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Chemistry, by its very nature, is concerned with change. Substances with well defined properties are converted by chemical reactions into other substances with different properties. For any chemical reaction, chemists try to find out

(a) the feasibility of a chemical reaction which can be predicted by thermodynamics ( as you know that a reaction with \(\angle\)G < 0, at constant temperature and pressure is feasible);

(b) extent to which a reaction will proceed can be determined from chemical equilibrium;

(c) speed of a reaction i.e. time taken by a reaction to reach equilibrium.

Along with feasibility and extent, it is equally important to know the rate and the factors controlling the rate of a chemical reaction for its complete 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 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.

Rate of  Chemical  Reaction

Some reactions such as ionic reactions occur very fast, for example, precipitation of silver chloride occurs instantaneously by mixing of aqueous solutions of silver nitrate and sodium chloride. On the other 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:

1. the rate of decrease in concentration of any one of the reactants, or

2. 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 \(\to\) 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,

\(\Delta\)t      = t₂ – t₁

\(\Delta\)[R] = [R]₂ – [R]₁

\(\Delta\) [P] = [P]₂ – [P]₁

The square brackets in the above expressions are used to express molar concentration.

Rate of disappearance of R

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqGH9aqpdaWcaaqaaiaadseacaWGLbGaam4yaiaadkhacaWGLbGa % amyyaiaadohacaWGLbGaaGPaVlaaykW7caWGPbGaamOBaiaaykW7ca % aMc8Uaam4yaiaad+gacaWGUbGaam4yaiaadwgacaWGUbGaamiDaiaa % dkhacaWGHbGaamiDaiaadMgacaWGVbGaamOBaiaaykW7caaMc8Uaam % 4BaiaadAgacaaMc8UaamOuaiaaykW7aeaacaWGubGaamyAaiaad2ga % caWGLbGaaGPaVlaadshacaWGHbGaam4AaiaadwgacaWGUbaaaiabg2 % da9iabgkHiTmaalaaabaGaeyiLdq0aamWaaeaacaWGsbaacaGLBbGa % ayzxaaaabaGaeyiLdqKaamiDaaaaaaa!6E04! = \frac{{Decrease\,\,in\,\,concentration\,\,of\,R\,}}{{Time\,taken}} = - \frac{{\Delta \left[ R \right]}}{{\Delta t}}\)      (4.1)

Rate of appearance of P

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqGH9aqpdaWcaaqaaiaadMeacaWGUbGaam4yaiaadkhacaWGLbGa % amyyaiaadohacaWGLbGaaGPaVlaaykW7caWGPbGaamOBaiaaykW7ca % aMc8Uaam4yaiaad+gacaWGUbGaam4yaiaadwgacaWGUbGaamiDaiaa % dkhacaWGHbGaamiDaiaadMgacaWGVbGaamOBaiaaykW7caaMc8Uaam % 4BaiaadAgacaaMc8UaamiuaiaaykW7aeaacaWGubGaamyAaiaad2ga % caWGLbGaaGPaVlaadshacaWGHbGaam4AaiaadwgacaWGUbaaaiabg2 % da9iabgUcaRmaalaaabaGaeyiLdq0aamWaaeaacaWGqbaacaGLBbGa % ayzxaaaabaGaeyiLdqKaamiDaaaaaaa!6E03! = \frac{{Increase\,\,in\,\,concentration\,\,of\,P\,}}{{Time\,taken}} = + \frac{{\Delta \left[ P \right]}}{{\Delta t}}\)    (4.2)

Since, \(\Delta\)[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 (4.1) and (4.2) given above represent the average rate of a reaction, r_av.

Average rate depends upon the change in concentration of reactants or products and the time taken for that change to occur (Fig. 4.1).

               Fig. 4.1: Instantaneous and average rate of a reaction

Units of rate of a reaction

From equations (4.1) and (4.2), 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.

