2.10 The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below: Concentration/M 0.001 0.010 0.020 0.050 0.100 102 × k/S m–1 1.237 11.85 23.15 55.53 106.74 Calculate Λm for all concentrations and draw a plot between Λm and c½. Find the value of 0m . 2.11 Conductivity of 0.00241 M acetic acid is 7.896 × 10–5 S cm–1. Calculate its molar conductivity. If 0m for acetic acid is 390.5 S cm2 mol–1, what is its dissociation constant? 2.12 How much charge is required for the following reductions: (i) 1 mol of Al3+ to Al? (ii) 1 mol of Cu2+ to Cu? (iii) 1 mol of MnO4– to Mn2+? 2.13 How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl2? (ii) 40.0 g of Al from molten Al2O3? 2.14 How much electricity is required in coulomb for the oxidation of (i) 1 mol of H2O to O2? (ii) 1 mol of FeO to Fe2O3? 2.15 A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode? 2.16 Three electrolytic cells A,B,C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited? 2.17 Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible: (i) Fe3+(aq) and I–(aq) (ii) Ag+ (aq) and Cu(s) (iii) Fe3+ (aq) and Br– (aq) (iv) Ag(s) and Fe 3+ (aq) (v) Br2 (aq) and Fe2+ (aq). 2.18 Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO3 with silver electrodes. (ii) An aqueous solution of AgNO3 with platinum electrodes. (iii) A dilute solution of H2SO4 with platinum electrodes. (iv) An aqueous solution of CuCl2 with platinum electrodes. Answers to Some Intext Questions 2.5 E(cell) = 0.91 V 2.6 ∆ rG o = −45.54 kJ mol −1 , Kc = 9.62 ×107 2.9 0.114, 3.67 × 10–4 mol L–1 Chemistry 60 Reprint 2025-26 UnitUnitUnit Unit33Unit Objectives ChemicalChemical KineticsKinetics After studying this Unit, you will be able to · define the average and Chemical Kinetics helps us to understand how chemical reactions instantaneous rate of a reaction; occur. · express the rate of a reaction in terms of change in concentration Chemistry, by its very nature, is concerned with change. of either of the reactants or Substances with well defined properties are converted products with time; by chemical reactions into other substances with · distinguish between elementary different properties. For any chemical reaction, chemists and complex reactions; try to find out · differentiate between the (a) the feasibility of a chemical reaction which can be molecularity and order of a reaction; predicted by thermodynamics ( as you know that a · define rate constant; reaction with DG < 0, at constant temperature and pressure is feasible);· discuss the dependence of rate of reactions on concentration, (b) extent to which a reaction will proceed can be temperature and catalyst; determined from chemical equilibrium; · derive integrated rate equations (c) speed of a reaction i.e. time taken by a reaction to for the zero and first order reach equilibrium. reactions; Along with feasibility and extent, it is equally · determine the rate constants for important to know the rate and the factors controlling zeroth and first order reactions; the rate of a chemical reaction for its complete · describe collision theory. understanding. For example, which parameters determine as to how rapidly food gets spoiled? How to design a rapidly setting material for dental filling? Or what controls the rate at which fuel burns in an auto engine? All these questions can be answered by the branch of chemistry, which deals with the study of reaction rates and their mechanisms, called chemical kinetics. The word kinetics is derived from the Greek word ‘kinesis’ meaning movement. Thermodynamics tells only about the feasibility of a reaction whereas chemical kinetics tells about the rate of a reaction. For example, thermodynamic data indicate that diamond shall convert to graphite but in reality the conversion rate is so slow that the change is not perceptible at all. Therefore, most people think Reprint 2025-26 that diamond is forever. Kinetic studies not only help us to determine the speed or rate of a chemical reaction but also describe the conditions by which the reaction rates can be altered. The factors such as concentration, temperature, pressure and catalyst affect the rate of a reaction. At the macroscopic level, we are interested in amounts reacted or formed and the rates of their consumption or formation. At the molecular level, the reaction mechanisms involving orientation and energy of molecules undergoing collisions, are discussed. In this Unit, we shall be dealing with average and instantaneous rate of reaction and the factors affecting these. Some elementary ideas about the collision theory of reaction rates are also given. However, in order to understand all these, let us first learn about the reaction rate. 3.13.13.13.13.1 RateRateRateRateRate ofofofofof aaaaa Some reactions such as ionic reactions occur very fast, for example, ChemicalChemicalChemicalChemicalChemical precipitation of silver chloride occurs instantaneously by mixing of aqueous solutions of silver nitrate and sodium chloride. On the other ReactionReactionReactionReactionReaction hand, some reactions are very slow, for example, rusting of iron in the presence of air and moisture. Also there are reactions like inversion of cane sugar and hydrolysis of starch, which proceed with a moderate speed. Can you think of more examples from each category? You must be knowing that speed of an automobile is expressed in terms of change in the position or distance covered by it in a certain period of time. Similarly, the speed of a reaction or the rate of a reaction can be defined as the change in concentration of a reactant or product in unit time. To be more specific, it can be expressed in terms of: (i) the rate of decrease in concentration of any one of the reactants, or (ii) the rate of increase in concentration of any one of the products. Consider a hypothetical reaction, assuming that the volume of the system remains constant. R ® P One mole of the reactant R produces one mole of the product P. If [R]1 and [P]1 are the concentrations of R and P respectively at time t1 and [R]2 and [P]2 are their concentrations at time t2 then, Dt = t2 – t1 D[R] = [R]2 – [R]1 D [P] = [P]2 – [P]1 The square brackets in the above expressions are used to express molar concentration. Rate of disappearance of R Decrease in concentration of R ∆ [ R ] = = − (3.1) Time taken ∆ t Chemistry 62 Reprint 2025-26 Rate of appearance of P Increase in concentration of P ∆ [ P ] = = + (3.2) Time taken ∆t Since, D[R] is a negative quantity (as concentration of reactants is decreasing), it is multiplied with –1 to make the rate of the reaction a positive quantity. Equations (3.1) and (3.2) given above represent the average rate of a reaction, rav. Average rate depends upon the change in concentration of reactants or products and the time taken for that change to occur (Fig. 3.1). { } Fig. 3.1: Instantaneous and average rate of a reaction Units of rate of a reaction From equations (3.1) and (3.2), it is clear that units of rate are concentration time–1. For example, if concentration is in mol L–1 and time is in seconds then the units will be mol L-1s–1. However, in gaseous reactions, when the concentration of gases is expressed in terms of their partial pressures, then the units of the rate equation will be atm s–1. From the concentrations of C4H9Cl (butyl chloride) at different times given ExampleExampleExampleExampleExample 3.13.13.13.13.1 below, calculate the average rate of the reaction: C4H9Cl + H2O ® C4H9OH + HCl during different intervals of time. t/s 0 50 100 150 200 300 400 700 800 [C4H9Cl]/mol L–1 0.100 0.0905 0.0820 0.0741 0.0671 0.0549 0.0439 0.0210 0.017 We can determine the difference in concentration over different intervals SolutionSolutionSolutionSolutionSolution of time and thus determine the average rate by dividing D[R] by Dt (Table 3.1). 63 Chemical Kinetics Reprint 2025-26 Table 3.1: Average rates of hydrolysis of butyl chloride [C4H9CI]t1 / [C4H9CI]t2 / t1/s t2/s rav × 104/mol L–1s–1 × 10 4 mol L–1 mol L–1 = – / ( t 2 − t1 ) C 4 H 9 Cl ]t 2 – [ C 4 H 9 Cl ]t1 {[ } 0.100 0.0905 0 50 1.90 0.0905 0.0820 50 100 1.70 0.0820 0.0741 100 150 1.58 0.0741 0.0671 150 200 1.40 0.0671 0.0549 200 300 1.22 0.0549 0.0439 300 400 1.10 0.0439 0.0335 400 500 1.04 0.0210 0.017 700 800 0.4 It can be seen (Table 3.1) that the average rate falls from 1.90 × 0-4 mol L-1s-1 to 0.4 × 10-4 mol L-1s-1. However, average rate cannot be used to predict the rate of a reaction at a particular instant as it would be constant for the time interval for which it is calculated. So, to express the rate at a particular moment of time we determine the instantaneous rate. It is obtained when we consider the average rate at the smallest time interval say dt ( i.e. when Dt approaches zero). Hence, mathematically for an infinitesimally small dt instantaneous rate is given by −∆ [ R ] ∆ [ P ] rav = = (3.3) ∆t ∆ t  d  R  d P As Dt ® 0 or rinst   d t d t Fig 3.2 Instantaneous rate of hydrolysis of butyl chloride(C4H9Cl) Chemistry 64 Reprint 2025-26 It can be determined graphically by drawing a tangent at time t on either of the curves for concentration of R and P vs time t and calculating its slope (Fig. 3.1). So in problem 3.1, rinst at 600s for example, can be calculated by plotting concentration of butyl chloride as a function of time. A tangent is drawn that touches the curve at t = 600 s (Fig. 3.2). The slope of this tangent gives the instantaneous