the surroundings together constitute the and matter between system and surroundings universe. [Fig. 5.2 (a)]. The presence of reactants in an open beaker is an example of an open system*.The universe = The system + The surroundings Here the boundary is an imaginary surface However, the entire universe other than enclosing the beaker and reactants. the system is not affected by the changes taking place in the system. Therefore, for all 2. Closed System practical purposes, the surroundings are that In a closed system, there is no exchange of portion of the remaining universe which can matter, but exchange of energy is possible interact with the system. Usually, the region between system and the surroundings of space in the neighbourhood of the system [Fig. 5.2 (b)]. The presence of reactants in a constitutes its surroundings. closed vessel made of conducting material For example, if we are studying the e.g., copper or steel is an example of a closed reaction between two substances A and B system. kept in a beaker, the beaker containing the reaction mixture is the system and the room where the beaker is kept is the surroundings (Fig. 5.1). Fig. 5.1 System and the surroundings Note that the system may be defined by physical boundaries, like beaker or test tube, or the system may simply be defined by a set of Cartesian coordinates specifying a particular volume in space. It is necessary to think of the system as separated from the surroundings by some sort of wall which may be real or imaginary. The wall that separates Fig. 5.2 Open, closed and isolated systems. * We could have chosen only the reactants as system then walls of the beakers will act as boundary. Reprint 2025-26 138 chemIstry 3. Isolated System a quantity which represents the total energy of the system. It may be chemical, electrical,In an isolated system, there is no exchange mechanical or any other type of energy youof energy or matter between the system and may think of, the sum of all these is the energythe surroundings [Fig. 5.2 (c)]. The presence of the system. In thermodynamics, we call itof reactants in a thermos flask or any other the internal energy, U of the system, whichclosed insulated vessel is an example of an may change, whenisolated system. • heat passes into or out of the system, 5.1.3 The State of the System • work is done on or by the system, The system must be described in order to • matter enters or leaves the system. make any useful calculations by specifying quantitatively each of the properties such as These systems are classified accordingly its pressure (p), volume (V), and temperature as you have already studied in section 5.1.2. (T ) as well as the composition of the system. (a) WorkWe need to describe the system by specifying it before and after the change. You would Let us first examine a change in internal energy recall from your Physics course that the by doing work. We take a system containing state of a system in mechanics is completely some quantity of water in a thermos flask or in an insulated beaker. This would notspecified at a given instant of time, by the allow exchange of heat between the systemposition and velocity of each mass point of and surroundings through its boundary andthe system. In thermodynamics, a different we call this type of system as adiabatic. Theand much simpler concept of the state of a manner in which the state of such a systemsystem is introduced. It does not need detailed may be changed will be called adiabaticknowledge of motion of each particle because, process. Adiabatic process is a process inwe deal with average measurable properties of which there is no transfer of heat betweenthe system. We specify the state of the system the system and surroundings. Here, the wallby state functions or state variables. separating the system and the surroundings The state of a thermodynamic system is is called the adiabatic wall (Fig. 5.3).described by its measurable or macroscopic (bulk) properties. We can describe the state of a gas by quoting its pressure (p), volume (V), temperature (T ), amount (n) etc. Variables like p, V, T are called state variables or state functions because their values depend only on the state of the system and not on how it is reached. In order to completely define the state of a system it is not necessary to define all the properties of the system; as only a certain number of properties can be varied independently. This number depends on the nature of the system. Once these minimum number of macroscopic properties are fixed, Fig. 5.3 An adiabatic system which does not others automatically have definite values. permit the transfer of heat through its The state of the surroundings can never boundary. be completely specified; fortunately it is not Let us bring the change in the internal necessary to do so. energy of the system by doing some work on it. Let us call the initial state of the system 5.1.4 The Internal Energy as a State as state A and its temperature as TA. Let Function the internal energy of the system in state A When we talk about our chemical system be called UA. We can change the state of the losing or gaining energy, we need to introduce system in two different ways. Reprint 2025-26 THERMODYNAMICS 139 One way: We do some mechanical work, say the route taken. Volume of water in a pond, for 1 kJ, by rotating a set of small paddles and example, is a state function, because change thereby churning water. Let the new state in volume of its water is independent of the be called B state and its temperature, as route by which water is filled in the pond, TB. It is found that TB > TA and the change either by rain or by tubewell or by both. in temperature, ∆T = TB–TA. Let the internal (b) Heat energy of the system in state B be UB and the We can also change the internal energychange in internal energy, ∆U =UB– UA. of a system by transfer of heat from theSecond way: We now do an equal amount surroundings to the system or vice-versa(i.e., 1kJ) electrical work with the help of an without expenditure of work. This exchangeimmersion rod and note down the temperature of energy, which is a result of temperaturechange. We find that the change in temperature difference is called heat, q. Let us consideris same as in the earlier case, say, TB – TA. bringing about the same change in temperature In fact, the experiments in the above (the same initial and final states as before manner were done by J. P. Joule between in section 5.1.4 (a) by transfer of heat 1840–50 and he was able to show that a through thermally conducting walls instead given amount of work done on the system, of adiabatic walls (Fig. 5.4). no matter how it was done (irrespective of path) produced the same change of state, as measured by the change in the temperature of the system. So, it seems appropriate to define a quantity, the internal energy U, whose value is characteristic of the state of a system, whereby the adiabatic work, wad required to bring about a change of state is equal to the difference between the value of U in one state and that in another state, ∆U i.e., ∆U =U2 –U1= wad Fig. 5.4 A system which allows heat transfer Therefore, internal energy, U, of the through its boundary. system is a state function. We take water at temperature, TA in a By conventions of IUPAC in chemical container having thermally conducting walls,thermodynamics. The positive sign expresses say made up of copper and enclose it in athat wad is positive when work is done on the huge heat reservoir at temperature, TB. Thesystem and the internal energy of system heat absorbed by the system (water), q can beincreases. Similarly, if the work is done by the measured in terms of temperature difference,system, wad will be negative because internal energy of the system decreases. TB – TA. In this case change in internal energy, ∆U = q, when no work is done at constant Can you name some other familiar state volume.functions? Some of other familiar state By conventions of IUPAC in chemicalfunctions are V, p, and T. For example, if we thermodynamics. The q is positive, when heatbring a change in temperature of the system is transferred from the surroundings to thefrom 25°C to 35°C, the change in temperature system and the internal energy of the systemis 35°C–25°C = +10°C, whether we go straight increases and q is negative when heat isup to 35°C or we cool the system for a few transferred from system to the surroundingsdegrees, then take the system to the final resulting in decrease of the internal energy oftemperature. Thus, T is a state function and the change in temperature is independent of the system. * Earlier negative sign was assigned when the work is done on the system and positive sign when the work is done by the system. This is still followed in physics books, although IUPAC has recommended the use of new sign convention. Reprint 2025-26 140 chemIstry (c) The general case SolutionLet us consider the general case in which a change of state is brought about both by (i) ∆ U = w ad, wall is adiabatic doing work and by transfer of heat. We write (ii) ∆ U = – q, thermally conducting change in internal energy for this case as: walls ∆U = q + w (5.1) (iii) ∆ U = q – w, closed system. For a given change in state, q and w can 5.2 Applications vary depending on how the change is carried Many chemical reactions involve the generation out. However, q +w = ∆U will depend only on of gases capable of doing mechanical work or initial and final state. It will be independent the generation of heat. It is important for us of the way the change is carried out. If there to quantify these changes and relate them is no transfer of energy as heat or as work to the changes in the internal energy. Let us (isolated system) i.e., if w = 0 and q = 0, then see how! ∆ U = 0. The equation 5.1 i.e., ∆U = q + w is 5.2.1 Work mathematical statement of the first law of First of all, let us concentrate on the nature of thermodynamics, which states that work a system can do. We will consider only The energy of an isolated system is mechanical work i.e., pressure-volume work. constant. For understanding pressure-volume work, let us consider a cylinder whichIt is commonly stated as the law of conservation contains one mole of an ideal gas fitted withof energy i.e., energy can neither be created a frictionless piston. Total volume of the gasnor be destroyed. is Vi and pressure of the gas inside is p. IfNote: There is considerable difference between external pressure is pex which is greater thanthe character of the thermodynamic property p, piston is moved inward till the pressureenergy and that of a mechanical property such as volume. We can specify an unambiguous (absolute) value for volume of a system in a particular state, but not the absolute value of the internal energy. However, we can measure only the changes in the internal energy, ∆U of the system. Problem 5.1 Express the change in internal energy of a system when (i) No heat is absorbed by the system from the surroundings, but work (w) is done on the system. What type of wall does the system have ? (ii) No work is done on the system, but q amount of heat is taken out from the system and given to the surroundings. What type of wall does the system have? Fig. 5.5 (a) Work done on an ideal gas in a (iii) w amount of work is done by the cylinder when it is compressed by system and q amount of heat is a constant external pressure, pex supplied to the system. What type (in single step) is equal to the shaded of system would it be? area. Reprint 2025-26 THERMODYNAMICS 141 inside becomes equal to pex. Let this change If the pressure is not constant but be achieved in a single step and the final changes during the process such that it volume be Vf . During this compression, is always infinitesimally greater than the suppose piston moves a distance, l and pressure of the gas, then, at each stage of is cross-sectional area of the piston is A compression, the volume decreases by an [Fig. 5.5(a)]. infinitesimal amount, dV. In such a case we then, volume change = l × A = ∆V = (Vf – Vi ) can calculate the work done on the gas by the relationWe also know, pressure = f Therefore, force on the piston = pex . A V ex dV (5.3) w pIf w is the work done on the system by Vi movement of the piston then w = force × distance = pex . A .l Here, pex at each stage is equal to (pin + dp) in case of compression [Fig. 5.5(c)]. In an = pex . (–∆V) = – pex ∆V = – pex (Vf – Vi ) (5.2) expansion process under similar conditions, The negative sign of this expression is the external pressure is always less than the required to obtain conventional sign for w, pressure of the system i.e., pex = (pin– dp). In which will be positive. It indicates that in case general case we can write, pex = (pin + dp). Such of compression work is done on the system. processes are called reversible processes. Here (Vf – Vi ) will be negative and negative A process or change is said to bemultiplied by negative will be positive. Hence reversible, if a change is brought out in such athe sign obtained for the work will be positive. way that the process could, at any moment, If the pressure is not constant at every be reversed by an infinitesimal change. stage of compression, but changes in number A reversible process proceeds infinitely of finite steps, work done on the gas will be slowly by a series of equilibrium states summed over all the steps and will be equal such that system and the surroundings are to – Σ р ∆V [Fig. 5.5 (b)] always in near equilibrium with each other. Fig. 5.5 (c) pV-plot when pressure is not constant Fig. 5.5 (b) pV-plot when pressure is not constant and changes in infinite steps (reversible and changes in finite steps during conditions) during compression from compression from initial volume, Vi to initial volume, Vi to final volume, Vf . final volume, Vf . Work done on the gas Work done on the gas is represented is represented by the shaded area. by the shaded area. Reprint 2025-26 142 chemIstry Processes other than reversible processes Isothermal and free expansion of an are known as irreversible processes. ideal gas In chemistry, we face problems that can For isothermal (T = constant) expansion of be solved if we relate the work term to the an ideal gas into vacuum; w = 0 since pex = 0. internal pressure of the system. We can Also, Joule determined experimentally that relate work to internal pressure of the system q = 0; therefore, ∆U = 0 under reversible conditions by writing Equation 5.1, can beequation 5.3 as follows: V f V f expressed for isothermal irreversible and reversible changes as follows: wrev  p ex dV  p in  dp ) dV  ( 1. For isothermal irreversible change V V i i q = – w = pex (Vf – Vi ) Since dp × dV is very small we can write 2. For isothermal reversible change V f in dV (5.4) q = – w = nRT ln wrev p Vi Now, the pressure of the gas (pin which we V f can write as p now) can be expressed in terms = 2.303 nRT log Vi of its volume through gas equation. For n mol For adiabatic change, q = 0, of an ideal gas i.e., pV =nRT ∆U = wad nRT  p  Problem 5.2 V Two litres of an ideal gas at a pressure of 10 Therefore, at constant temperature atm expands isothermally at 25 °C into a (isothermal process), vacuum until its total volume is 10 litres. V f How much heat is absorbed and how dV V f much work is done in the expansion ? RT n RT ln wrev  n V Vi V i Solution V f We have q = – w = pex (10 – 2) = 0(8) = 0= – 2.303 nRT log (5.5) No work is done; no heat is absorbed. Vi Problem 5.3 Free expansion: Expansion of a gas in Consider the same expansion, butvacuum (pex = 0) is called free expansion. this time against a constant externalNo work is done during free expansion of an pressure of 1 atm. ideal gas whether the process is reversible or irreversible (equation 5.2 and 5.3). Solution Now, we can write equation 5.1 in number We have q = – w = pex (8) = 8 litre-atm of ways depending on the type of processes. Problem 5.4 Let us substitute w = – pex∆V (eq. 5.2) in Consider the expansion given in problemequation 5.1, and we get 5.2, for 1 mol of an ideal gas conducted U  q  p ex V reversibly. If a process is carried out at constant volume Solution (∆V = 0), then V We have q = – w = 2.303 nRT log f s ∆U = qV V the subscript V in qV denotes that heat is = 2.303 × 1 × 0.8206 × 298 × log 10 supplied at constant volume. 