EXAMPLE 1

From the concentrations of C₄H₉Cl (butyl chloride) at different times given below, calculate the average rate of the reaction: C₄H₉Cl + H₂O \(\to\) 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

SOLUTION

We can determine the difference in concentration over different intervals of time and thus determine the average rate by dividing  \(\Delta\)[R] by \(\Delta\)t

(Table 4.1).

                                                 Table 4.1: Average rates of hydrolysis of butyl chloride

It can be seen (Table 4.1) 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  \(\Delta\)t approaches zero). Hence, mathematically for an infinitesimally small dt instantaneous rate is given by 

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGYbWaaSbaaSqaaiaadggacaWG2baabeaakiabg2da9maalaaa % baGaeyOeI0IaeyiLdq0aamWaaeaacaWGsbaacaGLBbGaayzxaaaaba % GaeyiLdqKaamiDaaaacqGH9aqpdaWcaaqaaiabgs5aenaadmaabaGa % amiuaaGaay5waiaaw2faaaqaaiabgs5aejaadsfaaaaaaa!493B! {r_{av}} = \frac{{ - \Delta \left[ R \right]}}{{\Delta t}} = \frac{{\Delta \left[ P \right]}}{{\Delta T}}\)

As \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqGHuoarcaWG0bGaeyOKH4QaaGimaiaaykW7caaMc8Uaam4Baiaa % dkhacaaMc8UaaGPaVlaaykW7caWGYbWaaSbaaSqaaiaadMgacaWGUb % Gaam4CaiaadshaaeqaaOGaeyypa0ZaaSaaaeaacqGHsislcaWGKbWa % amWaaeaacaWGsbaacaGLBbGaayzxaaaabaGaamizaiaadshaaaGaey % ypa0ZaaSaaaeaacaWGKbWaamWaaeaacaWGqbaacaGLBbGaayzxaaaa % baGaamizaiaadshaaaaaaa!57FD! \Delta t \to 0\,\,or\,\,\,{r_{inst}} = \frac{{ - d\left[ R \right]}}{{dt}} = \frac{{d\left[ P \right]}}{{dt}}\)

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. 4.1). So in problem 4.1, 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. 4.2).

Fig. 4.2 Instantaneous rate of hydrolysis of butyl chloride(C₄H₉Cl)

The slope of this tangent gives the instantaneous rate.

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGYbWaaSbaaSqaaiaadMgacaWGUbGaam4CaiaadshaaeqaaOGa % aGPaVlaaykW7caWGHbGaamiDaiaaykW7caaMc8UaaGOnaiaaicdaca % aIWaGaam4Caiabg2da9iabgkHiTmaabmaabaWaaSaaaeaacaaIWaGa % aiOlaiaaicdacaaIXaGaaGOnaiaaiwdacqGHsislcaaIWaGaaiOlai % aaicdacaaIZaGaaG4naaqaamaabmaabaGaaGioaiaaicdacaaIWaGa % eyOeI0IaaGinaiaaicdacaaIWaaacaGLOaGaayzkaaGaam4Caaaaai % aawIcacaGLPaaacaWGTbGaam4BaiaadYgacaWGmbWaaWbaaSqabeaa % cqGHsislcaaIXaaaaOGaeyypa0JaaGynaiaac6cacaaIXaGaaGOmai % abgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaI1aaaaOGa % amyBaiaad+gacaWGSbGaamitamaaCaaaleqabaGaeyOeI0IaaGymaa % aakiaadohadaahaaWcbeqaaiabgkHiTiaaigdaaaaaaa!71D7! {r_{inst}}\,\,at\,\,600s = - \left( {\frac{{0.0165 - 0.037}}{{\left( {800 - 400} \right)s}}} \right)mol{L^{ - 1}} = 5.12 \times {10^{ - 5}}mol{L^{ - 1}}{s^{ - 1}}\)

At t = 250 s r_inst = 1.22 × 10^–4 mol L^–1s^–1

t = 350 s   r_inst  = 1.0 × 10^–4 mol L^–1s^–1

t = 450 s   r_inst  = 6.4 × 10^–5 mol L^–1s^–1

Now consider a reaction

Hg(l) + Cl₂ (g) \(\to\) HgCl₂(s)