rate. So, rinst at 600 s = – mol L–1 = 5.12 × 10–5 mol L–1s–1 At t = 250 s rinst = 1.22 × 10–4 mol L–1s–1 t = 350 s rinst = 1.0 × 10–4 mol L–1s–1 t = 450 s rinst = 6.4 ×× 10–5 mol L–1s–1 Now consider a reaction Hg(l) + Cl2 (g) ® HgCl2(s) Where stoichiometric coefficients of the reactants and products are same, then rate of the reaction is given as ∆ [ Hg ] ∆ [ Cl 2 ] ∆ [ HgCl 2 ] Rate of reaction = – = – = ∆t ∆t ∆ t i.e., rate of disappearance of any of the reactants is same as the rate of appearance of the products. But in the following reaction, two moles of HI decompose to produce one mole each of H2 and I2, 2HI(g) ® H2(g) + I2(g) For expressing the rate of such a reaction where stoichiometric coefficients of reactants or products are not equal to one, rate of disappearance of any of the reactants or the rate of appearance of products is divided by their respective stoichiometric coefficients. Since rate of consumption of HI is twice the rate of formation of H2 or I2, to make them equal, the term D[HI] is divided by 2. The rate of this reaction is given by 1 ∆ [ HI ] ∆ [ H 2 ] ∆ [ I 2 ] Rate of reaction = − = = 2 ∆ t ∆ t ∆ t Similarly, for the reaction 5 Br- (aq) + BrO3– (aq) + 6 H+ (aq) ® 3 Br2 (aq) + 3 H2O (l) − − + 1 ∆ [ Br BrO 3 1 ∆ [ H ] 1 ∆ [ Br2 ] 1 ∆ [ H 2 O ] ] ∆  Rate = − = − = − = = 5 ∆ t ∆ t 6 ∆t 3 ∆ t 3 ∆t For a gaseous reaction at constant temperature, concentration is directly proportional to the partial pressure of a species and hence, rate can also be expressed as rate of change in partial pressure of the reactant or the product. 65 Chemical Kinetics Reprint 2025-26 ExampleExampleExampleExampleExample 3.23.23.23.23.2 The decomposition of N2O5 in CCl4 at 318K has been studied by monitoring the concentration of N2O5 in the solution. Initially the concentration of N2O5 is 2.33 mol L–1 and after 184 minutes, it is reduced to 2.08 mol L–1. The reaction takes place according to the equation 2 N2O5 (g) ® 4 NO2 (g) + O2 (g) Calculate the average rate of this reaction in terms of hours, minutes and seconds. What is the rate of production of NO2 during this period? 1  ( 2.08 − 2 .33 ) mol L−1  1 ∆ [ N 2 O5 ] SolutionSolutionSolutionSolutionSolution Average Rate = − = − 184 min 2  ∆t  2   = 6.79 × 10–4 mol L–1/min = (6.79 × 10–4 mol L–1 min–1) × (60 min/1h) = 4.07 × 10–2 mol L–1/h = 6.79 × 10–4 mol L–1 × 1min/60s = 1.13 × 10–5 mol L–1s–1 It may be remembered that 1 ∆ [ NO 2 ] Rate = 4  ∆t  ∆ [ NO 2 ] = 6.79 × 10–4 × 4 mol L–1 min–1 = 2.72 × 10–3 mol L–1min–1 ∆t IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 3.1 For the reaction R ® P, the concentration of a reactant changes from 0.03M to 0.02M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds. 3.2 In a reaction, 2A ® Products, the concentration of A decreases from 0.5 mol L–1 to 0.4 mol L–1 in 10 minutes. Calculate the rate during this interval? 3.23.23.23.23.2 FactorsFactorsFactorsFactorsFactors InfluencingInfluencingInfluencingInfluencingInfluencing Rate of reaction depends upon the experimental conditions such RateRateRateRateRate ofofofofof aaaaa ReactionReactionReactionReactionReaction as concentration of reactants (pressure in case of gases), temperature and catalyst. 3.2.1 Dependence The rate of a chemical reaction at a given temperature may depend on of Rate on the concentration of one or more reactants and products. The Concentration representation of rate of reaction in terms of concentration of the reactants is known as rate law. It is also called as rate equation or rate expression. 3.2.2 Rate The results in Table 3.1 clearly show that rate of a reaction decreases with Expression the passage of time as the concentration of reactants decrease. Conversely, and Rate rates generally increase when reactant concentrations increase. So, rate of Constant a reaction depends upon the concentration of reactants. Chemistry 66 Reprint 2025-26 Consider a general reaction aA + bB ® cC + dD where a, b, c and d are the stoichiometric coefficients of reactants and products. The rate expression for this reaction is Rate µ [A] x [B] y (3.4) where exponents x and y may or may not be equal to the stoichiometric coefficients (a and b) of the reactants. Above equation can also be written as Rate = k [A] x [B] y (3.4a) d [ R ] x y − = k [ A ] [ B ] (3.4b) d t This form of equation (3.4 b) is known as differential rate equation, where k is a proportionality constant called rate constant. The equation like (3.4), which relates the rate of a reaction to concentration of reactants is called rate law or rate expression. Thus, rate law is the expression in which reaction rate is given in terms of molar concentration of reactants with each term raised to some power, which may or may not be same as the stoichiometric coefficient of the reacting species in a balanced chemical equation. For example: 2NO(g) + O2(g) ® 2NO2 (g) We can measure the rate of this reaction as a function of initial concentrations either by keeping the concentration of one of the reactants constant and changing the concentration of the other reactant or by changing the concentration of both the reactants. The following results are obtained (Table 3.2). Table 3.2: Initial rate of formation of NO2 Experiment Initial [NO]/ mol L-1 Initial [O2]/ mol L-1 Initial rate of formation of NO2/ mol L-1s-1 1. 0.30 0.30 0.096 2. 0.60 0.30 0.384 3. 0.30 0.60 0.192 4. 0.60 0.60 0.768 It is obvious, after looking at the results, that when the concentration of NO is doubled and that of O2 is kept constant then the initial rate increases by a factor of four from 0.096 to 0.384 mol L–1s–1. This indicates that the rate depends upon the square of the concentration of NO. When concentration of NO is kept constant and concentration of O2 is doubled the rate also gets doubled indicating that rate depends on concentration of O2 to the first power. Hence, the rate equation for this reaction will be Rate = k [NO] 2[O2] 67 Chemical Kinetics Reprint 2025-26 The differential form of this rate expression is given as d [ R ] 2 − = k [ NO ] [ O 2 ] d t Now, we observe that for this reaction in the rate equation derived from the experimental data, the exponents of the concentration terms are the same as their stoichiometric coefficients in the balanced chemical equation. Some other examples are given below: Reaction Experimental rate expression 1. CHCl3 + Cl2 ® CCl4 + HCl Rate = k [CHCl3 ] [Cl2]1/2 2. CH3COOC2H5 + H2O ® CH3COOH + C2H5OH Rate = k [CH3COOC2H5]1 [H2O]0 In these reactions, the exponents of the concentration terms are not the same as their stoichiometric coefficients. Thus, we can say that: Rate law for any reaction cannot be predicted by merely looking at the balanced chemical equation, i.e., theoretically but must be determined experimentally. 3.2.3 Order of a In the rate equation (3.4) Reaction Rate = k [A]x [B]y x and y indicate how sensitive the rate is to the change in concentration of A and B. Sum of these exponents, i.e., x + y in (3.4) gives the overall order of a reaction whereas x and y represent the order with respect to the reactants A and B respectively. Hence, the sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction. Order of a reaction can be 0, 1, 2, 3 and even a fraction. A zero order reaction means that the rate of reaction is independent of the concentration of reactants. ExampleExampleExampleExampleExample 3.33.33.33.33.3 Calculate the overall order of a reaction which has the rate expression (a) Rate = k [A]1/2 [B]3/2 (b) Rate = k [A]3/2 [B]–1 SolutionSolutionSolutionSolutionSolution (a) Rate = k [A]x [B]y order = x + y So order = 1/2 + 3/2 = 2, i.e., second order (b) order = 3/2 + (–1) = 1/2, i.e., half order. A balanced chemical equation never gives us a true picture of how a reaction takes place since rarely a reaction gets completed in one step. The reactions taking place in one step are called elementary reactions. When a sequence of elementary reactions (called mechanism) gives us the products, the reactions are called complex reactions. Chemistry 68 Reprint 2025-26 These may be consecutive reactions (e.g., oxidation of ethane to CO2 and H2O passes through a series of intermediate steps in which alcohol, aldehyde and acid are formed), reverse reactions and side reactions (e.g., nitration of phenol yields o-nitrophenol and p-nitrophenol). Units of rate constant For a general reaction aA + bB ® cC + dD Rate = k [A]x [B]y Where x + y = n = order of the reaction Rate k = x [A] [B]y concentration 1 = × n ( where [A] = [B]) time ( concentration ) Taking SI units of concentration, mol L –1 and time, s, the units of k for different reaction order are listed in Table 3.3 Table 3.3: Units of rate constant Reaction Order Units of rate constant mol L−1 1 −1 − 1 × 0 = mol L s −1 Zero order reaction 0 s ( mol L ) −1 mol L 1 −1 × = s − 1 1 First order reaction 1 s ( mol L ) − 1 mol L 1 − 1 −1 × = mol L s −1 2 Second order reaction 2 s ( mol L ) Identify the reaction order from each of the following rate constants. ExampleExampleExampleExampleExample 3.43.43.43.43.4 (i) k = 2.3 × 10–5 L mol–1 s–1 (ii) k = 3 × 10–4 s–1 (i) The unit of second order rate constant is L mol–1 s–1, therefore SolutionSolutionSolutionSolutionSolution k = 2.3 × 10–5 L mol–1 s–1 represents a second order reaction. (ii) The unit of a first order rate constant is s–1 therefore k = 3 × 10–4 s–1 represents a first order reaction. 