2 Reprint 2025-26 THERMODYNAMICS 143 Remember ∆H = qp, heat absorbed by the = 2.303 x 0.8206 x 298 x log 5 system at constant pressure. = 2.303 x 0.8206 x 298 x 0.6990 ∆H is negative for exothermic reactions = 393.66 L atm which evolve heat during the reaction and ∆H is positive for endothermic reactions5.2.2 Enthalpy, H which absorb heat from the surroundings. (a) A Useful New State Function At constant volume (∆V = 0), ∆U = qV,We know that the heat absorbed at constant therefore equation 5.8 becomes volume is equal to change in the internal ∆H = ∆U = qVenergy i.e., ∆U = qV. But most of chemical reactions are carried out not at constant The difference between ∆H and ∆U is not volume, but in flasks or test tubes under usually significant for systems consisting constant atmospheric pressure. We need to of only solids and / or liquids. Solids and define another state function which may be liquids do not suffer any significant volume suitable under these conditions. changes upon heating. The difference, however, becomes significant when gases are We may write equation (5.1) as involved. Let us consider a reaction involving ∆U = qp – p∆V at constant pressure, where gases. If VA is the total volume of the gaseousqp is heat absorbed by the system and –p∆V reactants, VB is the total volume of the gaseousrepresent expansion work done by the system. products, nA is the number of moles of gaseous Let us represent the initial state by reactants and nB is the number of moles ofsubscript 1 and final state by 2 gaseous products, all at constant pressure We can rewrite the above equation as and temperature, then using the ideal gas U2–U1 = qp – p (V2 – V1) law, we write, On rearranging, we get pVA = nART qp = (U2 + pV2) – (U1 + pV1) (5.6) and pVB = nBRT Now we can define another thermodynamic Thus, pVB – pVA = nBRT – nART = (nB–nA)RTfunction, the enthalpy H [Greek word enthalpien, to warm or heat content] as : or p (VB – VA) = (nB – nA) RT H = U + pV (5.7) or p ∆V = ∆ngRT (5.9) so, equation (5.6) becomes qp= H2 – H1 = ∆H Here, ∆ng refers to the number of moles of gaseous products minus the number of moles Although q is a path dependent function, of gaseous reactants.H is a state function because it depends on U, p and V, all of which are state functions. Substituting the value of p∆V from Therefore, ∆H is independent of path. Hence, equation 5.9 in equation 5.8, we get qp is also independent of path. ∆H = ∆U + ∆ngRT (5.10) For finite changes at constant pressure, The equation 5.10 is useful for calculating we can write equation 5.7 as ∆H from ∆U and vice versa. ∆H = ∆U + ∆pV Problem 5.5 Since p is constant, we can write If water vapour is assumed to be a ∆H = ∆U + p∆V (5.8) perfect gas, molar enthalpy change for It is important to note that when heat is vapourisation of 1 mol of water at 1bar absorbed by the system at constant pressure, and 100°C is 41kJ mol–1. Calculate the we are actually measuring changes in the internal energy change, when enthalpy. Reprint 2025-26 144 chemIstry 1 mol of water is vapourised at 1 bar pressure and 100°C. Solution (i) The change H2O (l) → H2O (g) ∆H = ∆U + ∆ngRT Fig. 5.6(a) A gas at volume V and temperature T or ∆U = ∆H – ∆ngRT, substituting the values, we get ∆U = 41.00 kJ mol–1 – 1 × 8.3 J mol–1 K–1 × 373 K = 41.00 kJ mol-1 – 3.096 kJ mol-1 = 37.904 kJ mol–1 Fig. 5.6 (b) Partition, each part having half the volume of the gas (b) Extensive and Intensive Properties (c) Heat Capacity In thermodynamics, a distinction is made In this sub-section, let us see how to measure between extensive properties and intensive heat transferred to a system. This heat appears as a rise in temperature of the systemproperties. An extensive property is a in case of heat absorbed by the system.property whose value depends on the quantity or size of matter present in the system. For The increase of temperature is proportional example, mass, volume, internal energy, to the heat transferred enthalpy, heat capacity, etc. are extensive q  coeff Tproperties. The magnitude of the coefficient depends Those properties which do not depend on the size, composition and nature of the on the quantity or size of matter present system. We can also write it as q = C ∆T are known as intensive properties. For The coefficient, C is called the heatexample temperature, density, pressure etc. capacity.are intensive properties. A molar property, Thus, we can measure the heat suppliedχm, is the value of an extensive property χ of the system for 1 mol of the substance. If n is by monitoring the temperature rise, provided  we know the heat capacity.  mthe amount of matter, n is independent When C is large, a given amount of heat of the amount of matter. Other examples are results in only a small temperature rise. Water molar volume, Vm and molar heat capacity, has a large heat capacity i.e., a lot of energy Cm. Let us understand the distinction is needed to raise its temperature. between extensive and intensive properties by C is directly proportional to amount ofconsidering a gas enclosed in a container of substance. The molar heat capacity of avolume V and at temperature T [Fig. 5.6(a)]. Let us make a partition such that volume substance, Cm=  C  is the heat capacityis halved, each part [Fig. 5.6 (b)] now has  n , V for one mole of the substance and isone half of the original volume, , but the 2 the quantity of heat needed to raise the temperature will still remain the same i.e., T. temperature of one mole by one degree It is clear that volume is an extensive property celsius (or one kelvin). Specific heat, also and temperature is an intensive property. called specific heat capacity is the quantity Reprint 2025-26 THERMODYNAMICS 145 of heat required to raise the temperature of i) at constant volume, qV one unit mass of a substance by one degree ii) at constant pressure, qp celsius (or one kelvin). For finding out the heat, q, required to raise the temperatures (a) ∆U Measurements of a sample, we multiply the specific heat For chemical reactions, heat absorbed at of the substance, c, by the mass m, and constant volume, is measured in a bomb temperatures change, ∆T as calorimeter (Fig. 5.7). Here, a steel vessel (the bomb) is immersed in a water bath. The whole q  c  m  T  C T (5.11) device is called calorimeter. The steel vessel is (d) The Relationship between Cp and CV immersed in water bath to ensure that no heat for an Ideal Gas is lost to the surroundings. A combustible At constant volume, the heat capacity, C is denoted by CV and at constant pressure, this is denoted by Cp . Let us find the relationship between the two. We can write equation for heat, q at constant volume as qV = C V T  U at constant pressure as qp = C pT  H The difference between Cp and CV can be derived for an ideal gas as: For a mole of an ideal gas, ∆H = ∆U + ∆(pV ) = ∆U + ∆(RT ) = ∆U + R∆T  H  U  R T (5.12) On putting the values of ∆H and ∆U, we have C p T  C V T RT C p  C V R Fig. 5.7 Bomb calorimeter Cp – CV = R (5.13) substance is burnt in pure dioxygen supplied in the steel bomb. Heat evolved during the5.3 Measurement of ∆U and ∆H: reaction is transferred to the water around the Calorimetry bomb and its temperature is monitored. Since We can measure energy changes associated the bomb calorimeter is sealed, its volume with chemical or physical processes by an does not change i.e., the energy changes experimental technique called calorimetry. associated with reactions are measured at In calorimetry, the process is carried out in a constant volume. Under these conditions, no vessel called calorimeter, which is immersed work is done as the reaction is carried out in a known volume of a liquid. Knowing at constant volume in the bomb calorimeter. the heat capacity of the liquid in which Even for reactions involving gases, there is no calorimeter is immersed and the heat capacity work done as ∆V = 0. Temperature change of of calorimeter, it is possible to determine the the calorimeter produced by the completed heat evolved in the process by measuring reaction is then converted to qV, by using the temperature changes. Measurements are known heat capacity of the calorimeter with made under two different conditions: the help of equation 5.11. Reprint 2025-26 146 chemIstry (b) ∆H Measurements of the bomb calorimeter is 20.7kJ/K, Measurement of heat change at constant what is the enthalpy change for the above pressure (generally under atmospheric pressure) reaction at 298 K and 1 atm? can be done in a calorimeter shown in Fig. 5.8. SolutionWe know that ∆Η = qp (at constant p) and, therefore, heat absorbed or evolved, qp at Suppose q is the quantity of heat from constant pressure is also called the heat of the reaction mixture and CV is the reaction or enthalpy of reaction, ∆rH. heat capacity of the calorimeter, then the quantity of heat absorbed by the In an exothermic reaction, heat is evolved, calorimeter. and system loses heat to the surroundings. q = CV × ∆TTherefore, qp will be negative and ∆rH will also be negative. Similarly in an endothermic Quantity of heat from the reaction will reaction, heat is absorbed, qp is positive and have the same magnitude but opposite ∆rH will be positive. sign because the heat lost by the system (reaction mixture) is equal to the heat gained by the calorimeter. q = –CV × ∆T = – 20.7 kJ/K × (299 – 298) K = – 20.7 kJ (Here, negative sign indicates the exothermic nature of the reaction) Thus, ∆U for the combustion of the 1g of graphite = – 20.7 kJK–1 For combustion of 1 mol of graphite, 1 20 .7 kJ  12 .0 g mol  = 1 g = – 2.48 ×102 kJ mol–1 , Since ∆ ng = 0, ∆ H = ∆ U = – 2.48 ×102 kJ mol–1 5.4 Enthalpy change, ∆rH of a reaction – Reaction Enthalpy In a chemical reaction, reactants are converted into products and is represented by,Fig. 5.8 Calorimeter for measuring heat changes at constant pressure (atmospheric Reactants → Products pressure). The enthalpy change accompanying a reaction is called the reaction enthalpy. The Problem 5.6 enthalpy change of a chemical reaction, is 1g of graphite is burnt in a bomb given by the symbol ∆rH calorimeter in excess of oxygen at 298 K ∆rH = (sum of enthalpies of products) – (sum and 1 atmospheric pressure according of enthalpies of reactants) to the equation  H products b i H reactants → ai  (5.14) C (graphite) + O2 (g) CO2 (g) i i During the reaction, temperature rises Here symbol (sigma) is used for ∑ from 298 K to 299 K. If the heat capacity summation and ai and bi are the stoichiometric Reprint 2025-26 THERMODYNAMICS 147 coefficients of the products and reactants ethanol at 298 K is pure liquid ethanol at respectively in the balanced chemical 1 bar; standard state of solid iron at 500 K equation. For example, for the reaction is pure iron at 1 bar. Usually data are taken CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) at 298 K. b i H reac tan ts Standard conditions are denoted by∆r H = ∑ a i H Pr oducts − ∑ i i adding the superscript  to the symbol ∆H, = [Hm (CO2, g) + 2Hm (H2O, l)]– [Hm (CH4, g) e.g., ∆H + 2Hm (O2, g)] where Hm is the molar enthalpy. (b) Enthalpy Changes during Phase Transformations Enthalpy change is a very useful quantity. Knowledge of this quantity is required when Phase transformations also involve energy one needs to plan the heating or cooling changes. Ice, for example, requires heat for required to maintain an industrial chemical melting. Normally this melting takes place at reaction at constant temperature. It is also constant pressure (atmospheric pressure) and required to calculate temperature dependence during phase change, temperature remains of equilibrium constant. constant (at 273 K). H2O(s) → H2O(l); ∆fusH  = 6.00 kJ moI–1(a) Standard Enthalpy of Reactions Here ∆fusH is enthalpy of fusion in standardEnthalpy of a reaction depends on the state. If water freezes, then process is reversedconditions under which a reaction is carried and equal amount of heat is given off to theout. It is, therefore, necessary that we surroundings.must specify some standard conditions. The standard enthalpy of reaction is the The enthalpy change that accompanies enthalpy change for a reaction when all melting of one mole of a solid substance in the participating substances are in their standard state is called standard enthalpy standard states. of fusion or molar enthalpy of fusion, ∆fusH . The standard state of a substance at a specified temperature is its pure form at Melting of a solid is endothermic, so 1 bar. For