Where stoichiometric coefficients of the reactants and products are same, then rate of the reaction is given as

Rate of reaction = \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqGHsisldaWcaaqaaiabgs5aenaadmaabaGaamisaiaadEgaaiaa % wUfacaGLDbaaaeaacqGHuoarcaWG0baaaiabg2da9iabgkHiTmaala % aabaGaeyiLdq0aamWaaeaacaWGdbGaamiBamaaBaaaleaacaaIYaaa % beaaaOGaay5waiaaw2faaaqaaiabgs5aejaadshaaaGaeyypa0ZaaS % aaaeaacqGHuoardaWadaqaaiaadIeacaWGNbGaam4qaiaadYgadaWg % aaWcbaGaaGOmaaqabaaakiaawUfacaGLDbaaaeaacqGHuoarcaWG0b % aaaaaa!541F! - \frac{{\Delta \left[ {Hg} \right]}}{{\Delta t}} = - \frac{{\Delta \left[ {C{l_2}} \right]}}{{\Delta t}} = \frac{{\Delta \left[ {HgC{l_2}} \right]}}{{\Delta t}}\)

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) \(\to\) 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 \(\Delta\)[HI] is divided by 2. The rate of this reaction

is given by

Rate of reaction = \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqGHsisldaWcaaqaaiaaigdaaeaacaaIYaaaamaalaaabaGaeyiL % dq0aamWaaeaacaWGibGaamysaaGaay5waiaaw2faaaqaaiabgs5aej % aadshaaaGaeyypa0ZaaSaaaeaacqGHuoardaWadaqaaiaadIeadaWg % aaWcbaGaaGOmaaqabaaakiaawUfacaGLDbaaaeaacqGHuoarcaWG0b % aaaiabg2da9maalaaabaGaeyiLdq0aamWaaeaacaWGjbWaaSbaaSqa % aiaaikdaaeqaaaGccaGLBbGaayzxaaaabaGaeyiLdqKaamiDaaaaaa % a!510B! - \frac{1}{2}\frac{{\Delta \left[ {HI} \right]}}{{\Delta t}} = \frac{{\Delta \left[ {{H_2}} \right]}}{{\Delta t}} = \frac{{\Delta \left[ {{I_2}} \right]}}{{\Delta t}}\)

Similarly, for the reaction

5 Br^- (aq) + BrO \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaqhaaWcbaGaaG4maaqaaiabgkHiTaaaaaa!37ED! _3^ - \) (aq) + 6 H⁺ (aq)  \(\to\) 3 Br₂ (aq) + 3 H₂ O (l)

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGsbGaamyyaiaadshacaWGLbGaeyypa0JaeyOeI0YaaSaaaeaa % caaIXaaabaGaaGynaaaadaWcaaqaaiabgs5aenaadmaabaGaamOqai % aadkhadaahaaWcbeqaaiabgkHiTaaaaOGaay5waiaaw2faaaqaaiab % gs5aejaadshaaaGaeyypa0JaeyOeI0YaaSaaaeaacqGHuoardaWada % qaaiaadkeacaWGYbGaam4tamaaDaaaleaacaaIZaaabaGaeyOeI0ca % aaGccaGLBbGaayzxaaaabaGaeyiLdqKaamiDaaaacqGH9aqpcqGHsi % sldaWcaaqaaiaaigdaaeaacaaI2aaaamaalaaabaGaeyiLdq0aamWa % aeaacaWGcbGaamOCamaaBaaaleaacaaIYaaabeaaaOGaay5waiaaw2 % faaaqaaiabgs5aejaadshaaaGaeyypa0ZaaSaaaeaacaaIXaaabaGa % aG4maaaadaWcaaqaaiabgs5aenaadmaabaGaamisamaaBaaaleaaca % aIYaaabeaakiaad+eaaiaawUfacaGLDbaaaeaacqGHuoarcaWG0baa % aaaa!68EE! Rate = - \frac{1}{5}\frac{{\Delta \left[ {B{r^ - }} \right]}}{{\Delta t}} = - \frac{{\Delta \left[ {BrO_3^ - } \right]}}{{\Delta t}} = - \frac{1}{6}\frac{{\Delta \left[ {B{r_2}} \right]}}{{\Delta t}} = \frac{1}{3}\frac{{\Delta \left[ {{H_2}O} \right]}}{{\Delta t}}\)

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.