3.2.4 Molecularity Another property of a reaction called molecularity helps in of a understanding its mechanism. The number of reacting species Reaction (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction is called molecularity of a reaction. The reaction can be unimolecular when one reacting species is involved, for example, decomposition of ammonium nitrite. 69 Chemical Kinetics Reprint 2025-26 NH4NO2 ® N2 + 2H2O Bimolecular reactions involve simultaneous collision between two species, for example, dissociation of hydrogen iodide. 2HI ® H2 + I2 Trimolecular or termolecular reactions involve simultaneous collision between three reacting species, for example, 2NO + O2 ® 2NO2 The probability that more than three molecules can collide and react simultaneously is very small. Hence, reactions with the molecularity three are very rare and slow to proceed. It is, therefore, evident that complex reactions involving more than three molecules in the stoichiometric equation must take place in more than one step. KClO3 + 6FeSO4 + 3H2SO4 ® KCl + 3Fe2(SO4)3 + 3H2O This reaction which apparently seems to be of tenth order is actually a second order reaction. This shows that this reaction takes place in several steps. Which step controls the rate of the overall reaction? The question can be answered if we go through the mechanism of reaction, for example, chances to win the relay race competition by a team depend upon the slowest person in the team. Similarly, the overall rate of the reaction is controlled by the slowest step in a reaction called the rate determining step. Consider the decomposition of hydrogen peroxide which is catalysed by iodide ion in an alkaline medium. -I 2H2O2  2H2O + O2 Alkaline medium The rate equation for this reaction is found to be d  H 2 O 2   Rate  k  H2 O 2 I dt This reaction is first order with respect to both H2O2 and I–. Evidences suggest that this reaction takes place in two steps (1) H2O2 + I– ® H2O + IO– (2) H2O2 + IO– ® H2O + I– + O2 Both the steps are bimolecular elementary reactions. Species IO- is called as an intermediate since it is formed during the course of the reaction but not in the overall balanced equation. The first step, being slow, is the rate determining step. Thus, the rate of formation of intermediate will determine the rate of this reaction. Thus, from the discussion, till now, we conclude the following: (i) Order of a reaction is an experimental quantity. It can be zero and even a fraction but molecularity cannot be zero or a non integer. (ii) Order is applicable to elementary as well as complex reactions whereas molecularity is applicable only for elementary reactions. For complex reaction molecularity has no meaning. Chemistry 70 Reprint 2025-26 (iii) For complex reaction, order is given by the slowest step and molecularity of the slowest step is same as the order of the overall reaction. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 3.3 For a reaction, A + B ® Product; the rate law is given by, r = k [ A]1/2 [B]2. What is the order of the reaction? 3.4 The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y ? 3.33.33.33.33.3 IntegratedIntegratedIntegratedIntegratedIntegrated We have already noted that the concentration dependence of rate is RateRateRateRateRate called differential rate equation. It is not always convenient to determine the instantaneous rate, as it is measured by determination EquationsEquationsEquationsEquationsEquations of slope of the tangent at point ‘t’ in concentration vs time plot (Fig. 3.1). This makes it difficult to determine the rate law and hence the order of the reaction. In order to avoid this difficulty, we can integrate the differential rate equation to give a relation between directly measured experimental data, i.e., concentrations at different times and rate constant. The integrated rate equations are different for the reactions of different reaction orders. We shall determine these equations only for zero and first order chemical reactions. 3.3.1 Zero Order Zero order reaction means that the rate of the reaction is proportional Reactions to zero power of the concentration of reactants. Consider the reaction, R ® P d  R  Rate =   k  R 0 d t As any quantity raised to power zero is unity d  R  Rate =   k × 1 d t d[R] = – k dt Integrating both sides [R] = – k t + I (3.5) where, I is the constant of integration. At t = 0, the concentration of the reactant R = [R]0, where [R]0 is initial concentration of the reactant. Substituting in equation (3.5) [R]0 = –k × 0 + I [R]0 = I Substituting the value of I in the equation (3.5) [R] = -kt + [R]0 (3.6) 71 Chemical Kinetics Reprint 2025-26 Comparing (3.6) with equation of a straight line, y = mx + c, if we plot [R] against t, we get a straight [R0 ] line (Fig. 3.3) with slope = –k and intercept equal to [R]0. Further simplifying equation (3.6), we get the rateR k = -slope constant, k as of [ R ]0 − [ R ] k = (3.7) t Zero order reactions are relatively uncommon but they occur under special conditions. Some enzyme Concentration catalysed reactions and reactions which occur on metal surfaces are a few examples of zero order 0 Time reactions. The decomposition of gaseous ammonia on a hot platinum surface is a zero order reaction at Fig. 3.3: Variation in the concentration high pressure. vs time plot for a zero order 1130K reaction 2NH 3 ( g ) →Pt catalyst N 2 ( g ) +3H 2 ( g ) Rate = k [NH3]0 = k In this reaction, platinum metal acts as a catalyst. At high pressure, the metal surface gets saturated with gas molecules. So, a further change in reaction conditions is unable to alter the amount of ammonia on the surface of the catalyst making rate of the reaction independent of its concentration. The thermal decomposition of HI on gold surface is another example of zero order reaction. 3.3.2 First Order In this class of reactions, the rate of the reaction is proportional to the Reactions first power of the concentration of the reactant R. For example, R ® P d [ R ] Rate = − = k [ R ] d t d [ R ] or = – kdt [ R ] Integrating this equation, we get ln [R] = – kt + I (3.8) Again, I is the constant of integration and its value can be determined easily. When t = 0, R = [R]0, where [R]0 is the initial concentration of the reactant. Therefore, equation (3.8) can be written as ln [R]0 = –k × 0 + I ln [R]0 = I Substituting the value of I in equation (3.8) ln[R] = –kt + ln[R]0 (3.9) Chemistry 72 Reprint 2025-26 Rearranging this equation [ R ] ln = −kt [ R ]0 1 R 0 or k  ln (3.10) t R At time t1 from equation (3.8) *ln[R]1 = – kt1 + *ln[R]0 (3.11) At time t2 ln[R]2 = – kt2 + ln[R]0 (3.12) where, [R]1 and [R]2 are the concentrations of the reactants at time t1 and t2 respectively. Subtracting (3.12) from (3.11) ln[R]1– ln[R]2 = – kt1 – (–kt2) [ R ]1 ln = k (t 2 − t 1 ) [ R ]2 1 [ R ]1 k = ln (t 2 − t 1 ) [ R ]2 (3.13) Equation (3.9) can also be written as [ R ] ln = −kt [ R ]0 Taking antilog of both sides [R] = [R]0 e–kt (3.14) Comparing equation (3.9) with y = mx + c, if we plot ln [R] against t (Fig. 3.4) we get a straight line with slope = –k and intercept equal to ln [R]0 The first order rate equation (3.10) can also be written in the form 2.303 [ R ]0 k = log (3.15) t [ R ] [ R ]0 kt * log = [ R ] 2.303 If we plot a graph between log [R]0/[R] vs t, (Fig. 3.5), the slope = k/2.303 Hydrogenation of ethene is an example of first order reaction. C2H4(g) + H2 (g) ® C2H6(g) Rate = k [C2H4] All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics. * Refer to Appendix-IV for ln and log (logarithms). 73 Chemical Kinetics Reprint 2025-26 /[R]) 0] Slope = k /2.303 ([R log 0 Time Fig. 3.4: A plot between ln[R] and t Fig. 3.5: Plot of log [R]0/[R] vs time for a for a first order reaction first order reaction 226 88 Ra  24 He  22286 Rn Rate = k [Ra] Decomposition of N2O5 and N2O are some more examples of first order reactions. ExampleExampleExampleExampleExample 3.53.53.53.53.5 The initial concentration of N2O5 in the following first order reaction N2O5(g) ® 2 NO2(g) + 1/2O2 (g) was 1.24 × 10–2 mol L–1 at 318 K. The concentration of N2O5 after 60 minutes was 0.20 × 10–2 mol L–1. Calculate the rate constant of the reaction at 318 K. SolutionSolutionSolutionSolutionSolution For a first order reaction  R 1 k t 2  t 1  log =  R 2 2.303 2.303  R 1 log k =  t 2  t 1   R 2 2.303 1.24  10  2 mol L1 log  2  1 =  60 min  0 min  0.20  10 mol L 2.303  1 = log 6.2 min 60 k = 0.0304 min-1 Let us consider a typical first order gas phase reaction A(g) ® B(g) + C(g) Let pi be the initial pressure of A and pt the total pressure at time ‘t’. Integrated rate equation for such a reaction can be derived as Total pressure pt = pA + pB + pC (pressure units) Chemistry 74 Reprint 2025-26 pA, pB and pC are the partial pressures of A, B and C, respectively. If x atm be the decrease in pressure of A at time t and one mole each of B and C is being formed, the increase in pressure of B and C will also be x atm each. A(g) ® B(g) + C(g) At t = 0 pi atm 0 atm 0 atm At time t (pi–x) atm x atm x atm where, pi is the initial pressure at time t = 0. pt = (pi – x) + x + x = pi + x x = (pt - pi) where, pA = pi – x = pi – (pt – pi) = 2pi – pt  2.303   p i  k =   log  (3.16)  t  p A  2.303 p i log = t  2 p i  p t  The following data were obtained during the first order thermal ExampleExampleExampleExampleExample 3.63.63.63.63.6 decomposition of N2O5 (g) at constant volume: 2N 2 O5 ( g ) → 2N 2 O 4 ( g ) + O 2 ( g ) S.No. Time/s Total Pressure/(atm) 1. 