example, the standard state of liquid all enthalpies of fusion are positive. Water Table 5.1 Standard Enthalpy Changes of Fusion and Vaporisation (Tf and Tb are melting and boiling points, respectively) Reprint 2025-26 148 chemIstry requires heat for evaporation. At constant Solutiontemperature of its boiling point Tb and at constant pressure: We can represent the process of evaporation asH2O(l) → H2O(g); ∆vapH = + 40.79 kJ moI–1 ∆vapH is the standard enthalpy of vaporisation. H 2 O(1) →vaporisation H 2 O(g) 1mol 1mol Amount of heat required to vaporize one mole of a liquid at constant temperature No. of moles in 18 g H2O(l) is 18g and under standard pressure (1bar) is called = –1 =1 mol its standard enthalpy of vaporization or 18g mol molar enthalpy of vaporization, ∆vapH . Heat supplied to evaporate18g water at Sublimation is direct conversion of a 298 K = n × ∆vap H  solid into its vapour. Solid CO2 or ‘dry ice’ = (1 mol) × (44.01 kJ mol–1) sublimes at 195K with ∆subH=25.2 kJ mol–1; = 44.01 kJ naphthalene sublimes slowly and for this (assuming steam behaving as an ideal ∆sub H = 73.0 kJ mol–1 . gas). Standard enthalpy of sublimation, ∆vapU = ∆vapH – p∆V = ∆vapH – ∆ngRT∆subH is the change in enthalpy when one mole of a solid substance sublimes at a constant temperature and under standard ∆vapHV – ∆ngRT = 44.01 kJ pressure (1bar). –(1)(8.314 JK–1mol–1)(298K)(10–3kJ J–1) The magnitude of the enthalpy change ∆vapUV = 44.01 kJ – 2.48kJ depends on the strength of the intermolecular = 41.53 kJinteractions in the substance undergoing the phase transfomations. For example, Problem 5.8 the strong hydrogen bonds between water Assuming the water vapour to be a perfectmolecules hold them tightly in liquid phase. gas, calculate the internal energy changeFor an organic liquid, such as acetone, the when 1 mol of water at 100°C and 1 bar intermolecular dipole-dipole interactions are pressure is converted to ice at 0°C. Given significantly weaker. Thus, it requires less the enthalpy of fusion of ice is 6.00 kJ heat to vaporise 1 mol of acetone than it does mol-1 heat capacity of water is 4.2 J/g°C to vaporize 1 mol of water. Table 5.1 gives The change take place as follows:  1values of standard enthalpy changes of fusion Step - 1 1 mol H2O (l, 100°C) and vaporisation for some substances. mol (l, 0°C) Enthalpy change ∆H1 Problem 5.7  1 mol Step - 2 1 mol H2O (l, 0°C) A swimmer coming out from a pool is H2O( S, 0°C) Enthalpy covered with a film of water weighing change ∆H2 about 18g. How much heat must be Total enthalpy change will be - supplied to evaporate this water at ∆H = ∆H1 + ∆H2 298 K ? Calculate the internal energy of vaporisation at 298K. ∆H1 = - (18 x 4.2 x 100) J mol-1 ∆vap H  for water = - 7560 J mol-1 = - 7.56 k J mol-1 at 298K= 44.01kJ mol–1 ∆H2 = - 6.00 kJ mol-1 Reprint 2025-26 THERMODYNAMICS 149 Table 5.2 Standard Molar Enthalpies of Formation (∆f H ) at 298K of a Few Selected Substances of aggregation (also known as reference Therefore, states) is called Standard Molar Enthalpy ∆H = - 7.56 kJ mol-1 + (-6.00 kJ mol-1) of Formation. Its symbol is ∆fH , where the = -13.56 kJ mol-1 subscript ‘ f ’ indicates that one mole of the There is negligible change in the volume compound in question has been formed in its during the change form liquid to solid standard state from its elements in their most state. stable states of aggregation. The reference Therefore, p∆v = ∆ng RT = 0 state of an element is its most stable state ∆H = ∆U = - 13.56kJ mol-1 of aggregation at 25°C and 1 bar pressure. For example, the reference state of dihydrogen (c) Standard Enthalpy of Formation is H2 gas and those of dioxygen, carbon The standard enthalpy change for the and sulphur are O2 gas, Cgraphite and Srhombic formation of one mole of a compound from respectively. Some reactions with standard its elements in their most stable states molar enthalpies of formation are as follows. Reprint 2025-26 150 chemIstry H2(g) + ½O2 (g) → H2O(1); Here, we can make use of standard enthalpy of formation and calculate the enthalpy∆f H = –285.8 kJ mol–1 change for the reaction. The following general C (graphite, s) + 2H2(g) → Ch4 (g); equation can be used for the enthalpy change ∆f H = –74.81 kJ mol–1 calculation. 2C (graphite, s)+3H2 (g)+ ½O2(g) → C2H5OH(1); ∆rH = i∑ai ∆f H (products) – i∑bi ∆f H (reactants) ∆f H  = – 277.7kJ mol–1 (5.15) where a and b represent the coefficients of It is important to understand that a the products and reactants in the balancedstandard molar enthalpy of formation, ∆fH , equation. Let us apply the above equation foris just a special case of ∆rH , where one mole decomposition of calcium carbonate. Here,of a compound is formed from its constituent coefficients ‘a’ and ‘b’ are 1 each. Therefore,elements, as in the above three equations, where 1 mol of each, water, methane and ∆rH = ∆f H  = [CaO(s)]+ ∆f H  [CO2(g)] ethanol is formed. In contrast, the enthalpy – ∆f H  = [CaCO3(s)] change for an exothermic reaction: =1 (–635.1 kJ mol–1) + 1(–393.5 kJ mol–1) CaO(s) + CO2(g) → CaCo3(s); –1(–1206.9 kJ mol–1) ∆rH  = – 178.3kJ mol–1 = 178.3 kJ mol–1 is not an enthalpy of formation of calcium carbonate, since calcium carbonate has been Thus, the decomposition of CaCO3 (s) is formed from other compounds, and not from an endothermic process and you have to heat its constituent elements. Also, for the reaction it for getting the desired products. given below, enthalpy change is not standard (d) Thermochemical Equations enthalpy of formation, ∆fH  for HBr(g). A balanced chemical equation together with H2(g) + Br2(l) → 2HBr(g); the value of its ∆rH is called a thermochemical ∆r H  = – 178.3kJ mol–1 equation. We specify the physical state Here two moles, instead of one mole of the (alongwith allotropic state) of the substance product is formed from the elements, i.e., in an equation. For example: ∆r H  = 2∆f H C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l); Therefore, by dividing all coefficients in ∆rH = – 1367 kJ mol–1 the balanced equation by 2, expression for The above equation describes the enthalpy of formation of HBr (g) is written as combustion of liquid ethanol at constant ½H2(g) + ½Br2(1) → HBr(g); temperature and pressure. The negative sign ∆f H = – 36.4 kJ mol–1 of enthalpy change indicates that this is an exothermic reaction. Standard enthalpies of formation of some common substances are given in Table 5.2. It would be necessary to remember the following conventions regarding thermo- By convention, standard enthalpy for chemical equations.formation, ∆fH , of an element in reference 1. The coefficients in a balanced thermo-state, i.e., its most stable state of aggregation chemical equation refer to the number ofis taken as zero. moles (never molecules) of reactants and Suppose, you are a chemical engineer and products involved in the reaction.want to know how much heat is required to decompose calcium carbonate to lime and 2. The numerical value of ∆rH  refers to the carbon dioxide, with all the substances in number of moles of substances specified their standard state. by an equation. Standard enthalpy change CaCO3(s) → CaO(s) + CO2(g); ∆r H = ? ∆rH will have units as kJ mol–1. Reprint 2025-26 THERMODYNAMICS 151 To illustrate the concept, let us consider (e) Hess’s Law of Constant Heat the calculation of heat of reaction for the Summation following reaction : We know that enthalpy is a state function, Fe 2 O 3 s 3 H 2 g 2 Fe s 3 H 2 O ,l therefore the change in enthalpy is independent of the path between initial state (reactants) From the Table (5.2) of standard enthalpy of and final state (products). In other words, formation (∆f H ), we find : enthalpy change for a reaction is the same ∆f H (H2O,l) = –285.83 kJ mol–1; whether it occurs in one step or in a series of steps. This may be stated as follows in the∆f H (Fe2O3,s) = – 824.2 kJ mol–1; form of Hess’s Law. Also ∆f H (Fe, s) = 0 and If a reaction takes place in several ∆f H (H2, g) = 0 as per convention steps then its standard reaction enthalpy Then, is the sum of the standard enthalpies of ∆f H1 = 3(–285.83 kJ mol–1) the intermediate reactions into which the overall reaction may be divided at the same – 1(– 824.2 kJ mol–1) temperature. = (–857.5 + 824.2) kJ mol–1 Let us understand the importance of this = –33.3 kJ mol–1 law with the help of an example. Note that the coefficients used in these Consider the enthalpy change for thecalculations are pure numbers, which reactionare equal to the respective stoichiometric C (graphite,s) + O2 (g) → CO (g); ∆r H  = ?coefficients. The unit for ∆rH  is kJ mol–1, which means per mole of reaction. Although CO(g) is the major product, some Once we balance the chemical equation in a CO2 gas is always produced in this reaction. particular way, as above, this defines the mole Therefore, we cannot measure enthalpy of reaction. If we had balanced the equation change for the above reaction directly. differently, for example, However, if we can find some other reactions 1 3 3 involving related species, it is possible to Fe 2 O 3 s H 2 g Fe s H 2 O l2 2 2 calculate the enthalpy change for the above reaction.then this amount of reaction would be one mole of reaction and ∆rH  would be Let us consider the following reactions:  3 C (graphite,s) + O2 (g) → CO2 (g);∆f H 2 = (–285.83 kJ mol–1)  = – 393.5 kJ mol–1 (i) 2 ∆r H 1 – (–824.2 kJ mol–1) 1 2 CO (g) + O2 (g) → CO2 (g) 2  = (– 428.7 + 412.1) kJ mol–1 ∆r H = – 283.0 kJ mol–1 (ii) = –16.6 kJ mol–1 = ½ ∆r H 1 We can combine the above two reactions It shows that enthalpy is an extensive in such a way so as to obtain the desired quantity. reaction. To get one mole of CO(g) on the 3. When a chemical equation is reversed, right, we reverse equation (ii). In this, heat the value of ∆rH  is reversed in sign. For is absorbed instead of being released, so we example change sign of ∆rH  value N2(g) + 3H2 (g) → 2NH3 (g); CO2 (g) → CO (g) + O2 (g); ∆r H  = – 91.8 kJ. mol–1 ∆r H  = + 283.0 kJ mol–1 (iii) 2NH3(g) → N2(g) + 3H2 (g); ∆r H  = + 91.8 kJ mol–1 Reprint 2025-26 152 chemIstry Adding equation (i) and (iii), we get the Similarly, combustion of glucose gives out desired equation, 2802.0 kJ/mol of heat, for which the overall equation is : 1 C  graphite, s  O 2 g CO g ; 2 C 6 H12 O 6 ( g )  6 O 2 ( g )  6 CO 2 ( g )  6 H 2 O(1);  = (– 393.5 + 283.0) ∆C H = – 2802.0 kJ mol–1 for which ∆r H  Our body also generates energy from food = – 110.5 kJ mol–1 by the same overall process as combustion, In general, if enthalpy of an overall although the final products are produced after reaction A→B along one route is ∆rH and a series of complex bio-chemical reactions ∆rH1, ∆rH2, ∆rH3..... representing enthalpies involving enzymes. of reactions leading to same product, B along another route, then we have Problem 5.9 ∆rH = ∆rH1 + ∆rH2 + ∆rH3 ... (5.16) The combustion of one mole of benzene It can be represented as: takes place at 298 K and 1 atm. After combustion, CO2(g) and H2O (1) are ∆rH produced and 3267.0 kJ of heat is A B liberated. Calculate the standard ∆H1 ∆rH3 enthalpy of formation, ∆f H  of benzene. Standard enthalpies of formation of C D CO2(g) and H2O(l) are –393.5 kJ mol–1 ∆rH2 and – 285.83 kJ mol–1 respectively. Solution5.5 Enthalpies for different types of reactions The formation reaction of benezene is given by :It is convenient to give name to enthalpies specifying the types of reactions. 6 C  graphite  3 H 2 g C 6 H 6 ;l  = ? ... (i)(a) Standard Enthalpy of Combustion ∆f H (symbol : ∆cH ) The enthalpy of combustion of 1 mol Combustion reactions are exothermic in of benzene is : nature. These are important in industry, 15rocketry, and other walks of life. Standard C 6 H 6 l O 2  6 CO 2 g 3 H 2 O ;l 2enthalpy of combustion is defined as the enthalpy change per mole (or per unit amount) ∆C H  = – 3267 kJ mol–1... (ii) of a substance, when it undergoes combustion The enthalpy of formation of 1 mol of and all the reactants and products being CO2(g) : in their standard states at the specified C  graphite  O 2 g CO 2 g ; temperature. ∆f H  = – 393.5 kJ mol–1... (iii) Cooking gas in cylinders contains mostly butane (C4H10). During complete combustion The enthalpy of formation of 1 mol of of one mole of butane, 2658 kJ of heat is H2O(l) is : released. We can write the thermochemical 1 H 2 g O 2 g H 2 O ;lreactions for this as: 2 13 C 4 H10 ( g )  O 2 ( g )  4 CO 2 ( g )  5 H 2 O(1); ∆C H = – 285.83 kJ mol–1... (iv) 2 multiplying eqn. (iii) by 6 and eqn. (iv) ∆C H  = – 2658.0 kJ mol–1 by 3 we get: Reprint 2025-26 THERMODYNAMICS 153 In this case, the enthalpy of atomization is same as the enthalpy of sublimation. 