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

11.2 THERMAL EQUILIBRIUM independent variables. Let the pressure and Equilibrium in mechanics means that the net volume of the gases be (PA, VA) and (PB, VB) external force and torque on a system are zero. respectively. Suppose first that the two systems The term ‘equilibrium’ in thermodynamics appears are put in proximity but are separated by an * Thermodynamics may also involve other variables that are not so obvious to our senses e.g. entropy, enthalpy, etc., and they are all macroscopic variables. However, a thermodynamic state is specified by five state variables viz., pressure, volume, temperature, internal energy and entropy. Entropy is a measure of disorderness in the system. Enthalpy is a measure of total heat content of the system. Reprint 2025-26 228 PHYSICS adiabatic wall – an insulating wall (can be law in 1931 long after the first and second Laws movable) that does not allow flow of energy (heat) of thermodynamics were stated and so numbered. from one to another. The systems are insulated The Zeroth Law clearly suggests that when two from the rest of the surroundings also by similar systems A and B, are in thermal equilibrium, adiabatic walls. The situation is shown there must be a physical quantity that has the schematically in Fig. 11.1 (a). In this case, it is same value for both. This thermodynamic found that any possible pair of values (PA, VA) will variable whose value is equal for two systems in be in equilibrium with any possible pair of values thermal equilibrium is called temperature (T ). (PB, VB). Next, suppose that the adiabatic wall is Thus, if A and B are separately in equilibrium replaced by a diathermic wall – a conducting wall with C, TA = TC and TB = TC. This implies that that allows energy flow (heat) from one to another. TA = TB i.e. the systems A and B are also in It is then found that the macroscopic variables of thermal equilibrium. the systems A and B change spontaneously until We have arrived at the concept of temperature both the systems attain equilibrium states. After formally via the Zeroth Law. The next question that there is no change in their states. The is : how to assign numerical values to situation is shown in Fig. 11.1(b). The pressure temperatures of different bodies ? In other words, and volume variables of the two gases change to how do we construct a scale of temperature ? (PB ′, VB ′) and (PA ′, VA ′) such that the new states Thermometry deals with this basic question to of A and B are in equilibrium with each other*. which we turn in the next section. There is no more energy flow from one to another. We then say that the system A is in thermal equilibrium with the system B. What characterises the situation of thermal equilibrium between two systems ? You can guess the answer from your experience. In thermal equilibrium, the temperatures of the two systems are equal. We shall see how does one arrive at the concept of temperature in thermodynamics? The Zeroth law of thermodynamics provides the clue.

11.4 HEAT, INTERNAL ENERGY AND WORK associated with various random motions of its The Zeroth Law of Thermodynamics led us to molecules. We will see in the next chapter that the concept of temperature that agrees with our in a gas this motion is not only translational commonsense notion. Temperature is a marker (i.e. motion from one point to another in the of the ‘hotness’ of a body. It determines the volume of the container); it also includes rotational and vibrational motion of thedirection of flow of heat when two bodies are molecules (Fig. 11.3).placed in thermal contact. Heat flows from the body at a higher temperature to the one at lower temperature. The flow stops when the temperatures equalise; the two bodies are then in thermal equilibrium. We saw in some detail how to construct temperature scales to assign temperatures to different bodies. We now describe the concepts of heat and other relevant quantities like internal energy and work. Fig. 11.3 (a) Internal energy U of a gas is the sum of the kinetic and potential energies of its The concept of internal energy of a system is molecules when the box is at rest. Kinetic not difficult to understand. We know that every energy due to various types of motion bulk system consists of a large number of (translational, rotational, vibrational) is to molecules. Internal energy is simply the sum of be included in U. (b) If the same box is the kinetic energies and potential energies of moving as a whole with some velocity, these molecules. We remarked earlier that in the kinetic energy of the box is not to be thermodynamics, the kinetic energy of the included in U. system, as a whole, is not relevant. Internal energy is thus, the sum of molecular kinetic and potential energies in the frame of reference relative to which the centre of mass of the system is at rest. Thus, it includes only the (disordered) energy associated with the random motion of molecules of the system. We denote the internal energy of a system by U. Though we have invoked the molecular picture to understand the meaning of internal energy, as far as thermodynamics is concerned, U is simply a macroscopic variable of the system. The important thing about internal energy is that it depends only on the state of the system, not on how that state was achieved. Internal energy U of a system is an example of a thermodynamic ‘state variable’ – its value depends only on the given state of the system, not on history i.e. not on the ‘path’ taken to arrive at that state. Thus, the internal energy of a given mass of gas depends on its state described by Fig. 11.4 Heat and work are two distinct modes of specific values of pressure, volume and energy transfer to a system that results in temperature. It does not depend on how this change in its internal energy. (a) Heat is energy transfer due to temperaturestate of the gas came about. Pressure, volume, difference between the system and the temperature, and internal energy are surroundings. (b) Work is energy transfer thermodynamic state variables of the system brought about by means (e.g. moving the (gas) (see section 11.7). If we neglect the small piston by raising or lowering some weight intermolecular forces in a gas, the internal connected to it) that do not involve such