0 0.5 2. 100 0.512 Calculate the rate constant. Let the pressure of N2O5(g) decrease by 2x atm. As two moles of SolutionSolutionSolutionSolutionSolution N2O5 decompose to give two moles of N2O4(g) and one mole of O2 (g), the pressure of N2O4 (g) increases by 2x atm and that of O2 (g) increases by x atm. 2N 2 O5 ( g ) → 2N 2 O 4 ( g ) + O 2 ( g ) Start t = 0 0.5 atm 0 atm 0 atm At time t (0.5 – 2x) atm 2x atm x atm pt = p N 2 O 5  p N 2 O 4  p O 2 = (0.5 – 2x) + 2x + x = 0.5 + x x = tp − 0.5 p N 2 O5 = 0.5 – 2x = 0.5 – 2 (pt – 0.5) = 1.5 – 2pt At t = 100 s; pt = 0.512 atm 75 Chemical Kinetics Reprint 2025-26 p N 2 O 5 = 1.5 – 2 × 0.512 = 0.476 atm Using equation (3.16) 2.303 p i 2.303 0.5 atm k  log  log t p A 100s 0.476 atm 2.303  4 1   0.0216  4.98  10 s 100s 3.3.3 Half-Life of The half-life of a reaction is the time in which the concentration of a a Reaction reactant is reduced to one half of its initial concentration. It is represented as t1/2. For a zero order reaction, rate constant is given by equation 3.7. [ R ]0 − [ R ] k = t 1 [ R ]0 At t = t 1/2 , [ R ] = 2 The rate constant at t1/2 becomes [ R ]0 − 1/2 [ R ]0 k = t 1/2 [ R ]0 t 1/2 = 2k It is clear that t1/2 for a zero order reaction is directly proportional to the initial concentration of the reactants and inversely proportional to the rate constant. For the first order reaction, 2.303 [ R ]0 k = log (3.15) t [ R ] [ R ]0 at t1/2 [ R ] = (3.16) 2 So, the above equation becomes 2.303 [ R ]0 k = log t 1/2 [ R ] /2 2.303 or t1/2  log 2 k 2.303 t 1/2 = × 0.301 k 0.693 t 1/2 = (3.17) k Chemistry 76 Reprint 2025-26 It can be seen that for a first order reaction, half-life period is constant, i.e., it is independent of initial concentration of the reacting species. The half-life of a first order equation is readily calculated from the rate constant and vice versa. For zero order reaction t1/2 µ [R]0. For first order reaction t1/2 is independent of [R]0. A first order reaction is found to have a rate constant, k = 5.5 × 10-14 s-1. ExampleExampleExampleExampleExample 3.73.73.73.73.7 Find the half-life of the reaction. Half-life for a first order reaction is SolutionSolutionSolutionSolutionSolution 0.693 t 1/2 = k 0.693 t 1/2 = –14 –1 = 1.26 × 1013s 5.5×10 s Show that in a first order reaction, time required for completion of 99.9% is 10 times of half-life (t1/2) of the reaction. When reaction is completed 99.9%, [R]n = [R]0 – 0.999[R]0 ExampleExampleExampleExampleExample 3.83.83.83.83.8 2.303  R 0 log k = SolutionSolutionSolutionSolutionSolution t  R  2.303  R 0 2.303 3 log = = log10 t  R 0  0.999  R 0 t t = 6.909/k For half-life of the reaction t1/2 = 0.693/k t 6.909 k =   10 t1/2 k 0.693 Table 3.4 summarises the mathematical features of integrated laws of zero and first order reactions. Table 3.4: Integrated Rate Laws for the Reactions of Zero and First Order Order Reaction Differential Integrated Straight Half- Units of k type rate law rate law line plot life 0 R® P d[R]/dt = -k kt = [R]0-[R] [R] vs t [R]0/2k conc time-1 or mol L–1s–1 1 R® P d[R]/dt = -k[R] [R] = [R]0e-kt ln[R] vs t ln 2/k time-1 or s–1 or kt = ln{[R]0/[R]} 77 Chemical Kinetics Reprint 2025-26 The order of a reaction is sometimes altered by conditions. There are many reactions which obey first order rate law although they are higher order reactions. Consider the hydrolysis of ethyl acetate which is a chemical reaction between ethyl acetate and water. In reality, it is a second order reaction and concentration of both ethyl acetate and water affect the rate of the reaction. But water is taken in large excess for hydrolysis, therefore, concentration of water is not altered much during the reaction. Thus, the rate of reaction is affected by concentration of ethyl acetate only. For example, during the hydrolysis of 0.01 mol of ethyl acetate with 10 mol of water, amounts of the reactants and products at the beginning (t = 0) and completion (t) of the reaction are given as under. H CH3COOH + C2H5OH CH3COOC2H5 + H2O  t = 0 0.01 mol 10 mol 0 mol 0 mol t 0 mol 9.99 mol 0.01 mol 0.01 mol The concentration of water does not get altered much during the course of the reaction. So, the reaction behaves as first order reaction. Such reactions are called pseudo first order reactions. Inversion of cane sugar is another pseudo first order reaction. C12H22O11 + H2O →H+ C6H12O6 + C6H12O6 Cane sugar Glucose Fructose Rate = k [C12H22O11] IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 3.5 A first order reaction has a rate constant 1.15 × 10-3 s-1. How long will 5 g of this reactant take to reduce to 3 g? 3.6 Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction. 3.43.43.43.43.4 TemperatureTemperatureTemperatureTemperatureTemperature Most of the chemical reactions are accelerated by increase in temperature. For example, in decomposition of N2O5, the time taken for half of the DependenceDependenceDependenceDependenceDependence ofofofofof original amount of material to decompose is 12 min at 50oC, 5 h at thethethethethe RateRateRateRateRate ofofofofof aaaaa 25oC and 10 days at 0oC. You also know that in a mixture of potassium ReactionReactionReactionReactionReaction permanganate (KMnO4) and oxalic acid (H2C2O4), potassium permanganate gets decolourised faster at a higher temperature than that at a lower temperature. It has been found that for a chemical reaction with rise in temperature by 10°, the rate constant is nearly doubled. The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation (3.18). It was first proposed by Dutch chemist, J.H. van’t Hoff but Swedish chemist, Arrhenius provided its physical justification and interpretation. Chemistry 78 Reprint 2025-26 k = A e -Ea /RT (3.18) where A is the Arrhenius factor or the frequency factor. It is also called pre-exponential factor. It is a constant specific to a particular reaction. R is gas constant and Ea is activation energy measured in joules/mole (J mol –1). It can be understood clearly using the following simple reaction H 2 g  I 2 g  2HI g According to Arrhenius, this reaction can take place only when a molecule of hydrogen and a molecule of iodine Intermediate collide to form an unstable intermediate (Fig. 3.6). It exists for a very short time and then breaks up to form two Fig. 3.6: Formation of HI through molecules of hydrogen iodide. the intermediate The energy required to form this intermediate, called activated complex (C), is known as activation energy (Ea). Fig. 3.7 is obtained by plotting potential energy vs reaction coordinate. Reaction coordinate represents the profile of energy change when reactants change into products. Some energy is released when the complex decomposes to form products. So, the final enthalpy of the reaction depends upon the nature of reactants and products. All the molecules in the reacting species do not have the same kinetic Fig. 3.7: Diagram showing plot of potential energy. Since it is difficult to predict the energy vs reaction coordinate behaviour of any one molecule with precision, Ludwig Boltzmann and James Clark Maxwell used statistics to predict the behaviour of large number of molecules. According to them, the distribution of kinetic energy may be described by plotting the fraction of molecules (NE/NT) with a given kinetic energy (E) vs kinetic energy (Fig. 3.8). Here, NE is the number of molecules with energy E and NT is total number of molecules. The peak of the curve corresponds to the most probable kinetic energy, i.e., kinetic energy of maximum fraction of molecules. There are decreasing number Fig. 3.8: Distribution curve showing energies of molecules with energies higher or among gaseous molecules lower than this value. When the 79 Chemical Kinetics Reprint 2025-26 temperature is raised, the maximum of the curve moves to the higher energy value (Fig. 3.9) and the curve broadens out, i.e., spreads to the right such that there is a greater proportion of molecules with much higher energies. The area under the curve must be constant since total probability must be one at all times. We can mark the position of Ea on Fig. 3.9: Distribution curve showing temperature Maxwell Boltzmann distribution curve dependence of rate of a reaction (Fig. 3.9). Increasing the temperature of the substance increases the fraction of molecules, which collide with energies greater than Ea. It is clear from the diagram that in the curve at (t + 10), the area showing the fraction of molecules having energy equal to or greater than activation energy gets doubled leading to doubling the rate of a reaction. In the Arrhenius equation (3.18) the factor e -Ea /RT corresponds to the fraction of molecules that have kinetic energy greater than Ea. Taking natural logarithm of both sides of equation (3.18) E a ln k = – + ln A (3.19) RT The plot of ln k vs 1/T gives a straight line according to the equation (3.19) as shown in Fig. 3.10. Thus, it has been found from Arrhenius equation (3.18) that increasing the temperature or decreasing the activation energy will result in an increase in the rate of the reaction and an exponential increase in the rate constant. E a In Fig. 3.10, slope = – and intercept = ln R A. So we can calculate Ea and A using these values. At temperature T1, equation (3.19) is E a ln k1 = – RT1 + ln A (3.20) At temperature T2, equation (3.19) is E a ln k2 = – RT2 + ln A (3.21) (since A is constant for a given reaction) k1 and k2 are the values of rate constants at temperatures T1 and T2 respectively. Fig. 3.10: A plot between ln k and 1/T Chemistry 80 Reprint 2025-26 Subtracting equation (3.20) from (3.21), we obtain E a E a ln k2 – ln k1 = RT1 – RT2 k 2 E a  1 1  ln = − k1 R  T1 T2  k 2 E a  1 1  log = − (3.22) k1 2.303 R  T1 T2  k 2 E a  T2 − T1  log = k1 2. 