EXAMPLE 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 . The reaction takes place according to the equation 

2 N₂O₅ (g) \(\to\) 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?

SOLUTION

Average Rate = \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaWcaaqaaiaaigdaaeaacaaIYaaaamaacmaabaGaeyOeI0YaaSaa % aeaacqGHuoardaWadaqaaiaad6eadaWgaaWcbaGaaGOmaaqabaGcca % WGpbWaaSbaaSqaaiaaiwdaaeqaaaGccaGLBbGaayzxaaaabaGaeyiL % dqKaamiDaaaaaiaawUhacaGL9baacqGH9aqpcqGHsisldaWcaaqaai % aaigdaaeaacaaIYaaaamaadmaabaWaaSaaaeaadaqadaqaaiaaikda % caGGUaGaaGimaiaaiIdacqGHsislcaaIYaGaaiOlaiaaiodacaaIZa % aacaGLOaGaayzkaaGaamyBaiaad+gacaWGSbGaamitamaaCaaaleqa % baGaeyOeI0IaaGymaaaaaOqaaiaaigdacaaI4aGaaGinaiGac2gaca % GGPbGaaiOBaaaaaiaawUfacaGLDbaaaaa!5C6A! \frac{1}{2}\left\{ { - \frac{{\Delta \left[ {{N_2}{O_5}} \right]}}{{\Delta t}}} \right\} = - \frac{1}{2}\left[ {\frac{{\left( {2.08 - 2.33} \right)mol{L^{ - 1}}}}{{184\min }}} \right]\)

=    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

Rate = \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaWcaaqaaiaaigdaaeaacaaI0aaaamaacmaabaWaaSaaaeaacqGH % uoardaWadaqaaiaad6eacaWGpbWaaSbaaSqaaiaaikdaaeqaaaGcca % GLBbGaayzxaaaabaGaeyiLdqKaamiDaaaaaiaawUhacaGL9baaaaa!4232! \frac{1}{4}\left\{ {\frac{{\Delta \left[ {N{O_2}} \right]}}{{\Delta t}}} \right\}\)

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcqaaaaaaaaaWdbe % aadaWcaaqaaiabgs5aenaadmaabaGaamOtaiaad+eadaWgaaWcbaGa % aGOmaaqabaaakiaawUfacaGLDbaaaeaacqGHuoarcaWG0baaaaaa!3E78! {\frac{{\Delta \left[ {N{O_2}} \right]}}{{\Delta t}}}\) = 6.79 × 10^–4 × 4 mol L^–1 min^–1 = 2.72 × 10^–3 mol L^–1min^–1

INTEXT QUESTION

For the reaction R \(\to\) 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.

In a reaction, 2A ​\(\to\)​ 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?

Factors Influencing Rate of a Reaction

Rate of reaction depends upon the experimental conditions such as concentration of reactants (pressure in case of gases), temperature and catalyst.

Dependent of Rate on Concentration

The rate of a chemical reaction at a given temperature may depend on the concentration of one or more reactants and products. The 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.

RATE Expression and Rate Constant

The results in Table 4.1 clearly show that rate of a reaction decreases with the passage of time as the concentration of reactants decrease. Conversely, rates generally increase when reactant concentrations increase. So, rate of a reaction depends upon the concentration of reactants.

Consider a general reaction

aA + bB \(\to\)  cC + dD

where a, b, c and d are the stoichiometric coefficients of reactants and products.