6 C  graphite  6 O 2 g 6 CO 2 ;g ∆f H  = – 2361 kJ mol–1 (c) Bond Enthalpy (symbol: ∆bondH ) 3 Chemical reactions involve the breaking and 3 H 2 g O 2 g 3 H 2 O ;1 making of chemical bonds. Energy is required 2  to break a bond and energy is released when a ∆f H = – 857.49 kJ mol–1 bond is formed. It is possible to relate heat of Summing up the above two equations : reaction to changes in energy associated with 6 C  graphite  3 H 2 g 15 O 2 g 6 CO 2 g breaking and making of chemical bonds. With 2 reference to the enthalpy changes associated  3 H 2 O ;l with chemical bonds, two different terms are  used in thermodynamics. ∆f H = – 3218.49 kJ mol–1... (v) (i) Bond dissociation enthalpy Reversing equation (ii); (ii) Mean bond enthalpy 15 6 CO 2 g 3 H 2 O l C 6 H 6 l O 2 ; Let us discuss these terms with reference 2 to diatomic and polyatomic molecules. ∆f H  = – 3267.0 kJ mol–1... (vi) Diatomic Molecules: Consider the following Adding equations (v) and (vi), we get process in which the bonds in one mole of 6 C  graphite  3 H 2 g C 6 H 6 ;l dihydrogen gas (H2) are broken: H2(g) → 2H(g); ∆H–HH = 435.0 kJ mol–1 ∆f H  = – 48.51 kJ mol–1... (iv) The enthalpy change involved in this process is the bond dissociation enthalpy of H–H bond. The bond dissociation enthalpy is the (b) Enthalpy of Atomization change in enthalpy when one mole of covalent (symbol: ∆aH ) bonds of a gaseous covalent compound is Consider the following example of atomization broken to form products in the gas phase. of dihydrogen Note that it is the same as the enthalpy of H2(g) → 2H(g); ∆aH  = 435.0 kJ mol–1 atomization of dihydrogen. This is true for all diatomic molecules. For example:You can see that H atoms are formed by breaking H–H bonds in dihydrogen. The Cl2(g) → 2Cl(g); ∆Cl–ClH = 242 kJ mol–1 enthalpy change in this process is known O2(g) → 2O(g); ∆O=OH = 428 kJ mol–1as enthalpy of atomization, ∆aH . It is the enthalpy change on breaking one mole of In the case of polyatomic molecules, bond bonds completely to obtain atoms in the gas dissociation enthalpy is different for different bonds within the same molecule.phase. Polyatomic Molecules: Let us now consider In case of diatomic molecules, like a polyatomic molecule like methane, CH4.dihydrogen (given above), the enthalpy of The overall thermochemical equation for itsatomization is also the bond dissociation atomization reaction is given below:enthalpy. The other examples of enthalpy of atomization can be CH 4 (g) → C(g) + 4H(g);  = 1665 kJ mol–1CH4(g) → C(g) + 4H(g); ∆aH  = 1665 kJ mol–1 ∆a H Note that the products are only atoms of C In methane, all the four C – H bonds are and H in gaseous phase. Now see the following identical in bond length and energy. However, reaction: the energies required to break the individual Na(s) → Na(g); ∆aH = 108.4 kJ mol–1 C – H bonds in each successive step differ : Reprint 2025-26 154 chemIstry CH4(g) → CH3(g)+H(g);∆bond H = +427 kJ mol–1 given in Table 5.3. The reaction enthalpies are very important quantities as these arise fromCH3(g) → CH2(g)+H(g);∆bond H = +439 kJ mol–1 the changes that accompany the breaking of CH2(g) → CH(g)+H(g);∆bond H = +452 kJ mol–1 old bonds and formation of the new bonds. CH(g) → C(g)+H(g);∆bond H = +347 kJ mol–1 We can predict enthalpy of a reaction in gas phase, if we know different bond enthalpies.Therefore, The standard enthalpy of reaction, ∆rH isCH4(g) → C(g)+4H(g);∆a H = 1665 kJ mol–1 related to bond enthalpies of the reactants In such cases we use mean bond enthalpy and products in gas phase reactions as: of C – H bond. bond enthalpiesreactants ∆r H  For example in CH4, ∆C–HH  is calculated as: bond enthalpies products∆C–HH = ¼ (∆a H) = ¼ (1665 kJ mol–1)   (5.17)** = 416 kJ mol–1 This relationship is particularly more We find that mean C–H bond enthalpy useful when the required values of ∆f H are in methane is 416 kJ/mol. It has been not available. The net enthalpy change of a found that mean C–H bond enthalpies differ reaction is the amount of energy required slightly from compound to compound, as to break all the bonds in the reactant in CH3CH2Cl, CH3NO2, etc., but it does not molecules minus the amount of energy differ in a great deal*. Using Hess’s law, bond required to break all the bonds in the product enthalpies can be calculated. Bond enthalpy molecules. Remember that this relationship is values of some single and multiple bonds are approximate and is valid when all substances Table 5.3(a) Some Mean Single Bond Enthalpies in kJ mol–1 at 298 K H C N O F Si P S Cl Br I 435.8 414 389 464 569 293 318 339 431 368 297 H 347 293 351 439 289 264 259 330 276 238 C 159 201 272 - 209 - 201 243 - N 138 184 368 351 - 205 - 201 O 155 540 490 327 255 197 - F 176 213 226 360 289 213 Si 213 230 331 272 213 P 213 251 213 - S 243 218 209 CI 192 180 Br 151 I Table 5.3(b) Some Mean Multiple Bond Enthalpies in kJ mol–1 at 298 K N = N 418 C = C 611 O = O 498 N ≡ N 946 C ≡ C 837 C = N 615 C = O 741 C ≡ N 891 C ≡ O 1070 * Note that symbol used for bond dissociation enthalpy and mean bond enthalpy is the same. ** If we use enthalpy of bond formation, (∆f H bond), which is the enthalpy change when one mole of a particular type of bond is formed from gaseous atom, then ∆f H  = ∑∆f Hbonds of products – ∑∆f H bonds of reactants Reprint 2025-26 THERMODYNAMICS 155 (reactants and products) in the reaction are  1 2. Na( g )  Na ( g )  e ( g ) , the ionization ofin gaseous state. sodium atoms, ionization enthalpy (d) Lattice Enthalpy ∆iH = 496 kJ mol–1 1 The lattice enthalpy of an ionic compound is 3. Cl 2 ( g ) → Cl( g ), the dissociation of 2the enthalpy change which occurs when one mole of an ionic compound dissociates into chlorine, the reaction enthalpy is half the its ions in gaseous state. bond dissociation enthalpy.     1Na Cl s Na ( g )  Cl g ; ∆bondH = 121 kJ mol–1  2 ∆latticeH = +788 kJ mol–1 4. Cl( g )  e 1 ( g )  Cl( g ) electron gained Since it is impossible to determine lattice by chlorine atoms. The electron gainenthalpies directly by experiment, we use  enthalpy, ∆egH = – 348.6 kJ mol–1.an indirect method where we construct an You have learnt about ionization enthalpyenthalpy diagram called a Born-Haber Cycle and electron gain enthalpy in Unit 3.(Fig. 5.9). In fact, these terms have been taken Let us now calculate the lattice enthalpy from thermodynamics. Earlier terms, of Na+Cl–(s) by following steps given below : ionization energy and electron affinity 1. Na ( s ) → Na( g ) , sublimation of sodium were in practice in place of the above metal, ∆subH  = 108.4 kJ mol–1 terms (see the box for justification). Ionization Energy and Electron Affinity Ionization energy and electron affinity are defined at absolute zero. At any other temperature, heat capacities for the reactants and the products have to be taken into account. Enthalpies of reactions for M(g) → M+(g) + e– (for ionization) M(g) + e– → M–(g) (for electron gain) at temperature, T is T ∆rH(T ) = ∆rH(0) + ∆rCPdT 0∫ The value of Cp for each species in the above reaction is 5/2 R (CV = 3/2R) So, ∆rCp = + 5/2 R (for ionization) ∆rCp = – 5/2 R (for electron gain) Therefore, ∆rH (ionization enthalpy) = E0 (ionization energy) + 5/2 RT ∆rH (electron gain enthalpy) = – A( electron affinity) – 5/2 RT 5. Na + (g) + Cl − (g) → Na + Cl − (s) Fig. 5.9 Enthalpy diagram for lattice enthalpy of The sequence of steps is shown in NaCl Fig. 5.9, and is known as a Born-Haber Reprint 2025-26 156 chemIstry cycle. The importance of the cycle is that, The enthalpy of solution of AB(s), ∆solH, the sum of the enthalpy changes round a in water is, therefore, determined by the cycle is zero. Applying Hess’s law, we get, selective values of the lattice enthalpy, ∆latticeH ∆latticeH = 411.2 + 108.4 + 121 + 496 – 348.6 and enthalpy of hydration of ions, ∆hydH as ∆sol H  = ∆latticeH  + ∆hydH ∆latticeH = + 788kJ For most of the ionic compounds, ∆solfor NaCl(s)  Na+(g) + Cl–(g) H is positive and the dissociation processInternal energy is smaller by 2RT (because ∆ng is endothermic. Therefore the solubility of= 2) and is equal to + 783 kJ mol–1. most salts in water increases with rise of Now we use the value of lattice enthalpy temperature. If the lattice enthalpy is very to calculate enthalpy of solution from the high, the dissolution of the compound may expression: not take place at all. Why do many fluorides tend to be less soluble than the corresponding∆solH = ∆latticeH + ∆hydH chlorides? Estimates of the magnitudes ofFor one mole of NaCl(s), enthalpy changes may be made by usinglattice enthalpy = + 788 kJ mol–1  tables of bond energies (enthalpies) and latticeand ∆hydH = – 784 kJ mol–1( from the energies (enthalpies). literature) ∆sol H = + 788 kJ mol–1 – 784 kJ mol–1 (f) Enthalpy of Dilution = + 4 kJ mol–1 It is known that enthalpy of solution is the The dissolution of NaCl(s) is accompanied enthalpy change associated with the addition by very little heat change. of a specified amount of solute to the specified amount of solvent at a constant temperature(e) Enthalpy of Solution (symbol : ∆solH ) and pressure. This argument can be appliedEnthalpy of solution of a substance is the to any solvent with slight modification.enthalpy change when one mole of it dissolves Enthalpy change for dissolving one mole ofin a specified amount of solvent. The enthalpy gaseous hydrogen chloride in 10 mol of waterof solution at infinite dilution is the enthalpy can be represented by the following equation.change observed on dissolving the substance For convenience we will use the symbol aq.in an infinite amount of solvent when the for waterinteractions between the ions (or solute molecules) are negligible. HCl(g) + 10 aq. → HCl.10 aq. When an ionic compound dissolves in a ∆H = –69.01 kJ / mol solvent, the ions leave their ordered positions Let us consider the following set of on the crystal lattice. These are now more free in enthalpy changes: solution. But solvation of these ions (hydration (S-1) HCl(g) + 25 aq. → HCl.25 aq.in case solvent is water) also occurs at the ∆H = –72.03 kJ / molsame time. This is shown diagrammatically, for an ionic compound, AB (s) (S-2) HCl(g) + 40 aq. → HCl.40 aq. ∆H = –72.79 kJ / mol (S-3) HCl(g) + ∞ aq. → HCl. ∞ aq. ∆H = –74.85 kJ / mol The values of ∆H show general dependence of the enthalpy of solution on amount of solvent. As more and more solvent is used, the enthalpy of solution approaches a limiting value, i.e, the value in infinitely dilute solution. For hydrochloric acid this value of ∆H is given above in equation (S-3). Reprint 2025-26 THERMODYNAMICS 157 If we subtract the first equation (equation many years without observing any perceptible S-1) from the second equation (equation S-2) change. Although the reaction is taking place in the above set of equations, we obtain– between them, it is at an extremely slow rate. It is still called spontaneous reaction. HCl.25 aq. + 15 aq. → HCl.40 aq. So spontaneity means ‘having the potential ∆H = [ –72.79 – (–72.03)] kJ / mol to proceed without the assistance of external = – 0.76 kJ / mol agency’. However, it does not tell about the This value (–0.76kJ/mol) of ∆H is enthalpy rate of the reaction or process. Another aspect of dilution. It is the heat withdrawn from of spontaneous reaction or process, as we see the surroundings when additional solvent is is that these cannot reverse their direction on added to the solution. The enthalpy of dilution their own. We may summarise it as follows: of a solution is dependent on the original A spontaneous process is an concentration of the solution and the amount irreversible process and may only be of solvent added. reversed by some external agency. 5.6 spontaneity (a) Is Decrease in Enthalpy a Criterion The first law of thermodynamics tells us for Spontaneity ? about the relationship between the heat If we examine the phenomenon like flow of absorbed and the work performed on or water down hill or fall of a stone on to the by a system. It puts no restrictions on ground, we find that there is a net decrease the direction of heat flow. However, the in potential energy in the direction of change. flow of heat is unidirectional from higher By analogy, we may be tempted to state that temperature to lower temperature. In fact, a chemical reaction is spontaneous in a all naturally occurring processes whether given direction, because decrease in energy chemical or physical will tend to proceed has taken place, as in the case of exothermic spontaneously in one direction only. For reactions. For example: example, a gas expanding to fill the available 1 volume, burning carbon in dioxygen giving N2(g) + H2(g) = NH3(g); 2 carbon dioxide. ∆r H  = – 46.1 kJ mol–1 But heat will not flow from colder body to 1 1 H2(g) + Cl2(g) = HCl (g);warmer body on its own, the gas in a container 2 2 will not spontaneously contract into one ∆r H  = – 92.32 kJ mol–1 corner or carbon dioxide will not form carbon 1 H2(g) + O2(g) → H2O(l) ;and dioxygen spontaneously. These and many 2 ∆r H  = –285.8 kJ mol–1other spontaneously occurring changes show unidirectional change. We may ask ‘what is The decrease in enthalpy in passing from the driving force of spontaneously occurring reactants to products may be shown for any changes ? What determines the direction of a exothermic reaction on an enthalpy diagram spontaneous change ? In this section, we shall as shown in Fig. 5.10(a). establish some criterion for these processes Thus, the postulate that driving force for whether these will take place or not. a chemical reaction may be due to decrease Let us first understand what do we mean in energy sounds ‘reasonable’ as the basis of by spontaneous reaction or change ? You evidence so far ! may think by your common observation that Now let us examine the following reactions: spontaneous reaction is one which occurs 1 N2(g) + O2(g) → NO2(g);immediately when contact is made between 2 the reactants. Take the case of combination ∆r H = +33.2 kJ mol–1 of hydrogen and oxygen. These gases may C(graphite, s) + 2 S(l) → CS2(l); be mixed at room temperature and left for ∆r H = +128.5 kJ mol–1 Reprint 2025-26 158 chemIstry Fig. 5.10 (a) Enthalpy diagram for exothermic reactions These reactions though endothermic, are spontaneous. The increase in enthalpy may be represented on an enthalpy diagram as shown in Fig. 5.10(b). Fig. 5.11 Diffusion of two gases respectively and separated by a movable partition [Fig. 5.11 (a)]. When the partition is withdrawn [Fig. 5.11(b)], the gases begin to diffuse into each other and after a period of time, diffusion will be complete. Let us examine the process. Before partition, if we were to pick up the gas molecules from left container, we would be Fig. 5.10 (b) Enthalpy diagram for endothermic sure that these will be molecules of gas A reactions and similarly if we were to pick up the gas molecules from right container, we would Therefore, it becomes obvious that while be sure that these will be molecules of gasdecrease in enthalpy may be a contributory B. But, if we were to pick up molecules fromfactor for spontaneity, but it is not true for container when partition is removed, we areall cases. not sure whether the molecules picked are of (b) Entropy and Spontaneity gas A or gas B. We say that the system has Then, what drives the spontaneous process become less predictable or more chaotic. in a given direction ? Let us examine such a We may now formulate another postulate: case in which ∆H = 0 i.e., there is no change in in an isolated system, there is always a enthalpy, but still the process is spontaneous. tendency for the systems’ energy to become Let us consider diffusion of two gases more disordered or chaotic and this could be into each other in a closed container which a criterion for spontaneous change ! is isolated from the surroundings as shown At this point, we introduce another in Fig. 5.11. thermodynamic function, entropy denoted The two gases, say, gas A and gas B are as S. The above mentioned disorder is the represented by black dots and white dots manifestation of entropy. To form a mental Reprint 2025-26 THERMODYNAMICS 159 picture, one can think of entropy as a measure qrev of the degree of randomness or disorder in