303R  T1T2  ExampleExampleExampleExampleExample 3.93.93.93.93.9 The rate constants of a reaction at 500K and 700K are 0.02s–1 and 0.07s–1 respectively. Calculate the values of Ea and A. k 2 E a  T2  T1 SolutionSolutionSolutionSolutionSolution log =   k1 2.303 R  T1T2  0.07  E a   700  500  log =      700  500 0.02  2.303  8.314 JK  1 mol  1  0.544 = Ea × 5.714 × 10-4/19.15 Ea = 0.544 × 19.15/5.714 × 10–4 = 18230.8 J Since k = Ae-Ea/RT × 500 0.02 = Ae-18230.8/8.314 A = 0.02/0.012 = 1.61 ExampleExampleExampleExampleExample 3.103.103.103.103.10 The first order rate constant for the decomposition of ethyl iodide by the reaction C2H5I(g) ® C2H4 (g) + HI(g) at 600K is 1.60 × 10–5 s–1. Its energy of activation is 209 kJ/mol. Calculate the rate constant of the reaction at 700K. SolutionSolutionSolutionSolutionSolution We know that E a  1 1     log k2 – log k1 = 2.303R  T1 T2  81 Chemical Kinetics Reprint 2025-26 E a  1 1  log k2 = log k1     2.303R  T1 T2  209000 J mol L 1  1 1  5 = log 1.60  10     1  1    2.303  8.314 J mol L K  600 K 700K  log k2 = – 4.796 + 2.599 = – 2.197 k2 = 6.36 × 10–3 s–1 3.4.1 Effect of A catalyst is a substance which increases the rate of a reaction without Catalyst itself undergoing any permanent chemical change. For example, MnO2 catalyses the following reaction so as to increase its rate considerably. 2KClO3 MnO2 2 KCl + 3O2 The word catalyst should not be used when the added substance reduces the rate of raction. The substance is then called inhibitor. The action of the catalyst can be explained by intermediate complex theory. According to this theory, a catalyst participates in a chemical reaction by forming temporary bonds with the reactants resulting in an intermediate complex. This has a transitory existence and decomposes to yield products and the catalyst. It is believed that the catalyst provides an alternate pathway or reaction mechanism by reducing the activation energy between reactants and products and hence lowering the potential energy barrier as shown in Fig. 3.11. It is clear from Arrhenius equation (3.18) that lower the value of activation energy faster will be the rate of a reaction. A small amount of the catalyst can catalyse a large amount of reactants. A catalyst does Fig. 3.11: Effect of catalyst on activation energy not alter Gibbs energy, DG of a reaction. It catalyses the spontaneous reactions but does not catalyse non-spontaneous reactions. It is also found that a catalyst does not change the equilibrium constant of a reaction rather, it helps in attaining the equilibrium faster, that is, it catalyses the forward as well as the backward reactions to the same extent so that the equilibrium state remains same but is reached earlier. 3.53.53.53.53.5 CollisionCollisionCollisionCollisionCollision Though Arrhenius equation is applicable under a wide range of TheoryTheoryTheoryTheoryTheory ofofofofof circumstances, collision theory, which was developed by Max Trautz ChemicalChemicalChemicalChemicalChemical and William Lewis in 1916 -18, provides a greater insight into the energetic and mechanistic aspects of reactions. It is based on kinetic ReactionsReactionsReactionsReactionsReactions theory of gases. According to this theory, the reactant molecules are Chemistry 82 Reprint 2025-26 assumed to be hard spheres and reaction is postulated to occur when molecules collide with each other. The number of collisions per second per unit volume of the reaction mixture is known as collision frequency (Z). Another factor which affects the rate of chemical reactions is activation energy (as we have already studied). For a bimolecular elementary reaction A + B ® Products rate of reaction can be expressed as a / RT (3.23) Rate = Z AB e − E where ZAB represents the collision frequency of reactants, A and B and e-Ea /RT represents the fraction of molecules with energies equal to or greater than Ea. Comparing (3.23) with Arrhenius equation, we can say that A is related to collision frequency. Equation (3.23) predicts the value of rate constants fairly accurately for the reactions that involve atomic species or simple molecules but for complex molecules significant deviations are observed. The reason could be that all collisions do not lead to the formation of products. The collisions in which molecules collide with sufficient kinetic energy (called threshold energy*) and proper orientation, so as to facilitate breaking of bonds between reacting species and formation of new bonds to form products are called as effective collisions. For example, formation of methanol from bromoethane depends upon the orientation of reactant molecules as shown in Fig. 3.12. The proper orientation of reactant molecules lead to bond formation whereas improper orientation makes them simply bounce back and no products are formed. Fig. 3.12: Diagram showing molecules having proper and To account for effective collisions, improper orientation another factor P, called the probability or steric factor is introduced. It takes into account the fact that in a collision, molecules must be properly oriented i.e., − E a / RT Rate = PZ AB e Thus, in collision theory activation energy and proper orientation of the molecules together determine the criteria for an effective collision and hence the rate of a chemical reaction. Collision theory also has certain drawbacks as it considers atoms/ molecules to be hard spheres and ignores their structural aspect. You will study details about this theory and more on other theories in your higher classes. * Threshold energy = Activation Energy + energy possessed by reacting species. 83 Chemical Kinetics Reprint 2025-26 IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 3.7 What will be the effect of temperature on rate constant ? 3.8 The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate Ea. 3.9 The activation energy for the reaction 2 HI(g) ® H2 + I2 (g) is 209.5 kJ mol–1 at 581K.Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy? SummarySummarySummarySummarySummary Chemical kinetics is the study of chemical reactions with respect to reaction rates, effect of various variables, rearrangement of atoms and formation of intermediates. The rate of a reaction is concerned with decrease in concentration of reactants or increase in the concentration of products per unit time. It can be expressed as instantaneous rate at a particular instant of time and average rate over a large interval of time. A number of factors such as temperature, concentration of reactants, catalyst, affect the rate of a reaction. Mathematical representation of rate of a reaction is given by rate law. It has to be determined experimentally and cannot be predicted. Order of a reaction with respect to a reactant is the power of its concentration which appears in the rate law equation. The order of a reaction is the sum of all such powers of concentration of terms for different reactants. Rate constant is the proportionality factor in the rate law. Rate constant and order of a reaction can be determined from rate law or its integrated rate equation. Molecularity is defined only for an elementary reaction. Its values are limited from 1 to 3 whereas order can be 0, 1, 2, 3 or even a fraction. Molecularity and order of an elementary reaction are same. Temperature dependence of rate constants is described by Arrhenius equation (k = Ae–Ea/RT). Ea corresponds to the activation energy and is given by the energy difference between activated complex and the reactant molecules, and A (Arrhenius factor or pre-exponential factor) corresponds to the collision frequency. The equation clearly shows that increase of temperature or lowering of Ea will lead to an increase in the rate of reaction and presence of a catalyst lowers the activation energy by providing an alternate path for the reaction. According to collision theory, another factor P called steric factor which refers to the orientation of molecules which collide, is important and contributes to effective collisions, thus, modifying the Arrhenius equation to k  P Z AB e  E a / RT . Chemistry 84 Reprint 2025-26 ExercisesExercisesExercisesExercisesExercises 3.1 From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants. (i) 3NO(g) ® N2O (g) Rate = k[NO]2 (ii) H2O2 (aq) + 3I– (aq) + 2H+ ® 2H2O (l) + 3I Rate = k[H2O2][I-] (iii) CH3CHO (g) ® CH4 (g) + CO(g) Rate = k [CH3CHO]3/2 (iv) C2H5Cl (g) ® C2H4 (g) + HCl (g) Rate = k [C2H5Cl] 3.2 For the reaction: 2A + B ® A2B the rate = k[A][B]2 with k = 2.0 × 10–6 mol–2 L2 s–1. Calculate the initial rate of the reaction when [A] = 0.1 mol L–1, [B] = 0.2 mol L–1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L–1. 3.3 The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10–4 mol–1 L s–1? 3.4 The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by Rate = k [CH3OCH3]3/2 The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e., 3/2 Rate = k ( p CH 3 OCH 3 ) If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants? 3.5 Mention the factors that affect the rate of a chemical reaction. 3.6 A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half ? 3.7 What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively? 3.8 In a pseudo first order reaction in water, the following results were obtained: t/s 0 30 60 90 [A]/ mol L–1 0.55 0.31 0.17 0.085 Calculate the average rate of reaction between the time interval 30 to 60 seconds.