The rate expression for this reaction is

Rate \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqGHDisTaaa!3796! \propto \) [A]^x [B]^y                                            (4.4)

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                                         (4.4a)

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqGHsisldaWcaaqaaiaadsgadaWadaqaaiaadkfaaiaawUfacaGL % DbaaaeaacaWGKbGaamiDaaaacaqGGaGaaeypaiabeQ7aRnaadmaaba % GaamyqaaGaay5waiaaw2faamaaCaaaleqabaGaamiEaaaakmaadmaa % baGaamOqaaGaay5waiaaw2faamaaCaaaleqabaGaamyEaaaaaaa!478C! - \frac{{d\left[ R \right]}}{{dt}}{\rm{ = }}\kappa {\left[ A \right]^x}{\left[ B \right]^y}\)                                         (4.4b)
This form of equation (4.4 b) is known as differential rate equation, where k is a proportionality constant called rate constant. The equation like (4.4), 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) \(\to\) 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 4.2).

                                     Table 4.2: Initial rate of formation of NO₂

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]

The differential form of this rate expression is given as

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0YaaS % aaaeaacaWGKbWaamWaaeaacaWGsbaacaGLBbGaayzxaaaabaGaamiz % aiaadshaaaGaeyypa0Jaam4AamaadmaabaGaamOtaiaad+eaaiaawU % facaGLDbaadaahaaWcbeqaaiaaikdaaaGcdaWadaqaaiaad+eadaWg % aaWcbaGaaGOmaaqabaaakiaawUfacaGLDbaaaaa!46C1! - \frac{{d\left[ R \right]}}{{dt}} = k{\left[ {NO} \right]^2}\left[ {{O_2}} \right]\)

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

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGymaiaac6 % cacaWGdbGaamisaiaadoeacaWGSbWaaSbaaSqaaiaaiodaaeqaaOGa % ey4kaSIaam4qaiaadYgadaWgaaWcbaGaaGOmaaqabaGccqGHsgIRca % WGdbGaam4qaiaadYgadaWgaaWcbaGaaGinaaqabaGccqGHRaWkcaWG % ibGaam4qaiaadYgaaaa!47FB! 1.CHC{l_3} + C{l_2} \to CC{l_4} + HCl\)                                                       \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuaiaadg % gacaWG0bGaamyzaiabg2da9iaadUgadaWadaqaaiaadoeacaWGibGa % am4qaiaadYgadaWgaaWcbaGaaG4maaqabaaakiaawUfacaGLDbaada % WadaqaaiaadoeacaWGSbWaaSbaaSqaaiaaikdaaeqaaaGccaGLBbGa % ayzxaaWaaWbaaSqabeaacaaIXaGaai4laiaaikdaaaaaaa!48B3! Rate = k\left[ {CHC{l_3}} \right]{\left[ {C{l_2}} \right]^{1/2}}\)

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiaac6 % cacaWGdbGaamisamaaBaaaleaacaaIZaaabeaakiaadoeacaWGpbGa % am4taiaadoeadaWgaaWcbaGaaGOmaaqabaGccaWGibWaaSbaaSqaai % aaiwdaaeqaaOGaey4kaSIaamisamaaBaaaleaacaaIYaaabeaakiaa % d+eacqGHsgIRcaWGdbGaamisamaaBaaaleaacaaIZaaabeaakiaado % eacaWGpbGaam4taiaadIeacqGHRaWkcaWGdbWaaSbaaSqaaiaaikda % aeqaaOGaamisamaaBaaaleaacaaI1aaabeaakiaad+eacaWGibaaaa!50FE! 2.C{H_3}COO{C_2}{H_5} + {H_2}O \to C{H_3}COOH + {C_2}{H_5}OH\)                  \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuaiaadg % gacaWG0bGaamyzaiabg2da9iaadUgadaWadaqaaiaadoeacaWGibWa % aSbaaSqaaiaaiodaaeqaaOGaam4qaiaad+eacaWGpbGaam4qamaaBa % aaleaacaaIYaaabeaakiaadIeadaWgaaWcbaGaaGynaaqabaaakiaa % wUfacaGLDbaadaahaaWcbeqaaiaaigdaaaGcdaWadaqaaiaadIeada % WgaaWcbaGaaGOmaaqabaGccaWGpbaacaGLBbGaayzxaaWaaWbaaSqa % beaacaaIWaaaaaaa!4C50! Rate = k{\left[ {C{H_3}COO{C_2}{H_5}} \right]^1}{\left[ {{H_2}O} \right]^0}\)

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.