the ∆S = (5.18) T system. The greater the disorder in an isolated The total entropy change (∆Stotal) for the system, the higher is the entropy. As far as a system and surroundings of a spontaneous chemical reaction is concerned, this entropy process is given by change can be attributed to rearrangement of S total  S system  S surr 0 (5.19)atoms or ions from one pattern in the reactants to another (in the products). If the structure When a system is in equilibrium, the of the products is very much disordered than entropy is maximum, and the change in that of the reactants, there will be a resultant entropy, ∆S = 0. increase in entropy. The change in entropy We can say that entropy for a spontaneousaccompanying a chemical reaction may be process increases till it reaches maximumestimated qualitatively by a consideration of and at equilibrium the change in entropy isthe structures of the species taking part in the zero. Since entropy is a state property, we canreaction. Decrease of regularity in structure calculate the change in entropy of a reversiblewould mean increase in entropy. For a given process bysubstance, the crystalline solid state is the state of lowest entropy (most ordered), The qsys ,rev gaseous state is state of highest entropy. ∆Ssys = T Now let us try to quantify entropy. One way We find that both for reversible and to calculate the degree of disorder or chaotic irreversible expansion for an ideal gas, under distribution of energy among molecules isothermal conditions, ∆U = 0, but ∆Stotalwould be through statistical method which i.e., S sys  S surr is not zero for irreversible is beyond the scope of this treatment. Other process. Thus, ∆U does not discriminate way would be to relate this process to the between reversible and irreversible process, heat involved in a process which would make whereas ∆S does. entropy a thermodynamic concept. Entropy, like any other thermodynamic property such as internal energy U and enthalpy H is a state Problem 5.10 function and ∆S is independent of path. Predict in which of the following, entropy Whenever heat is added to the system, increases/decreases : it increases molecular motions causing (i) A liquid crystallizes into a solid. increased randomness in the system. Thus (ii) Temperature of a crystalline solidheat (q) has randomising influence on the is raised from 0 K to 115 K.system. Can we then equate ∆S with q ? Wait ! Experience suggests us that the distribution  iii  2 NaHCO 3 s Na 2 CO 3 s of heat also depends on the temperature at CO 2 g H 2 O gwhich heat is added to the system. A system at higher temperature has greater randomness (iv) H 2 g 2 H g in it than one at lower temperature. Thus, Solutiontemperature is the measure of average chaotic motion of particles in the system. (i) After freezing, the molecules attain an ordered state and therefore,Heat added to a system at lower temperature entropy decreases.causes greater randomness than when the same quantity of heat is added to it at higher (ii) At 0 K, the contituent particles are temperature. This suggests that the entropy static and entropy is minimum. change is inversely proportional to the If temperature is raised to 115 K, temperature. ∆S is related with q and T for a these begin to move and oscillate reversible reaction as : Reprint 2025-26 160 chemIstry = 4980.6 JK–1 mol–1 about their equilibrium positions This shows that the above reaction is in the lattice and system becomes spontaneous. more disordered, therefore entropy increases. (c) Gibbs Energy and Spontaneity (iii) Reactant, NaHCO3 is a solid and it has low entropy. Among products We have seen that for a system, it is the there are one solid and two gases. total entropy change, ∆Stotal which decides Therefore, the products represent a the spontaneity of the process. But most of condition of higher entropy. the chemical reactions fall into the category (iv) Here one molecule gives two atoms of either closed systems or open systems. i.e., number of particles increases Therefore, for most of the chemical reactions leading to more disordered state. there are changes in both enthalpy and Two moles of H atoms have entropy. It is clear from the discussion in higher entropy than one mole of previous sections that neither decrease in dihydrogen molecule. enthalpy nor increase in entropy alone can determine the direction of spontaneous Problem 5.11 change for these systems. For oxidation of iron, For this purpose, we define a new 4 Fe s 3O 2 g 2 Fe 2 O 3 s thermodynamic function the Gibbs energy or Gibbs function, G, as entropy change is – 549.4 JK–1 mol–1 at 298 K. Inspite of negative entropy G = H – TS (5.20) change of this reaction, why is the Gibbs function, G is an extensive property reaction spontaneous? and a state function. (∆rH for this reaction is The change in Gibbs energy for the –1648 × 103 J mol–1) system, ∆Gsys can be written as Solution G sys = H sys  T S sys  S sys T One decides the spontaneity of a reaction by considering At constant temperature,  G sys = H sys  T S sys S total  S sys  S surr . For calculating ∆Ssurr, we have to consider the heat Usually the subscript ‘system’ is dropped absorbed by the surroundings which and we simply write this equation as is equal to – ∆rH . At temperature T, G  H  T S (5.21) entropy change of the surroundings is Thus, Gibbs energy change = enthalpy change – temperature × entropy change, and is referred to as the Gibbs equation, one of 1648  10 3 J mol 1   the most important equations in chemistry.  Here, we have considered both terms together 298 K for spontaneity: energy (in terms of ∆H) = 5530 JK–1mol–1 and entropy (∆S, a measure of disorder) as Thus, total entropy change for this indicated earlier. Dimensionally if we analyse, reaction we find that ∆G has units of energy because, r S total  5530 JK –1 mol –1  both ∆H and the T∆S are energy terms, since –1 –1 T∆S = (K) (J/K) = J. 549 .4 JK mol   Now let us consider how ∆G is related to reaction spontaneity. Reprint 2025-26 THERMODYNAMICS 161 We know, be small. The former is one of the reasons ∆Stotal = ∆Ssys + ∆Ssurr why reactions are often carried out at high temperature. Table 5.4 summarises the effect If the system is in thermal equilibrium with of temperature on spontaneity of reactions.the surrounding, then the temperature of the surrounding is same as that of the system. (d) Entropy and Second Law of Also, increase in enthalpy of the surrounding Thermodynamics is equal to decrease in the enthalpy of the We know that for an isolated systemsystem. the change in energy remains constant. Therefore, entropy change of surroundings, Therefore, increase in entropy in such H surr H sys systems is the natural direction of a Ssurr =  T T spontaneous change. This, in fact is the second law of thermodynamics. Like first  H sys  law of thermodynamics, second law can also S total  S sys   T  be stated in several ways. The second law of thermodynamics explains why spontaneous Rearranging the above equation: exothermic reactions are so common. T∆Stotal = T∆Ssys – ∆Hsys In exothermic reactions heat released For spontaneous process, Stotal 0 , so by the reaction increases the disorder T∆Ssys – ∆Hsys > Ο of the surroundings and overall entropy change is positive which makes the reaction  0 spontaneous.  H sys  T S sys  Using equation 5.21, the above equation can (e) Absolute Entropy and Third Law of be written as Thermodynamics –∆G > O Molecules of a substance may move in a straight line in any direction, they mayG  H  T S 0 (5.22) spin like a top and the bonds in the ∆Hsys is the enthalpy change of a reaction, molecules may stretch and compress. T∆Ssys is the energy which is not available to do These motions of the molecule are called useful work. So ∆G is the net energy available translational, rotational and vibrational to do useful work and is thus a measure of the motion respectively. When temperature of ‘free energy’. For this reason, it is also known the system rises, these motions become as the free energy of the reaction. more vigorous and entropy increases. On the ∆G gives a criteria for spontaneity at other hand when temperature is lowered, the entropy decreases. The entropy of any pureconstant pressure and temperature. crystalline substance approaches zero as (i) If ∆G is negative (< 0), the process is the temperature approaches absolute zero. spontaneous. This is called third law of thermodynamics. (ii) If ∆G is positive (> 0), the process is non This is so because there is perfect order in spontaneous. a crystal at absolute zero. The statement is confined to pure crystalline solids becauseNote : If a reaction has a positive enthalpy theoretical arguments and practical evidenceschange and positive entropy change, it can have shown that entropy of solutions andbe spontaneous when T∆S is large enough to super cooled liquids is not zero at 0 K. The outweigh ∆H. This can happen in two ways; importance of the third law lies in the fact that (a) The positive entropy change of the system it permits the calculation of absolute values can be ‘small’ in which case T must be of entropy of pure substance from thermal large. (b) The positive entropy change of the data alone. For a pure substance, this can system can be ‘large’, in which case T may Reprint 2025-26 162 chemIstry q rev is minimum. If it is not, the system wouldbe done by summing increments from 0 T spontaneously change to configuration of K to 298 K. Standard entropies can be used lower free energy. to calculate standard entropy changes by a So, the criterion for equilibrium Hess’s law type of calculation. A + B  C + D; is ∆rG = 0 5.7 Gibbs energy change and Gibbs energy for a reaction in which all equilibrium reactants and products are in standard state, We have seen how a knowledge of the sign ∆rG  is related to the equilibrium constant of and magnitude of the free energy change of a the reaction as follows:  + RT ln Kchemical reaction allows: 0 = ∆rG (i) Prediction of the spontaneity of the or ∆rG = – RT ln K chemical reaction.  or ∆rG = – 2.303 RT log K (5.23) (ii) Prediction of the useful work that could We also know that be extracted from it. So far we have considered free energy (5.24) changes in irreversible reactions. Let us now For strongly endothermic reactions, the examine the free energy changes in reversible value of ∆rH may be large and positive. In reactions. such a case, value of K will be much smaller ‘Reversible’ under strict thermodynamic than 1 and the reaction is unlikely to sense is a special way of carrying out form much product. In case of exothermic a process such that system is at all reactions, ∆rH is large and negative, and ∆rG times in perfect equilibrium with its is likely to be large and negative too. In such surroundings. When applied to a chemical cases, K will be much larger than 1. We may reaction, the term ‘reversible’ indicates expect strongly exothermic reactions to have a that a given reaction can proceed in either large K, and hence can go to near completion. direction simultaneously, so that a dynamic ∆rG also depends upon ∆rS, if the changes equilibrium is set up. This means that the in the entropy of reaction is also taken into reactions in both the directions should account, the value of K or extent of chemical proceed with a decrease in free energy, which reaction will also be affected, depending upon seems impossible. It is possible only if at whether ∆rS is positive or negative. equilibrium the free energy of the system Using equation (5.24), Table 5.4 Effect of Temperature on Spontaneity of Reactions ∆rH ∆rS ∆rG Description* – + – Reaction spontaneous at all temperatures – – – (at low T ) Reaction spontaneous at low temperature – – + (at high T ) Reaction nonspontaneous at high temperature + + + (at low T ) Reaction nonspontaneous at low temperature + + – (at high T ) Reaction spontaneous at high temperature + – + (at all T ) Reaction nonspontaneous at all temperatures * The term low temperature and high temperature are relative. For a particular reaction, high temperature could even mean room temperature. Reprint 2025-26 THERMODYNAMICS 163 (i) It is possible to obtain an estimate of ∆G –13 .6  10 3 J mol –1   from the measurement of ∆H and ∆S, = –1 –1 2 .303 8 .314 JK mol 298 K    and then calculate K at any temperature for economic yields of the products. = 2.38 Hence K = antilog 2.38 = 2.4 × 102. (ii) If K is measured directly in the laboratory, value of ∆G at any other temperature Problem 5.14 can be calculated. At 60°C, dinitrogen tetroxide is 50 Using equation (5.24), per cent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere. Problem 5.12 Calculate ∆rG for conversion of oxygen Solution to ozone, 3/2 O2(g) → O3(g) at 298 K. if Kp N2O4(g) 2NO2(g) for this conversion is 2.47 × 10–29. If N2O4 is 50% dissociated, the mole Solution fraction of both the substances is given We know ∆rG = – 2.303 RT log Kp and by R = 8.314 JK–1 mol–1 1  0 .5 2  0 .5 x  : x NO 2  N 2 O 4 Therefore, ∆rG = 1  0 .5 1  0 .5 – 2.303 (8.314 J K–1 mol–1) 0 .5 × (298 K) (log 2.47 × 10–29) p N 2 O 4   1 atm, p NO 2  1 .5 = 163000 J mol–1 1 = 163 kJ mol–1.  