2.9 The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 W. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10–3 S cm–1. 59 Electrochemistry Reprint 2025-26

1.8 Mole concept and Molar Masses molecules Atoms and molecules are extremely small 1 mol of sodium chloride = 6.022 ×1023 formula in size and their numbers in even a small units of sodium chloride amount of any substance is really very large. Having defined the mole, it is easier toTo handle such large numbers, a unit of convenient magnitude is required. know the mass of one mole of a substance Just as we denote one dozen for 12 items, or the constituent entities. The mass of one score for 20 items, gross for 144 items, we mole of a substance in grams is called its use the idea of mole to count entities at the molar mass. The molar mass in grams is microscopic level (i.e., atoms, molecules, numerically equal to atomic/molecular/ particles, electrons, ions, etc). formula mass in u. In SI system, mole (symbol, mol) was Molar mass of water = 18.02 g mol–1introduced as seventh base quantity for the amount of a substance. Molar mass of sodium chloride = 58.5 g mol–1 The mole, symbol mol, is the SI unit of 1.9 Percentage Compositionamount of substance. One mole contains exactly 6.02214076 × 1023 elementary entities. So far, we were dealing with the number of This number is the fixed numerical value of entities present in a given sample. But many the Avogadro constant, NA, when expressed a time, information regarding the percentage in the unit mol–1 and is called the Avogadro of a particular element present in a compound number. The amount of substance, symbol is required. Suppose, an unknown or new n, of a system is a measure of the number of compound is given to you, the first questionspecified elementary entities. An elementary entity may be an atom, a molecule, an ion, an electron, any other particle or specified group of particles. It may be emphasised that the mole of a substance always contains the same number of entities, no matter what the substance may be. In order to determine this number precisely, the mass of a carbon–12 atom was determined by a mass spectrometer and found to be equal to 1.992648 × 10–23 g. Knowing that one mole of carbon weighs 12 g, the number of atoms in it is equal to: 12 g / mol 12 C 1 .992648  10 23 g /12 Catom  6 .0221367  1023 atoms/mol Fig. 1.11 One mole of various substances Reprint 2025-26 Some Basic Concepts of Chemistry 19 you would ask is: what is its formula or what 1.9.1 Empirical Formula for Molecular are its constituents and in what ratio are they Formula present in the given compound? For known An empirical formula represents the simplest compounds also, such information provides a whole number ratio of various atoms present check whether the given sample contains the in a compound, whereas, the molecular same percentage of elements as present in a formula shows the exact number of different pure sample. In other words, one can check types of atoms present in a molecule of a the purity of a given sample by analysing this compound. data. If the mass per cent of various elements Let us understand it by taking the example present in a compound is known, its empirical of water (H2O). Since water contains hydrogen formula can be determined. Molecular formula and oxygen, the percentage composition of can further be obtained if the molar mass is both these elements can be calculated as known. The following example illustrates follows: this sequence. Mass % of an element = mass of that element in the compound × 100 molar mass of the compound Problem 1.2 A compound contains 4.07% hydrogen,Molar mass of water = 18.02 g 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are itsMass % of hydrogen = empirical and molecular formulas? = 11.18 Solution 16.00Mass % of oxygen = × 100 Step 1. Conversion of mass per cent 18.02 to grams = 88.79 Since we are having mass per cent, it is Let us take one more example. What is the convenient to use 100 g of the compound percentage of carbon, hydrogen and oxygen as the starting material. Thus, in the in ethanol? 100 g sample of the above compound, 4.07g hydrogen, 24.27g carbon andMolecular formula of ethanol is: C2H5OH 71.65g chlorine are present.Molar mass of ethanol is: Step 2. Convert into number moles of(2×12.01 + 6×1.008 + 16.00) g = 46.068 g each element Mass per cent of carbon Divide the masses obtained above by 24.02g = × 100 = 52.14% respective atomic masses of various 46.068g elements. This gives the number of Mass per cent of hydrogen moles of constituent elements in the compound 6.048g = × 100 = 13.13% 46.068g 4.07 g Moles of hydrogen = = 4.04 1.008gMass per cent of oxygen 16.00g = × 100 = 34.73% 24.27 g Moles of carbon = = 2 .021 46.068g 12 .01 g After understanding the calculation of 71.65g per cent of mass, let us now see what Moles of chlorine = = 2 .021 35 .453 ginformation can be obtained from the per cent composition data. Reprint 2025-26 20 chemistry equation of a given reaction. Let us consider Step 3. Divide each of the mole values the combustion of methane. A balanced obtained above by the smallest number equation for this reaction is as given below: amongst them CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (g) Since 2.021 is smallest value, division Here, methane and dioxygen are called by it gives a ratio of 2:1:1 for H:C:Cl. reactants and carbon dioxide and water are In case the ratios are not whole numbers, called products. Note that all the reactants then they may be converted into whole and the products are gases in the above number by multiplying by the suitable reaction and this has been indicated by coefficient. letter (g) in the brackets next to its formula. Step 4. Write down the empirical Similarly, in case of solids and liquids, (s) and formula by mentioning the numbers (l) are written respectively. after writing the symbols of respective The coefficients 2 for O2 and H2O are elements called stoichiometric coefficients. Similarly CH2Cl is, thus, the empirical formula the coefficient for CH4 and CO2 is one in each of the above compound. case. They represent the number of molecules (and moles as well) taking part in the reaction Step 5. Writing molecular formula or formed in the reaction. (a) Determine empirical formula mass by Thus, according to the above chemical adding the atomic masses of various reaction, atoms present in the empirical formula. For CH2Cl, empirical formula mass is • One mole of CH4(g) reacts with two moles 12.01 + (2 ×1.008) + 35.453 of O2(g) to give one mole of CO2(g) and = 49.48 g two moles of H2O(g) (b) Divide Molar mass by empirical • One molecule of CH4(g) reacts with formula mass 2 molecules of O2(g) to give one molecule of CO2(g) and 2 molecules of H2O(g) • 22.7 L of CH4(g) reacts with 45.4 L of O2 = 2 = (n) (g) to give 22.7 L of CO2 (g) and 45.4 L of (c) Multiply empirical formula by n H2O(g) obtained above to get the molecular • 16 g of CH4 (g) reacts with 2×32 g of O2 formula (g) to give 44 g of CO2 (g) and 2×18 g of Empirical formula = CH2Cl, n = 2. Hence H2O (g). molecular formula is C2H4Cl2. From these relationships, the given data can be interconverted as follows: 1.10 Stoichiometry and mass Stoichiometric Calculations The word ‘stoichiometry’ is derived from two Greek words — stoicheion (meaning, Mass = Density element) and metron (meaning, measure). Volume Stoichiometry, thus, deals with the calculation of masses (sometimes volumes also) of the 1.10.1 Limiting Reagent reactants and the products involved in a Many a time, reactions are carried out with chemical reaction. Before understanding how the amounts of reactants that are different to calculate the amounts of reactants required than the amounts as required by a balanced or the products produced in a chemical chemical reaction. In such situations, one reaction, let us study what information reactant is in more amount than the amount is available from the balanced chemical required by balanced chemical reaction. The Reprint 2025-26 Some Basic Concepts of Chemistry 21 reactant which is present in the least amount important to understand as how the amount gets consumed after sometime and after that of substance is expressed when it is present in further reaction does not take place whatever the solution. The concentration of a solution be the amount of the other reactant. Hence, or the amount of substance present in its the reactant, which gets consumed first, given volume can be expressed in any of the limits the amount of product formed and is, following ways. therefore, called the limiting reagent. 