a energy of a gas is just the sum of kinetic energies temperature difference. Reprint 2025-26 230 PHYSICS What are the ways of changing internal 11.5 FIRST LAW OF THERMODYNAMICS energy of a system ? Consider again, for We have seen that the internal energy U of a simplicity, the system to be a certain mass of system can change through two modes of energy gas contained in a cylinder with a movable transfer : heat and work. Let piston as shown in Fig. 11.4. Experience shows ∆Q = Heat supplied to the system by thethere are two ways of changing the state of the surroundingsgas (and hence its internal energy). One way is to put the cylinder in contact with a body at a ∆W = Work done by the system on the higher temperature than that of the gas. The surroundings temperature difference will cause a flow of ∆U = Change in internal energy of the system energy (heat) from the hotter body to the gas, The general principle of conservation of thus increasing the internal energy of the gas. energy then implies that The other way is to push the piston down i.e. to ∆Q = ∆U + ∆W (11.1) do work on the system, which again results in i.e. the energy (∆Q) supplied to the system goesincreasing the internal energy of the gas. Of in partly to increase the internal energy of thecourse, both these things could happen in the system (∆U) and the rest in work on thereverse direction. With surroundings at a lower environment (∆W). Equation (11.1) is known astemperature, heat would flow from the gas to the First Law of Thermodynamics. It is simplythe surroundings. Likewise, the gas could push the general law of conservation of energy applied the piston up and do work on the surroundings. to any system in which the energy transfer from In short, heat and work are two different modes or to the surroundings is taken into account. of altering the state of a thermodynamic system Let us put Eq. (11.1) in the alternative form and changing its internal energy. The notion of heat should be carefully ∆Q – ∆W = ∆U (11.2) distinguished from the notion of internal energy. Heat is certainly energy, but it is the energy in Now, the system may go from an initial state transit. This is not just a play of words. The to the final state in a number of ways. For distinction is of basic significance. The state of example, to change the state of a gas from a thermodynamic system is characterised by its (P1, V1) to (P2, V2), we can first change the volume of the gas from V1 to V2, keeping itsinternal energy, not heat. A statement like ‘a pressure constant i.e. we can first go the stategas in a given state has a certain amount of heat’ is as meaningless as the statement that (P1, V2) and then change the pressure of the gas from P1 to P2, keeping volume constant, to‘a gas in a given state has a certain amount take the gas to (P2, V2). Alternatively, we canof work’. In contrast, ‘a gas in a given state first keep the volume constant and then keep has a certain amount of internal energy’ is a the pressure constant. Since U is a state perfectly meaningful statement. Similarly, the variable, ∆U depends only on the initial and statements ‘a certain amount of heat is final states and not on the path taken by the supplied to the system’ or ‘a certain amount gas to go from one to the other. However, ∆Q of work was done by the system’ are perfectly and ∆W will, in general, depend on the path meaningful. taken to go from the initial to final states. From To summarise, heat and work in the First Law of Thermodynamics, Eq. (11.2), thermodynamics are not state variables. They it is clear that the combination ∆Q – ∆W, is are modes of energy transfer to a system however, path independent. This shows that resulting in change in its internal energy, if a system is taken through a process in which ∆U = 0 (for example, isothermal expansion ofwhich, as already mentioned, is a state variable. an ideal gas, see section 11.8), In ordinary language, we often confuse heat with internal energy. The distinction between ∆Q = ∆W them is sometimes ignored in elementary physics books. For proper understanding of i.e., heat supplied to the system is used up thermodynamics, however, the distinction is entirely by the system in doing work on the crucial. environment. Reprint 2025-26 THERMODYNAMICS 231 If the system is a gas in a cylinder with a If the amount of substance is specified in movable piston, the gas in moving the piston does terms of moles µ (instead of mass m in kg ), we work. Since force is pressure times area, and can define heat capacity per mole of the area times displacement is volume, work done substance by by the system against a constant pressure P is S 1 ∆Q C = = (11.6) ∆W = P ∆V µ µ ∆T C is known as molar specific heat capacity of where ∆V is the change in volume of the gas. the substance. Like s, C is independent of the Thus, for this case, Eq. (11.1) gives amount of substance. C depends on the nature ∆Q = ∆U + P ∆V (11.3) of the substance, its temperature and the conditions under which heat is supplied. The As an application of Eq. (11.3), consider the unit of C is J mo1–1 K–1. As we shall see later (in change in internal energy for 1 g of water when connection with specific heat capacity of gases), we go from its liquid to vapour phase. The additional conditions may be needed to define measured latent heat of water is 2256 J/g. i.e., C or s. The idea in defining C is that simple for 1 g of water ∆Q = 2256 J. At atmospheric predictions can be made in regard to molar pressure, 1 g of water has a volume 1 cm3 in specific heat capacities. liquid phase and 1671 cm3 in vapour phase. Table 11.1 lists measured specific and molar heat capacities of solids at atmospheric pressure Therefore, and ordinary room temperature. ∆W =P (Vg –Vl ) = 1.013 ×105 × (1671 × 10–6) =169.2 J We will see in Chapter 12 that predictions of specific heats of gases generally agree with Equation (11.3) then gives experiment. We can use the same law of equipartition of energy that we use there to ∆U = 2256 – 169.2 = 2086.8 J predict molar specific heat capacities of solids We see that most of the heat goes to increase (See Section 12.5 and 12.6). Consider a solid of the internal energy of water in transition from N atoms, each vibrating about its mean the liquid to the vapour phase. position. An oscillator in one dimension has