1 atm. 1 .5 Problem 5.13 The equilibrium constant Kp is given by Find out the value of equilibrium constant 2 for the following reaction at 298 K.  p NO 2  1 .5 Kp =  2 p N 2 O 4 (1 .5 ) ( 0 .5 ) = 1.33 atm. Standard Gibbs energy change, ∆rG at Since the given temperature is –13.6 kJ mol–1. ∆rG = –RT ln Kp Solution ∆rG = (– 8.314 JK–1 mol–1) × (333 K) We know, log K = × (2.303) × (0.1239) = – 763.8 kJ mol–1 Reprint 2025-26 164 chemIstry Summary Thermodynamics deals with energy changes in chemical or physical processes and enables us to study these changes quantitatively and to make useful predictions. For these purposes, we divide the universe into the system and the surroundings. Chemical or physical processes lead to evolution or absorption of heat (q), part of which may be converted into work (w). These quantities are related through the first law of thermodynamics via ∆U = q + w. ∆U, change in internal energy, depends on initial and final states only and is a state function, whereas q and w depend on the path and are not the state functions. We follow sign conventions of q and w by giving the positive sign to these quantities when these are added to the system. We can measure the transfer of heat from one system to another which causes the change in temperature. The magnitude of rise in temperature depends on the heat capacity (C) of a substance. Therefore, heat absorbed or evolved is q = C∆T. Work can be measured by w = –pex∆V, in case of expansion of gases. Under reversible process, we can put pex = p for infinitesimal changes in the volume making wrev = – p dV. In this condition, we can use gas equation, pV = nRT. At constant volume, w = 0, then ∆U = qV , heat transfer at constant volume. But in study of chemical reactions, we usually have constant pressure. We define another state function enthalpy. Enthalpy change, ∆H = ∆U + ∆ngRT, can be found directly from the heat changes at constant pressure, ∆H = qp. There are varieties of enthalpy changes. Changes of phase such as melting, vaporization and sublimation usually occur at constant temperature and can be characterized by enthalpy changes which are always positive. Enthalpy of formation, combustion and other enthalpy changes can be calculated using Hess’s law. Enthalpy change for chemical reactions can be determined by b i  f H reactions   r H    a i  f H products    f i and in gaseous state by ∆rH = Σ bond enthalpies of the reactants – Σ bond enthalpies of the products First law of thermodynamics does not guide us about the direction of chemical reactions i.e., what is the driving force of a chemical reaction. For isolated systems, ∆U = 0. We define another state function, S, entropy for this purpose. Entropy is a measure of disorder or randomness. For a spontaneous change, total entropy change is positive. Therefore, for an isolated system, ∆U = 0, ∆S > 0, so entropy change distinguishes a spontaneous change, while energy change does not. Entropy changes can be measured by the equation qrev qrev ∆S = for a reversible process. is independent of path. T T Chemical reactions are generally carried at constant pressure, so we define another state function Gibbs energy, G, which is related to entropy and enthalpy changes of the system by the equation: ∆rG = ∆rH – T ∆rS For a spontaneous change, ∆Gsys < 0 and at equilibrium, ∆Gsys = 0. Standard Gibbs energy change is related to equilibrium constant by ∆rG = – RT ln K. K can be calculated from this equation, if we know ∆rG which can be found from . Temperature is an important factor in the equation. Many reactions which are non-spontaneous at low temperature, are made spontaneous at high temperature for systems having positive entropy of reaction. Reprint 2025-26 THERMODYNAMICS 165 Exercises 5.1 Choose the correct answer. A thermodynamic state function is a quantity (i) used to determine heat changes (ii) whose value is independent of path (iii) used to determine pressure volume work (iv) whose value depends on temperature only. 5.2 For the process to occur under adiabatic conditions, the correct condition is: (i) ∆T = 0 (ii) ∆p = 0 (iii) q = 0 (iv) w = 0 5.3 The enthalpies of all elements in their standard states are: (i) unity (ii) zero (iii) < 0 (iv) different for each element 5.4 ∆U  of combustion of methane is – X kJ mol–1. The value of ∆H is (i) = ∆U  (ii) > ∆U  (iii) < ∆U  (iv) = 0 5.5 The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol–1 –393.5 kJ mol–1, and –285.8 kJ mol–1 respectively. Enthalpy of formation of CH4(g) will be (i) –74.8 kJ mol–1 (ii) –52.27 kJ mol–1 (iii) +74.8 kJ mol–1 (iv) +52.26 kJ mol–1. 5.6 A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be (i) possible at high temperature (ii) possible only at low temperature (iii) not possible at any temperature (v) possible at any temperature 5.7 In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process? 5.8 The reaction of cyanamide, NH2CN (s), with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be –742.7 kJ mol–1 at 298 K. Calculate enthalpy change for the reaction at 298 K. 3 NH2CN(g) + O2(g) → N2(g) + CO2(g) + H2O(l) 2 Reprint 2025-26 166 chemIstry 5.9 Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol–1 K–1. 5.10 Calculate the enthalpy change on freezing of 1.0 mol of water at10.0°C to ice at –10.0°C. ∆fusH = 6.03 kJ mol–1 at 0°C. Cp [H2O(l)] = 75.3 J mol–1 K–1 Cp [H2O(s)] = 36.8 J mol–1 K–1 5.11 Enthalpy of combustion of carbon to CO2 is –393.5 kJ mol–1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas. 5.12 Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110, – 393, 81 and 9.7 kJ mol–1 respectively. Find the value of ∆rH for the reaction: N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g) 5.13 Given N2(g) + 3H2(g) → 2NH3(g); ∆rH = –92.4 kJ mol–1 What is the standard enthalpy of formation of NH3 gas? 5.14 Calculate the standard enthalpy of formation of CH3OH(l) from the following data: 3 CH3OH (l) + O2(g) → CO2(g) + 2H2O(l) ; ∆rH = –726 kJ mol–1 2 C(graphite) + O2(g) → CO2(g) ; ∆cH = –393 kJ mol–1 1 H2(g) + O2(g) → H2O(l); ∆f H = –286 kJ mol–1. 2 5.15 Calculate the enthalpy change for the process CCl4(g) → C(g) + 4 Cl(g) and calculate bond enthalpy of C – Cl in CCl4(g). ∆vapH(CCl4) = 30.5 kJ mol–1. ∆fH (CCl4) = –135.5 kJ mol–1. ∆aH (C) = 715.0 kJ mol–1, where ∆aH is enthalpy of atomisation ∆aH (Cl2) = 242 kJ mol–1 5.16 For an isolated system, ∆U = 0, what will be ∆S ? 5.17 For the reaction at 298 K, 2A + B → C ∆H = 400 kJ mol–1 and ∆S = 0.2 kJ K–1 mol–1 At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range. 5.18 For the reaction, 2 Cl(g) → Cl2(g), what are the signs of ∆H and ∆S ? 5.19 For the reaction 2 A(g) + B(g) → 2D(g) ∆U  = –10.5 kJ and ∆S = –44.1 JK–1. Calculate ∆G for the reaction, and predict whether the reaction may occur spontaneously. Reprint 2025-26 THERMODYNAMICS 167 5.20 The equilibrium constant for a reaction is 10. What will be the value of ∆G ? R = 8.314 JK–1 mol–1, T = 300 K. 5.21 Comment on the thermodynamic stability of NO(g), given 1 1 N2(g) + O2(g) → NO(g); ∆rH = 90 kJ mol–1 2 2 1 NO(g) + O2(g) → NO2(g): ∆rH= –74 kJ mol–1 2 5.22 Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ∆f H = –286 kJ mol–1. Reprint 2025-26 Unit 6 Equilibrium Chemical equilibria are important in numerous biological and environmental processes. For example, equilibria involving O2 molecules and the protein hemoglobin play After studying this unit you will be a crucial role in the transport and delivery of O2 fromable to our lungs to our muscles. Similar equilibria involving CO • identify dynamic nature of molecules and hemoglobin account for the toxicity of CO. equilibrium involved in physical and chemical processes; When a liquid evaporates in a closed container, • state the law of equilibrium; molecules with relatively higher kinetic energy escape • explain characteristics of the liquid surface into the vapour phase and number of equilibria involved in physical liquid molecules from the vapour phase strike the liquid and chemical processes; surface and are retained in the liquid phase. It gives rise • write expressions for equilibrium to a constant vapour pressure because of an equilibrium in constants; which the number of molecules leaving the liquid equals the• establish a relationship between number returning to liquid from the vapour. We say that Kp and Kc; the system has reached equilibrium state at this stage.• explain various factors that affect the equilibrium state of a However, this is not static equilibrium and there is a lot of reaction; activity at the boundary between the liquid and the vapour. • classify substances as acids or Thus, at equilibrium, the rate of evaporation is equal to the bases according to Arrhenius, rate of condensation. It may be represented by Bronsted-Lowry and Lewis concepts; H2O (l) H2O (vap) • classify acids and bases as The double half arrows indicate that the processes weak or strong in terms of their in both the directions are going on simultaneously. The ionization constants; • explain the dependence of degree mixture of reactants and products in the equilibrium state of ionization on concentration is called an equilibrium mixture. of the electrolyte and that of the Equilibrium can be established for both physical common ion; processes and chemical reactions. The reaction may be• describe pH scale for representing hydrogen ion concentration; fast or slow depending on the experimental conditions and • explain ionisation of water and the nature of the reactants. When the reactants in a closed its duel role as acid and base; vessel at a particular temperature react to give products, • describe ionic product (Kw ) and the concentrations of the reactants keep on decreasing, pKw for water; while those of products keep on increasing for some time • appreciate use of buffer after which there is no change in the concentrations solutions; of either of the reactants or products. This stage of the• calculate solubility product constant. system is the dynamic equilibrium and the rates of the forward and reverse reactions become equal. It is due to Reprint 2025-26 EQUILIBRIUM 169 this dynamic equilibrium stage that there is characteristic features. We observe that the no change in the concentrations of various mass of ice and water do not change with species in the reaction mixture. Based on the time and the temperature remains constant. extent to which the reactions proceed to reach However, the equilibrium is not static. the state of chemical equilibrium, these may The intense activity can be noticed at the be classified in three groups. boundary between ice and water. Molecules from the liquid water collide against ice and(i) The reactions that proceed nearly adhere to it and some molecules of ice escape to completion and only negligible into liquid phase. There is no change of mass concentrations of the reactants are of ice and water, as the rates of transfer of left. In some cases, it may not be even molecules from ice into water and of reverse possible to detect these experimentally. transfer from water into ice are equal at(ii) The reactions in which only small atmospheric pressure and 273 K. amounts of products are formed and It is obvious that ice and water are in most of the reactants remain unchanged equilibrium only at particular temperature at equilibrium stage. and pressure. For any pure substance at(iii) The reactions in which the concentrations atmospheric pressure, the temperature at of the reactants and products are which the solid and liquid phases are at comparable, when the system is in equilibrium is called the normal melting point equilibrium. or normal freezing point of the substance. The The extent of a reaction in equilibrium system here is in dynamic equilibrium and we varies with the experimental conditions such can infer the following: as concentrations of reactants, temperature, (i) Both the opposing processes occur etc. Optimisation of the operational conditions simultaneously. is very important in industry and laboratory (ii) Both the processes occur at the same so that equilibrium is favorable in the rate so that the amount of ice and water direction of the desired product. Some remains constant. important aspects of equilibrium involving physical and chemical processes are dealt in 6.1.2 Liquid-Vapour Equilibrium this unit along with the equilibrium involving This equilibrium can be better understood if ions in aqueous solutions which is called as we consider the example of a transparent box ionic equilibrium. carrying a U-tube with mercury (manometer). Drying agent like anhydrous calcium chloride6.1 E Q U I L I B R I U M I N P H Y S I C A L (or phosphorus penta-oxide) is placed for PROCESSES a few hours in the box. After removing the The characteristics of system at equilibrium drying agent by tilting the box on one side, a are better understood if we examine some watch glass (or petri dish) containing water physical processes. The most familiar examples is quickly placed inside the box. It will be are phase transformation processes, e.g., observed that the mercury level in the right solid liquid limb of the manometer slowly increases and liquid gas finally attains a constant value, that is, the solid gas pressure inside the box increases and reaches a constant value. Also the volume of water in6.1.1 Solid-Liquid Equilibrium the watch glass decreases (Fig. 6.1). Initially Ice and water kept in a perfectly insulated there was no water vapour (or very less) inside thermos flask (no exchange of heat between its the box. As water evaporated the pressure in contents and the surroundings) at 273K and the box increased due to addition of water the atmospheric pressure are in equilibrium molecules into the gaseous phase inside state and the system shows interesting the box. The rate of evaporation is constant. Reprint 2025-26 170 chemistry Fig. 6.1 Measuring equilibrium vapour pressure of water at a constant temperature However, the rate of increase in pressure vapour to liquid state is much less than the decreases with time due to condensation rate of evaporation. These are open systems of vapour into water. Finally it leads to an and it is not possible to reach equilibrium in equilibrium condition when there is no net an open system. evaporation. This implies that the number Water and water vapour are in equilibriumof water molecules from the gaseous state position at atmospheric pressure (1.013 bar)into the liquid state also increases till the and at 100°C in a closed vessel. The boilingequilibrium is attained i.e., point of water is 100°C at 1.013 bar pressure. rate of evaporation= rate of condensation For any pure liquid at one atmospheric H2O(l) H2O (vap) pressure (1.013 bar), the temperature At equilibrium the pressure exerted by at which the liquid and vapours are at the water molecules at a given temperature equilibrium is called normal boiling point of remains constant and is called the equilibrium the liquid. Boiling point of the liquid depends vapour pressure of water (or just vapour on the atmospheric pressure. It depends on pressure of water); vapour pressure of water the altitude of the place; at high altitude the increases with temperature. If the above boiling point decreases. experiment is repeated with methyl alcohol, acetone and ether, it is observed that different 6.1.3 Solid – Vapour Equilibrium liquids have different equilibrium vapour Let us now consider the systems where solids pressures at the same temperature, and the sublime to vapour phase. If we place solidliquid which has a higher vapour pressure is iodine in a closed vessel, after sometimemore volatile and has a lower boiling point. the vessel gets filled up with violet