1. Mass per cent or weight per cent (w/w %) In performing stoichiometric calculations, 2. Mole fraction this aspect is also to be kept in mind. 3. Molarity 1.10.2 Reactions in Solutions 4. Molality A majority of reactions in the laboratories Let us now study each one of them in are carried out in solutions. Therefore, it is detail. Balancing a chemical equation According to the law of conservation of mass, a balanced chemical equation has the same number of atoms of each element on both sides of the equation. Many chemical equations can be balanced by trial and error. Let us take the reactions of a few metals and non-metals with oxygen to give oxides 4 Fe(s) + 3O2(g) → 2Fe2O3(s) (a) balanced equation 2 Mg(s) + O2(g) → 2MgO(s) (b) balanced equation P4(s) + O2 (g) → P4O10(s) (c) unbalanced equation Equations (a) and (b) are balanced, since there are same number of metal and oxygen atoms on each side of the equations. However equation (c) is not balanced. In this equation, phosphorus atoms are balanced but not the oxygen atoms. To balance it, we must place the coefficient 5 on the left of oxygen on the left side of the equation to balance the oxygen atoms appearing on the right side of the equation. P4(s) + 5O2(g) → P4O10(s) balanced equation Now, let us take combustion of propane, C3H8. This equation can be balanced in steps. Step 1 Write down the correct formulas of reactants and products. Here, propane and oxygen are reactants, and carbon dioxide and water are products. C3H8(g) + O2(g) → CO2 (g) + H2O(l) unbalanced equation Step 2 Balance the number of C atoms: Since 3 carbon atoms are in the reactant, therefore, three CO2 molecules are required on the right side. C3H8 (g) + O2 (g) → 3CO2 (g) + H2O (l) Step 3 Balance the number of H atoms: on the left there are 8 hydrogen atoms in the reactants however, each molecule of water has two hydrogen atoms, so four molecules of water will be required for eight hydrogen atoms on the right side. C3H8 (g) +O2 (g) → 3CO2 (g)+4H2O (l) Step 4 Balance the number of O atoms: There are 10 oxygen atoms on the right side (3 × 2 = 6 in CO2 and 4×1= 4 in water). Therefore, five O2 molecules are needed to supply the required 10 CO2 and 4×1= 4 in water). Therefore, five O2 molecules are needed to supply the required 10 oxygen atoms. C3H8 (g) +5O2 (g) → 3CO2 (g) + 4H2O (l) Step 5 Verify that the number of atoms of each element is balanced in the final equation. The equation shows three carbon atoms, eight hydrogen atoms, and 10 oxygen atoms on each side. All equations that have correct formulas for all reactants and products can be balanced. Always remember that subscripts in formulas of reactants and products cannot be changed to balance an equation. Reprint 2025-26 22 chemistry Problem 1.3 the limiting reagent in the production of NH3 in this situation. Calculate the amount of water (g) produced by the combustion of 16 g Solution of methane. A balanced equation for the above Solution reaction is written as follows : The balanced equation for the combustion of methane is : CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g) Calculation of moles : (i) 16 g of CH4 corresponds to one mole. Number of moles of N2 (ii) From the above equation, 1 mol of 1000 g N 2 1 mol N 2 × 2 × CH4 (g) gives 2 mol of H2O (g). = 50 .0 kg N 1 kg N 2 28 .0 g N 2 2 mol of water (H2O) = 2×(2+16) = 17.86×102 mol = 2×18 = 36 g Number of moles of H2 H 2 O ⇒18g 1000 g H 2 1 mol H 2 = 1 1 mol H2O = 18 g H2O × = 10 .00 kg H 2 × 1mol H 2 O 1 kg H 2 2 .016 g H 2 18g H 2 O = 4.96 × 103 mol Hence, 2 mol H2O× 1mol H 2 O According to the above equation, 1 mol N2 (g) requires 3 mol H2 (g), for the = 2×18 g H2O = 36 g H2O reaction. Hence, for 17.86×102 mol of Problem 1.4 N2, the moles of H2 (g) required would be How many moles of methane are 2 3 mol H 2 g 2  required to produce 22g CO2 (g) after 17 .86  10 mol N 1 mol N 2 g combustion? = 5.36 × 103 mol H2 Solution But we have only 4.96×103 mol H2. According to the chemical equation, Hence, dihydrogen is the limiting CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g) reagent in this case. So, NH3(g) would be formed only from that amount of 44g CO2 (g) is obtained from 16 g CH4 (g). available dihydrogen i.e., 4.96 × 103 mol [∴1 mol CO2(g) is obtained from 1 mol Since 3 mol H2(g) gives 2 mol NH3(g) of CH4(g)] Number of moles of CO2 (g) 2 mol NH 3 g 2 g 4.96 ×103 mol H2 (g) × 3 mol H 2 g = 22 g CO2 (g) ×1 molCO 44 gCO 2 g = 3.30 × 103 mol NH3 (g) = 0.5 mol CO2 (g) 3.30 × 103 mol NH3 (g) is obtained. Hence, 0.5 mol CO2 (g) would be obtained from 0.5 mol CH4 (g) or 0.5 mol If they are to be converted to grams, it of CH4 (g) would be required to produce is done as follows : 22 g CO2 (g). 1 mol NH3 (g) = 17.0 g NH3 (g) Problem 1.5 .0 g NH 3 g 50.0 kg of N2 (g) and 10.0 kg of H2 (g) 3.30 ×103 mol NH3 (g) ×17 3 g are mixed to produce NH3 (g). Calculate 1 mol NH the amount of NH3 (g) formed. Identify Reprint 2025-26 Some Basic Concepts of Chemistry 23 3. Molarity = 3.30×103×17 g NH3 (g) It is the most widely used unit and is denoted = 56.1×103 g NH3 by M. It is defined as the number of moles of = 56.1 kg NH3 the solute in 1 litre of the solution. Thus, No. of moles of solute 1. Mass per cent Molarity (M) = Volume of solution in litres It is obtained by using the following relation: Suppose, we have 1 M solution of a substance, say NaOH, and we want to prepare a 0.2 M solution from it. 1 M NaOH means 1 mol of NaOH present in 1 litre of the solution. For 0.2 M solution, Problem 1.6 we require 0.2 moles of NaOH dissolved in 1 litre solution. A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate Hence, for making 0.2M solution from 1M the mass per cent of the solute. solution, we have to take that volume of 1M NaOH solution, which contains 0.2 mol of NaOH Solution and dilute the solution with water to 1 litre. Now, how much volume of concentrated (1M) NaOH solution be taken, which contains 0.2 moles of NaOH can be calculated as follows: If 1 mol is present in 1L or 1000 mL solution then, 0.2 mol is present in 1000 mL × 0 .2 mol solution 1 mol 2. Mole Fraction = 200 mL solutionIt is the ratio of number of moles of a particular component to the total number Thus, 200 mL of 1M NaOH are taken and enough water is added to dilute it to make it 1 litre.of moles of the solution. If a substance ‘A’ dissolves in substance ‘B’ and their number In fact for such calculations, a general of moles are nA and nB, respectively, then the formula, M1×V1 = M2 × V2 where M and V are molarity and volume, respectively, can be used.mole fractions of A and B are given as: In this case, M1 is equal to 0.2M; V1 = 1000 mL and, M2 = 1.0M; V2 is to be calculated. Substituting the values in the formula: 0.2 M × 1000 mL = 1.0 M × V2 Note that the number of moles of solute (NaOH) was 0.2 in 200 mL and it has remained the same, i.e., 0.2 even after dilution ( in 1000 mL) as we have changed just the amount of solvent (i.e., water) and have not done anything with respect to NaOH. But keep in mind the concentration. Reprint 2025-26 24 chemistry Problem 1.7 Problem 1.8 Calculate the molarity of NaOH in the The density of 3 M solution of NaCl is solution prepared by dissolving its 4 g 1.25 g mL–1. Calculate the molality of in enough water to form 250 mL of the the solution. solution. Solution Solution M = 3 mol L–1 Since molarity (M) Mass of NaCl in 1 L solution = 3 × 58.5 = 175.5 g Mass of 1L solution = 1000 × 1.25 = 1250 g (since density = 1.25 g mL–1) Mass of water in solution = 1250 –75.5 = 1074.5 g No. of moles of solute Molality = Mass of solvent in kg 3 mol = = 2.79 m Note that molarity of a solution depends 1 .0745 kg upon temperature because volume of a Often in a chemistry laboratory, a solution is temperature dependent. solution of a desired concentration is prepared by diluting a solution of known 4. Molality higher concentration. The solution of higher concentration is also known as It is defined as the number of moles of solute stock solution. Note that the molality present in 1 kg of solvent. It is denoted by m. of a solution does not change with temperature since mass remains No. of moles of solute Thus, Molality (m) = unaffected with temperature. Mass of solvent in kg Summary Chemistry, as we understand it today is not a very old discipline. People in ancient India, already had the knowledge of many scientific phenomenon much before the advent of modern science. They applied the knowledge in various walks of life. The study of chemistry is very important as its domain encompasses every sphere of life. Chemists study the properties and structure of substances and the changes undergone by them. All substances contain matter, which can exist in three states – solid, liquid or gas. The constituent particles are held in different ways in these states of matter and they exhibit their characteristic properties. Matter can also be classified into elements, compounds or mixtures. An element contains particles of only one type, which may be atoms or molecules. The compounds are formed where atoms of two or more elements combine in a fixed ratio to each other. Mixtures occur widely and many of the substances present around us are mixtures. When the properties of a substance are studied, measurement is inherent. The quantification of properties requires a system of measurement and units in which the quantities are to be expressed. Many systems of measurement exist, of which the English Reprint 2025-26 Some Basic Concepts of Chemistry 25 and the Metric Systems are widely used. The scientific community, however, has agreed to have a uniform and common system throughout the world, which is abbreviated as SI units (International System of Units). Since measurements involve recording of data, which are always associated with a certain amount of u is very important. The measurements of quantities