vapour If we expose three watch glasses containing and the intensity of colour increases with separately 1mL each of acetone, ethyl alcohol, time. After certain time the intensity of and water to atmosphere and repeat the colour becomes constant and at this stage experiment with different volumes of the equilibrium is attained. Hence solid iodine liquids in a warmer room, it is observed sublimes to give iodine vapour and the iodinethat in all such cases the liquid eventually vapour condenses to give solid iodine. Thedisappears and the time taken for complete equilibrium can be represented as,evaporation depends on (i) the nature of the liquid, (ii) the amount of the liquid and (iii) the I2(solid) I2 (vapour) temperature. When the watch glass is open Other examples showing this kind of to the atmosphere, the rate of evaporation equilibrium are, remains constant but the molecules are Camphor (solid) Camphor (vapour)dispersed into large volume of the room. As a consequence the rate of condensation from NH4Cl (solid) NH4Cl (vapour) Reprint 2025-26 EQUILIBRIUM 171 6.1.4 Equilibrium Involving Dissolution pressure of the gas above the solvent. of Solid or Gases in Liquids This amount decreases with increase of Solids in liquids temperature. The soda water bottle is sealed under pressure of gas when its solubility inWe know from our experience that we can water is high. As soon as the bottle is opened,dissolve only a limited amount of salt or some of the dissolved carbon dioxide gassugar in a given amount of water at room escapes to reach a new equilibrium conditiontemperature. If we make a thick sugar syrup required for the lower pressure, namely itssolution by dissolving sugar at a higher partial pressure in the atmosphere. This istemperature, sugar crystals separate out if we how the soda water in bottle when left open cool the syrup to the room temperature. We call to the air for some time, turns ‘flat’. It can be it a saturated solution when no more of solute generalised that: can be dissolved in it at a given temperature. (i) For solid liquid equilibrium, there isThe concentration of the solute in a saturated only one temperature (melting point) atsolution depends upon the temperature. In 1 atm (1.013 bar) at which the twoa saturated solution, a dynamic equilibrium phases can coexist. If there is noexits between the solute molecules in the solid exchange of heat with the surroundings,state and in the solution: the mass of the two phases remainsSugar (solution) Sugar (solid), and constant. the rate of dissolution of sugar = rate of (ii) For liquid vapour equilibrium, thecrystallisation of sugar. vapour pressure is constant at a given Equality of the two rates and dynamic temperature. nature of equilibrium has been confirmed with (iii) For dissolution of solids in liquids,the help of radioactive sugar. If we drop some the solubility is constant at a givenradioactive sugar into saturated solution of temperature.non-radioactive sugar, then after some time radioactivity is observed both in the solution (iv) For dissolution of gases in liquids, and in the solid sugar. Initially there were no the concentration of a gas in liquid radioactive sugar molecules in the solution is proportional to the pressure but due to dynamic nature of equilibrium, (concentration) of the gas over the liquid. there is exchange between the radioactive These observations are summarised in and non-radioactive sugar molecules between Table 6.1 the two phases. The ratio of the radioactive Table 6.1 Some Features of Physical Equilibria to non-radioactive molecules in the solution increases till it attains a constant value. Process Conclusion Liquid Vapour pH2Oconstant at givenGases in liquids H2O (l) H2O (g) temperature When a soda water bottle is opened, some of Solid Liquid Melting point is fixed atthe carbon dioxide gas dissolved in it fizzes constant pressure out rapidly. The phenomenon arises due H2O (s) H2O (l) to difference in solubility of carbon dioxide Solute(s) Solute Concentration of solute at different pressures. There is equilibrium (solution) in solution is constant between the molecules in the gaseous state Sugar(s) Sugar at a given temperature and the molecules dissolved in the liquid (solution) under pressure i.e., Gas(g) Gas (aq) [gas(aq)]/[gas(g)] is CO2 (gas) CO2 (in solution) constant at a given This equilibrium is governed by Henry’s temperature CO2(g) CO2(aq)law, which states that the mass of a gas [CO2(aq)]/[CO2(g)] is dissolved in a given mass of a solvent at constant at a given temperatureany temperature is proportional to the Reprint 2025-26 172 chemistry 6.1.5 General Characteristics of Equilibria Involving Physical Processes For the physical processes discussed above, following characteristics are common to the system at equilibrium: (i) Equilibrium is possible only in a closed system at a given temperature. (ii) Both the opposing processes occur at the same rate and there is a dynamic but stable condition. (iii) All measurable properties of the system remain constant. (iv) When equilibrium is attained for a Fig. 6.2 Attainment of chemical equilibrium. physical process, it is characterised by constant value of one of its parameters at a given temperature. Table 6.1 lists Eventually, the two reactions occur at the such quantities. same rate and the system reaches a state of equilibrium.(v) The magnitude of such quantities at any stage indicates the extent to which the Similarly, the reaction can reach the state physical process has proceeded before of equilibrium even if we start with only C and reaching equilibrium. D; that is, no A and B being present initially, as the equilibrium can be reached from either 6.2 EQUILIBRIUM IN CHEMICAL direction. PROCESSES – DYNAMIC The dynamic nature of chemical EQUILIBRIUM equilibrium can be demonstrated in the Analogous to the physical systems chemical synthesis of ammonia by Haber’s process. reactions also attain a state of equilibrium. In a series of experiments, Haber started These reactions can occur both in forward with known amounts of dinitrogen and and backward directions. When the rates of dihydrogen maintained at high temperature the forward and reverse reactions become and pressure and at regular intervals equal, the concentrations of the reactants determined the amount of ammonia present. and the products remain constant. This He was successful in determining also the is the stage of chemical equilibrium. This concentration of unreacted dihydrogen and equilibrium is dynamic in nature as it consists dinitrogen. Fig. 6.4 (page 174) shows that after of a forward reaction in which the reactants a certain time the composition of the mixture give product(s) and reverse reaction in which remains the same even though some of the product(s) gives the original reactants. reactants are still present. This constancy in For a better comprehension, let us consider composition indicates that the reaction has a general case of a reversible reaction, reached equilibrium. In order to understand the dynamic nature of the reaction, synthesis A + B C + D of ammonia is carried out with exactly the With passage of time, there is same starting conditions (of partial pressure accumulation of the products C and D and and temperature) but using D2 (deuterium) depletion of the reactants A and B (Fig. 6.2). in place of H2. The reaction mixtures starting This leads to a decrease in the rate of either with H2 or D2 reach equilibrium with forward reaction and an increase in the rate the same composition, except that D2 and of the reverse reaction, ND3 are present instead of H2 and NH3. After Reprint 2025-26 EQUILIBRIUM 173 Dynamic Equilibrium – A Student’s Activity Equilibrium whether in a physical or in a chemical system, is always of dynamic nature. This can be demonstrated by the use of radioactive isotopes. This is not feasible in a school laboratory. However this concept can be easily comprehended by performing the following activity. The activity can be performed in a group of 5 or 6 students. Take two 100mL measuring cylinders (marked as 1 and 2) and two glass tubes each of 30 cm length. Diameter of the tubes may be same or different in the range of 3-5mm. Fill nearly half of the measuring cylinder-1 with coloured water (for this purpose add a crystal of potassium permanganate to water) and keep second cylinder (number 2) empty. Put one tube in cylinder 1 and second in cylinder 2. Immerse one tube in cylinder 1, close its upper tip with a finger and transfer the coloured water contained in its lower portion to cylinder 2. Using second tube, kept in 2nd cylinder, transfer the coloured water in a similar manner from cylinder 2 to cylinder 1. In this way keep on transferring coloured water using the two glass tubes from cylinder 1 to 2 and from 2 to 1 till you notice that the level of coloured water in both the cylinders becomes constant. If you continue intertransferring coloured solution between the cylinders, there will not be any further change in the levels of coloured water in two cylinders. If we take analogy of ‘level’ of coloured water with ‘concentration’ of reactants and products in the two cylinders, we can say the process of transfer, which continues even after the constancy of level, is indicative of dynamic nature of the process. If we repeat the experiment taking two tubes of different diameters we find that at equilibrium the level of coloured water in two cylinders is different. How far diameters are responsible for change in levels in two cylinders? Empty cylinder (2) is an indicator of no product in it at the beginning. 1 2 1 2 (a) (b) Fig.6.3 Demonstrating dynamic nature of equilibrium. (a) initial stage (b) final stage after the equilibrium is attained. Reprint 2025-26 174 chemistry 2NH3(g) N2(g) + 3H2(g) Similarly let us consider the reaction, H2(g) + I2(g) 2HI(g). If we start with equal initial concentration of H2 and I2, the reaction proceeds in the forward direction and the concentration of H2 and I2 decreases while that of HI increases, until all of these become constant at equilibrium (Fig. 6.5). We can also start with HI alone and make the reaction to proceed in the reverse direction; the concentration of HI will decrease and concentration of H2 and I2 will increase until they all become constant when equilibrium is reached (Fig. 6.5). If total number of H and I atoms are same in a given volume, the same equilibrium mixture is obtained whether we Fig. 6.4 Depiction of equilibrium for the reaction start it from pure reactants or pure product. N 2 g 3 H 2 g 2 NH 3 g equilibrium is attained, these two mixtures (H2, N2, NH3 and D2, N2, ND3) are mixed together and left for a while. Later, when this mixture is analysed, it is found that the concentration of ammonia is just the same as before. However, when this mixture is analysed by a mass spectrometer, it is found that ammonia and all deuterium containing forms of ammonia (NH3, NH2D, NHD2 and ND3) and dihydrogen and its deutrated forms (H2, HD and D2) are present. Thus one can conclude that scrambling of H and D atoms in the molecules must result from a continuation of the forward and reverse reactions in the mixture. If the reaction had simply stopped Fig. 6.5 Chemical equilibrium in the reaction H2(g)when they reached equilibrium, then there + I2(g) 2HI(g) can be attained fromwould have been no mixing of isotopes in either direction this way. 6.3 LAW OF CHEMICAL EQUILIBRIUM Use of isotope (deuterium) in the formation AND EQUILIBRIUM CONSTANT of ammonia clearly indicates that chemical A mixture of reactants and products in thereactions reach a state of dynamic equilibrium state is called an equilibriumequilibrium in which the rates of forward mixture. In this section we shall address aand reverse reactions are equal and there number of important questions about theis no net change in composition. composition of equilibrium mixtures: What is Equilibrium can be attained from both the relationship between the concentrations sides, whether we start reaction by taking, of reactants and products in an equilibrium H2(g) and N2(g) and get NH3(g) or by taking mixture? How can we determine equilibrium NH3(g) and decomposing it into N2(g) and H2(g). concentrations from initial concentrations? What factors can be exploited to alter the N2(g) + 3H2(g) 2NH3(g) Reprint 2025-26 EQUILIBRIUM 175 composition of an equilibrium mixture? The Six sets of experiments with varying initial last question in particular is important when conditions were performed, starting with only choosing conditions for synthesis of industrial gaseous H2 and I2 in a sealed reaction vessel chemicals such as H2, NH3, CaO etc. in first four experiments (1, 2, 3 and 4) and To answer these questions, let us consider only HI in other two experiments (5 and 6). a general reversible reaction: Experiment 1, 2, 3 and 4 were performed taking different concentrations of H2 and / or A + B C + D where A and B are the reactants, C and D I2, and with time it was observed that intensity of the purple colour remained constant andare the products in the balanced chemical equilibrium was attained. Similarly, forequation. On the basis of experimental studies of many reversible reactions, the Norwegian experiments 5 and 6, the equilibrium was chemists Cato Maximillian Guldberg and attained from the opposite direction. Peter Waage proposed in 1864 that the Data obtained from all six sets of concentrations in an equilibrium mixture experiments are given in Table 6.2. are related by the following equilibrium It is evident from the experiments 1, 2,equation, 3 and 4 that number of moles of dihydrogen C D Kc  (6.1) reacted = number of moles of iodine reacted A B = ½ (number of moles of HI formed). Also, (6.1) where Kc is the equilibrium constant and experiments 5 and 6 indicate that, the expression on the right side is called the [H2(g)]eq = [I2(g)]eqequilibrium constant expression. Knowing the above facts, in order The equilibrium equation is also known as the law of mass action because in the early to establish a relationship between days of chemistry, concentration was called concentrations of the reactants and products, “active mass”. In order to appreciate their several combinations can be tried. Let us work better, let us consider reaction between consider the simple expression, gaseous H2 and I2 carried out in a sealed vessel [HI(g)]eq / [H2(g)]eq [I2(g)]eqat 731K. H2(g) + I2(g)   2HI(g) It can be seen from Table 6.3 that if we put the equilibrium concentrations of the 1 mol 1 mol 2 mol reactants and products, the above expression Table 6.2 Initial and Equilibrium Concentrations of H2, I2 and HI Reprint 2025-26 176 chemistry Table 6.3 E x p r e s s i o n I n v o l v i n g t h e The equilibrium constant for a general Equilibrium Concentration of reaction, Reactants a A + b B c C + d D H2(g) + I2(g) 2HI(g) is expressed as, Kc = [C]c[D]d / [A]a[B]b (6.4) where [A], [B], [C] and [D] are the equilibrium concentrations of the reactants and products. Equilibrium constant for the reaction, 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) is written as Kc = [NO]4[H2O]6 / [NH3]4 [O2]5 Molar concentration of different species is is far from constant. However, if we consider indicated by enclosing these in square bracket the expression, and, as mentioned above, it is implied that these are equilibrium concentrations. While [HI(g)]2 eq / [H2(g)]eq [I2(g)]eq writing expression for equilibrium constant, we find that this expression gives constant symbol for phases (s, l, g) are generally ignored. value (as shown in Table 6.3) in all the six Let us write equilibrium constant for thecases. It can be seen that in this expression the power of the concentration for reactants reaction, H2(g) + I2(g) 2HI(g) (6.5) and products are actually the stoichiometric as, Kc = [HI]2 / [H2] [I2] = x (6.6)coefficients in the equation for the chemical reaction. Thus, for the reaction H2(g) + I2(g) The equilibrium constant for the reverse 2HI(g), following equation 6.1, the equilibrium reaction, 2HI(g) H2(g) + I2(g), at the same constant Kc is written as, temperature is, Kc = [HI(g)]eq2 / [H2(g)]eq [I2(g)]eq (6.2) K′c = [H2] [I2] / [HI]2 = 1/ x = 1 / Kc (6.7) Thus, K′c = 1 / Kc (6.8) Generally the subscript ‘eq’ (used for equilibrium) is omitted from the concentration Equilibrium constant for the reverse terms. It is taken for granted that the reaction is the inverse of the equilibrium concentrations in the expression for Kc are constant for the reaction in the forward equilibrium values. We, therefore, write, direction. Kc = [HI(g)]2 / [H2(g)] [I2(g)] (6.3) If we change the stoichiometric coefficients in a chemical equation by multiplying The subscript ‘c’ indicates that Kc is throughout by a factor then we must makeexpressed in concentrations of mol L–1. sure that the expression for equilibrium At a given temperature, the product of constant also reflects that change. For concentrations of the reaction products example, if the reaction (6.5) is written as, raised to the respective stoichiometric coefficient in the balanced chemical ½ H2(g) + ½ I2(g) HI(g) (6.9) equation divided by the product of the equilibrium constant for the above concentrations of the reactants raised to reaction is given by their individual stoichiometric coefficients K″c = [HI] / [H2]1/2[I2]1/2 = {[HI]2 / [H2][I2]}1/2 has a constant value. This is known as 1/2 = x1/2 = Kc (6.10)the Equilibrium Law or Law of Chemical Equilibrium. On multiplying the equation (6.5) by n, we get Reprint 2025-26 EQUILIBRIUM 177 nH2(g) + nI2(g) D 2nHI(g) (6.11) Therefore, equilibrium constant for the N2(g) + O2(g) 2NO(g) reaction is equal to Kc n. These findings are Solution summarised in Table 6.4. It should be noted For the reaction equilibrium constant, Kcthat because the equilibrium constants Kc can be written as, and K′c have different numerical values, it is 2 important to specify the form of the balanced NO Kc =chemical equation when quoting the value of N 2 O 2 an equilibrium constant. –3 2 2.8 10 M Table 6.4 Relations between Equilibrium = 3 3 3 .0 10 M 4 .2 10 M Constants for a General Reaction and its Multiples. = 0.622 Chemical equation Equilibrium constant a A + b B c C + dD Kc 6.4 HOMOGENEOUS EQUILIBRIA c C + d D a A + b B K′c =(1/Kc ) In a homogeneous system, all the reactants and products are in the same phase. na A + nb B ncC + ndD K′″c = (Kcn) For example, in the gaseous reaction, N2(g) + 3H2(g) 2NH3(g), reactants and products are in the homogeneous phase. Problem 6.1 Similarly, for the reactions, The following concentrations were CH3COOC2H5 (aq) + H2O (l) CH3COOH (aq) obtained for the formation of NH3 from + C2H5OH (aq) N 2 and H 2 at equilibrium at 500K. and, Fe3+ (aq) + SCN–(aq) Fe(SCN)2+ (aq) [N2] = 1.5 × 10–2M. [H2] = 3.0 ×10–2 M and [NH3] = 1.2 ×10–2M. Calculate equilibrium all the reactants and products are in constant. homogeneous solution phase. We shall now consider equilibrium constant for some Solution homogeneous reactions. The equilibrium constant for the reaction, N2(g) + 3H2(g) 2NH3(g) can be written as, 6.4.1 Equilibrium Constant in Gaseous Systems  NH 3  g  2 So far we have expressed equilibrium Kc = 3 constant of the reactions in terms of molar  N 2  g   H 2  g  concentration of the reactants and products, 2 2 and used symbol, Kc for it. For reactions 1 .2  10   involving gases, however, it is usually more = 2 2 3 1 .5  10 3 .0  10 convenient to express the equilibrium     constant in terms of partial pressure. = 0.355 × 103 = 3.55 × 102 The ideal gas equation is written as, pV = nRT Problem 6.2 n At equilibrium, the concentrations of ⇒ p = RT V N2=3.0 × 10–3M, O2 = 4.2 × 10–3M and NO= 2.8 × 10–3M in a sealed vessel at 800K. Here, p is the pressure in Pa, n is the number What will be Kc for the reaction of moles of the gas, V is the volume in m3 and T is the temperature in Kelvin Reprint 2025-26 178 chemistry Therefore, n/V is concentration expressed in mol/m3 2 −2 RT ] −2  NH 3 ( g )[ If concentration c, is in mol/L or mol/dm3, = 3 = K c ( RT ) and p is in bar then  N 2 ( g )  H 2 ( g ) p = cRT, −2 or K p = K c ( RT ) (6.14)We can also write p = [gas]RT. Here, R= 0.0831 bar litre/mol K Similarly, for a general reaction At constant temperature, the pressure of a A + b B c C + d D the gas is proportional to its concentration i.e., c d c d (c + d ) p p T ) C )( D ) [ C ] [ D ] ( Rp ∝ [gas] ( K p = a b = a b (a + b ) p p For reaction in equilibrium T ) ( A )( B ) [ A ] [ B ] ( R H2(g) + I2(g) 2HI(g) We can write either [ C ]c [ D ]d ( c + d )− (a + b ) 2 = a b ( RT ) A ] [ B ]  HI ( g ) [Kc =  H 2 ( g )  I 2 ( g ) c d [ C ] [ D ] ∆n ∆n 2 = a b ( RT ) = K c ( RT ) (6.15) ( p HI ) [ A ] [ B ]or K c = (6.12) where ∆n = (number of moles of gaseous ( p H 2 )( p I 2 ) products) – (number of moles of gaseous reactants) in the balanced chemical equation. RTFurther, since p HI =  HI ( g ) It is necessary that while calculating the value RT p I 2 =  I 2 ( g ) of Kp, pressure should be expressed in bar because standard state for pressure is 1 bar. RT p H 2 =  H 2 ( g ) We know from Unit 1 that : Therefore, 1pascal, Pa=1Nm–2, and 1bar = 105 Pa RT ]2 Kp values for a few selected reactions at ( p HI )2  HI ( g )[2 K p = = different temperatures are given in Table 6.5 RT H 2 ( g ) RT .  I 2 ( g ) ( p H 2 )( p I 2 )  2 Table 6.5 Equilibrium Constants, Kp for  HI ( g ) a Few Selected Reactions = = Kc (6.13)  H 2 ( g )  I 2 ( g ) In this example, Kp = Kc i.e., both equilibrium constants are equal. However, this is not always the case. For example in reaction N2(g) + 3H2(g) 2NH3(g) 2 ( p NH 3 )K p = 3 ( p N 2 )( p H 2 ) 2 2 Problem 6.3 RT ]  NH 3 ( g )[ = 3 3 PCl5, PCl3 and Cl2 are at equilibrium at 500 RT ) K and having concentration 1.59M PCl3,  N 2 ( g ) RT .  H 2 ( g )( 1.59M Cl2 and 1.41 M PCl5. Reprint 2025-26 EQUILIBRIUM 179 Calculate Kc for the reaction, the value 0.194 should be neglected because PCl5 PCl3 + Cl2 it will give concentration of the reactant Solution which is more than initial concentration. Hence the equilibrium concentrations are, The equilibrium constant Kc for the above reaction can be written as, [CO2] = [H2-] = x = 0.067 M Cl 2  1 .59 2 [CO] = [H2O] = 0.1 – 0.067 = 0.033 M PCl 3  Kc    1 .79 1 .41 Problem 6.5  PCl 5   Problem 6.4 For the equilibrium, The value of Kc = 4.24 at 800K for the 2NOCl(g) 2NO(g) + Cl2(g) reaction, the value of the equilibrium constant, Kc is CO (g) + H2O (g) CO2 (g) + H2 (g) 3.75 × 10–6 at 1069 K. Calculate the Kp for Calculate equilibrium concentrations of CO2, the reaction at this temperature? H2, CO and H2O at 800 K, if only CO and H2O are present initially at concentrations Solution of 0.10M each. We know that, Solution Kp = Kc(RT)∆n For the reaction, For the above reaction, ∆n = (2+1) – 2 = 1 CO (g) + H2O (g) CO2 (g) + H2 (g) Kp = 3.75 ×10–6 (0.0831 × 1069) Initial concentration: Kp = 0.033 0.1M 0.1M 0 0 Let x mole per litre of each of the product be formed. 6.5 HETEROGENEOUS EQUILIBRIA At equilibrium: Equilibrium in a system having more than one (0.1-x) M (0.1-x) M x M x M phase is called heterogeneous equilibrium. The equilibrium between water vapour and where x is the amount of CO2 and H2 at liquid water in a closed container is an equilibrium. example of heterogeneous equilibrium. Hence, equilibrium constant can be written as, H2O(l) H2O(g) Kc = x2/(0.1-x)2 = 4.24 In this example, there is a gas phase and x2 = 4.24(0.01 + x2-0.2x) a liquid phase. In the same way, equilibrium between a solid and its saturated solution, x2 = 0.0424 + 4.24x2-0.848x 3.24x2 – 0.848x + 0.0424 = 0 Ca(OH)2 (s) + (aq) Ca2+ (aq) + 2OH–(aq) a = 3.24, b = – 0.848, c = 0.0424 is a heterogeneous equilibrium. (for quadratic equation ax2 + bx + c = 0, Heterogeneous equilibria often involve 2 pure solids or liquids. We can simplify  b  b  4 ac   equilibrium expressions for the heterogeneous x  2 a equilibria involving a pure liquid or a pure solid, as the molar concentration of a pure x = 0.848±√(0.848)2– 4(3.24)(0.0424)/ solid or liquid is constant (i.e., independent (3.24×2) of the amount present). In other words if a x = (0.848 ± 0.4118)/ 6.48 substance ‘X’ is involved, then [X(s)] and [X(l)] x1 = (0.848 – 0.4118)/6.48 = 0.067 are constant, whatever the amount of ‘X’ is x2 = (0.848 + 0.4118)/6.48 = 0.194 taken. Contrary to this, [X(g)] and [X(aq)] will Reprint 2025-26 180 chemistry vary as the amount of X in a given volume This shows that at a particular varies. Let us take thermal dissociation of temperature, there is a constant concentration calcium carbonate which is an interesting and or pressure of CO2 in equilibrium with CaO(s) important example of heterogeneous chemical and CaCO3(s). Experimentally it has been equilibrium. found that at 1100 K, the pressure of CO2 in equilibrium with CaCO3(s) and CaO(s), is CaCO3 (s) CaO (s) + CO2 (g) (6.16) 2.0 ×105 Pa. Therefore, equilibrium constant On the basis of the stoichiometric equation, at 1100K for the above reaction is: we can write, Kp = PCO2 = 2 × 105 Pa/105 Pa = 2.00  CaO  s   CO 2  g  Similarly, in the equilibrium between Kc   CaCO 3  s  nickel, carbon monoxide and nickel carbonyl (used in the purification of nickel), Since [CaCO3(s)] and [CaO(s)] are both constant, therefore modified equilibrium Ni (s) + 4 CO (g) Ni(CO)4 (g), constant for the thermal decomposition of the equilibrium constant is written as calcium carbonate will be  Ni  CO 4  K´c = [CO2(g)] (6.17) Kc  4  CO  or Kp = pCO2 (6.18) It must be remembered that for the existence of heterogeneous equilibrium pure Units of Equilibrium Constant solids or liquids must also be present (however small the amount may be) at equilibrium, but The value of equilibrium constant Kc can be calculated by substituting the concentration their concentrations or partial pressures do terms in mol/L and for Kp partial pressure not appear in the expression of the equilibrium is substituted in Pa, kPa, bar or atm. This constant. In the reaction, results in units of equilibrium constant based Ag2O(s) + 2HNO3(aq) 2AgNO3(aq) +H2O(l) on molarity or pressure, unless the exponents of both the numerator and denominator are  AgNO3  2 same. Kc = 2 For the reactions,  HNO 3  H2(g) + I2(g) 2HI, Kc and Kp have no unit. N2O4(g) 2NO2 (g), Kc has unit mol/L and Kp has unit bar Problem 6.6 Equilibrium constants can also be The value of Kp for the reaction, expressed as dimensionless quantities if the CO2 (g) + C (s) 2CO (g) standard state of reactants and products is 3.0 at 1000 K. If initially PCO2 = 0.48 are specified. For a pure gas, the standard bar and PCO = 0 bar and pure graphite is state is 1bar. Therefore a pressure of 4 bar present, calculate the equilibrium partial in standard state can be expressed as 4 pressures of CO and CO2. bar/1 bar = 4, which is a dimensionless Solution number. Standard state (c0) for a solute is 1 molar solution and all concentrations can be For the reaction, measured with respect to it. The numerical let ‘x’ be the decrease in pressure of CO2, value of equilibrium constant depends on the then standard state chosen. Thus, in this system CO2(g) + C(s) 2CO(g) both Kp and Kc are dimensionless quantities Initial but have different numerical values due to pressure: 0.48 bar 0 different standard states. Reprint 2025-26 EQUILIBRIUM 181 5. The equilibrium constant K for a reaction At equilibrium: is related to the equilibrium constant (0.48 – x)bar 2x bar of the corresponding reaction, whose 2 equation is obtained by multiplying or pCO K p = dividing the equation for the original pCO 2 reaction by a small integer. Let us consider applications of equilibrium Kp = (2x)2/(0.48 – x) = 3 constant to: 4x2 = 3(0.48 – x) • predict the extent of a reaction on the 4x2 = 1.44 – x basis of its magnitude, 4x2 + 3x – 1.44 = 0 • predict the direction of the reaction, and a = 4, b = 3, c = –1.44 • calculate equilibrium concentrations.  b  b 2  4 ac   x  6.6.1 Predicting the Extent of a 2 a Reaction = [–3 ± √(3)2– 4(4)(–1.44)]/2 × 4 The numerical value of the equilibrium constant for a reaction indicates the extent = (–3 ± 5.66)/8 of the reaction. But it is important to note = (–3 + 5.66)/8 (as value of x cannot be that an equilibrium constant does not give negative hence we neglect that value) any information about the rate at which x = 2.66/8 = 0.33 the equilibrium is reached. The magnitude The equilibrium partial pressures are, of Kc or Kp is directly proportional to the concentrations of products (as these appear pCO2= 2x = 2 × 0.33 = 0.66 bar in the numerator of equilibrium constant pCO2= 0.48 – x = 0.48 – 0.33 = 0.15 bar expression) and inversely proportional to the concentrations of the reactants (these appear

3.20 For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained. t (sec) P(mm of Hg) 0 35.0 360 54.0 720 63.0 Calculate the rate constant.