in chemistry are spread over a wide range of 10–31 to 10+23. Hence, a convenient system of expressing the numbers in scientific notation is used. The u figures, in which the observations are reported. The dimensional analysis helps to express the measured quantities in different systems of units. Hence, it is possible to interconvert the results from one system of units to another. The combination of different atoms is governed by basic laws of chemical combination — these being the Law of Conservation of Mass, Law of Definite Proportions, Law of Multiple Proportions, Gay Lussac’s Law of Gaseous Volumes and Avogadro Law. All these laws led to the Dalton’s atomic theory, which states that atoms are building blocks of matter. The atomic mass of an element is expressed relative to 12C isotope of carbon, which has an exact value of 12u. Usually, the atomic mass used for an element is the average atomic mass obtained by taking into account the natural abundance of different isotopes of that element. The molecular mass of a molecule is obtained by taking sum of the atomic masses of different atoms present in a molecule. The molecular formula can be calculated by determining the mass per cent of different elements present in a compound and its molecular mass. The number of atoms, molecules or any other particles present in a given system are expressed in the terms of Avogadro constant (6.022 × 1023). This is known as 1 mol of the respective particles or entities. Chemical reactions represent the chemical changes undergone by different elements and compounds. A balanced chemical equation provides a lot of information. The coefficients indicate the molar ratios and the respective number of particles taking part in a particular reaction. The quantitative study of the reactants required or the products formed is called stoichiometry. Using stoichiometric calculations, the amount of one or more reactant(s) required to produce a particular amount of product can be determined and vice-versa. The amount of substance present in a given volume of a solution is expressed in number of ways, e.g., mass per cent, mole fraction, molarity and molality. exerciseS 1.1 Calculate the molar mass of the following: (i) H2O (ii) CO2 (iii) CH4 1.2 Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4). 1.3 Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass. 1.4 Calculate the amount of carbon dioxide that could be produced when (i) 1 mole of carbon is burnt in air. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. (iii) 2 moles of carbon are burnt in 16 g of dioxygen. 1.5 Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–1. Reprint 2025-26 26 chemistry 1.6 Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL–1 and the mass per cent of nitric acid in it being 69%. 1.7 How much copper can be obtained from 100 g of copper sulphate (CuSO4)? 1.8 Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively. 1.9 Calculate the atomic mass (average) of chlorine using the following data: % Natural Abundance Molar Mass 35Cl 75.77 34.9689 37Cl 24.23 36.9659 1.10 In three moles of ethane (C2H6), calculate the following: (i) Number of moles of carbon atoms. (ii) Number of moles of hydrogen atoms. (iii) Number of molecules of ethane. 1.11 What is the concentration of sugar (C12H22O11) in mol L–1 if its 20 g are dissolved in enough water to make a final volume up to 2L? 1.12 If the density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its 0.25 M solution? 1.13 Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below: 1Pa = 1N m–2 If mass of air at sea level is 1034 g cm–2, calculate the pressure in pascal. 1.14 What is the SI unit of mass? How is it defined? 1.15 Match the following prefixes with their multiples: Prefixes Multiples (i) micro 106 (ii) deca 109 (iii) mega 10–6 (iv) giga 10–15 (v) femto 10 1.16 What do you mean by significant figures? 1.17 A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass). (i) Express this in per cent by mass. (ii) Determine the molality of chloroform in the water sample. 1.18 Express the following in the scientific notation: (i) 0.0048 (ii) 234,000 (iii) 8008 (iv) 500.0 (v) 6.0012 1.19 How many significant figures are present in the following? (i) 0.0025 (ii) 208 (iii) 5005 Reprint 2025-26 Some Basic Concepts of Chemistry 27 (iv) 126,000 (v) 500.0 (vi) 2.0034 1.20 Round up the following upto three significant figures: (i) 34.216 (ii) 10.4107 (iii) 0.04597 (iv) 2808 1.21 The following data are obtained when dinitrogen and dioxygen react together to form different compounds: Mass of dinitrogen Mass of dioxygen (i) 14 g 16 g (ii) 14 g 32 g (iii) 28 g 32 g (iv) 28 g 80 g (a) Which law of chemical combination is obeyed by the above experimental data? Give its statement. (b) Fill in the blanks in the following conversions: (i) 1 km = ...................... mm = ...................... pm (ii) 1 mg = ...................... kg = ...................... ng (iii) 1 mL = ...................... L = ...................... dm3 1.22 If the speed of light is 3.0 × 108 m s–1, calculate the distance covered by light in 2.00 ns. 1.23 In a reaction A + B2  AB2 Identify the limiting reagent, if any, in the following reaction mixtures. (i) 300 atoms of A + 200 molecules of B (ii) 2 mol A + 3 mol B (iii) 100 atoms of A + 100 molecules of B (iv) 5 mol A + 2.5 mol B (v) 2.5 mol A + 5 mol B 1.24 Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: N2 (g) + H2 (g)  2NH3 (g) (i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 × 103 g of dihydrogen. (ii) Will any of the two reactants remain unreacted? (iii) If yes, which one and what would be its mass? 1.25 How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different? 1.26 If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced? 1.27 Convert the following into basic units: (i) 28.7 pm (ii) 15.15 pm (iii) 25365 mg Reprint 2025-26 28 chemistry 1.28 Which one of the following will have the largest number of atoms? (i) 1 g Au (s) (ii) 1 g Na (s) (iii) 1 g Li (s) (iv) 1 g of Cl2(g) 1.29 Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one). 1.30 What will be the mass of one 12C atom in g? 1.31 How many significant figures should be present in the answer of the following calculations? 0.02856 × 298.15 × 0.112 (i) (ii) 5 × 5.364 0 .5785 (iii) 0.0125 + 0.7864 + 0.0215 1.32 Use the data given in the following table to calculate the molar mass of naturally occuring argon isotopes: Isotope Isotopic molar mass Abundance 36Ar 35.96755 g mol–1 0.337% 38Ar 37.96272 g mol–1 0.063% 40Ar 39.9624 g mol–1 99.600% 1.33 Calculate the number of atoms in each of the following (i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He. 1.34 A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula. 1.35 Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3 (s) + 2 HCl (aq) → CaCl2 (aq) + CO2(g) + H2O(l) What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl? 1.36 Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction 4 HCl (aq) + MnO2(s) → 2H2O (l) + MnCl2(aq) + Cl2 (g) How many grams of HCl react with 5.0 g of manganese dioxide? Reprint 2025-26 Unit 2 structure of atom The rich diversity of chemical behaviour of different Objectives elementsstructure canof atomsbe tracedof theseto theelements.differences in the internal After studying this unit you will be able to • know about the discovery of The existence of atoms has been proposed since the time electron, proton and neutron and of early Indian and Greek philosophers (400 B.C.) who their characteristics; were of the view that atoms are the fundamental building • describe Thomson, Rutherford blocks of matter. According to them, the continued and Bohr atomic models; subdivisions of matter would ultimately yield atoms which would not be further divisible. The word ‘atom’ • understand the important features has been derived from the Greek word ‘a-tomio’ which of the quantum mechanical model means ‘uncut-able’ or ‘non-divisible’. These earlier ideas of atom; were mere speculations and there was no way to test • u n d e r s t a n d n a t u r e o f them experimentally. These ideas remained dormant for electromagnetic radiation and a very long time and were revived again by scientists in Planck’s quantum theory; the nineteenth century. • explain the photoelectric effect The atomic theory of matter was first proposed and describe features of atomic on a firm scientific basis by John Dalton, a British spectra; school teacher in 1808. His theory, called Dalton’s • state the de Broglie relation and atomic theory, regarded the atom as the ultimate Heisenberg u able to explain the law of conservation of mass, law of • define an atomic orbital in terms constant composition and law of multiple proportion of quantum numbers; very successfully. However, it failed to explain the results • state aufbau principle, Pauli of many experiments, for example, it was known that exclusion principle and Hund’s substances like glass or ebonite when rubbed with silk rule of maximum multiplicity; and or fur get electrically charged. • write the electronic configurations In this unit we start with the experimental observations of atoms. made by scientists towards the end of nineteenth and beginning of twentieth century. These established that atoms are made of sub-atomic particles, i.e., electrons, protons and neutrons — a concept very different from that of Dalton. Reprint 2025-26 30 chemistry