Order of A Reaction

In the rate equation (4.4)

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 (4.4) 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.

EXAMPLE 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

SOLUTION

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.

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  \(\to\) cC + dD Rate = k [A]^x [B]^y

Where x + y = n = order of the reaction

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqaH6o % WAcqGH9aqpdaWcaaqaaiaadkfacaWGHbGaamiDaiaadwgaaeaadaWa % daqaaiaadgeaaiaawUfacaGLDbaadaahaaWcbeqaaiaadIhaaaGcda % WadaqaaiaadkeaaiaawUfacaGLDbaadaahaaWcbeqaaiaadMhaaaaa % aaGcbaGaeyypa0ZaaSaaaeaacaWGJbGaam4Baiaad6gacaWGJbGaam % yzaiaad6gacaWG0bGaamOCaiaadggacaWG0bGaamyAaiaad+gacaWG % UbaabaGaamiDaiaadMgacaWGTbGaamyzaaaacqGHxdaTdaWcaaqaai % aaigdaaeaadaqadaqaaiaadogacaWGVbGaamOBaiaadogacaWGLbGa % amOBaiaadshacaWGYbGaamyyaiaadshacaWGPbGaam4Baiaad6gaai % aawIcacaGLPaaadaahaaWcbeqaaiaad6gaaaaaaOGaaGPaVlaaykW7 % caaMc8+aaeWaaeaacaWG3bGaamiAaiaadwgacaWGYbGaamyzaiaayk % W7caaMc8+aamWaaeaacaWGbbaacaGLBbGaayzxaaGaeyypa0ZaamWa % aeaacaWGcbaacaGLBbGaayzxaaaacaGLOaGaayzkaaaaaaa!7B88! \begin{array}{l} \kappa = \frac{{Rate}}{{{{\left[ A \right]}^x}{{\left[ B \right]}^y}}}\\ = \frac{{concentration}}{{time}} \times \frac{1}{{{{\left( {concentration} \right)}^n}}}\,\,\,\left( {where\,\,\left[ A \right] = \left[ B \right]} \right) \end{array}\)

Taking SI units of concentration, mol L^–1 and time, s, the units of k for different reaction order are listed in Table 4.3

         Table 4.3: Units of rate constant

EXAMPLE 1

Identify the reaction order from each of the following rate constants. (i) k = 2.3 × 10^–5 L mol^–1 s^–1 (ii) k = 3 × 10^–4 s^–1

SOLUTION

(1) The unit of second order rate constant is L mol^–1 s^–1, therefore k = 2.3 × 10^–5 L mol^–1 s^–1 represents a second order reaction.

(2) The unit of a first order rate constant is s^–1 therefore k = 3 × 10^–4 s^–1 represents a first order reaction.

Molecularity OF A Reaction

Another property of a reaction called molecularity helps in understanding its mechanism. The number of reacting species (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.

NH₄NO₂ \(\to\) N₂ + 2H₂O

Bimolecular reactions involve simultaneous collision between two species, for example, dissociation of hydrogen iodide.

2HI \(\to\) H₂ + I₂

Trimolecular or termolecular reactions involve simultaneous collision between three reacting species, for example,

2NO + O₂ \(\to\)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 \(\to\)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.

The rate equation for this reaction is found to be

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuaiaadg % gacaWG0bGaamyzaiabg2da9maalaaabaGaeyOeI0Iaamizamaadmaa % baGaamisamaaBaaaleaacaaIYaaabeaakiaad+eadaWgaaWcbaGaaG % OmaaqabaaakiaawUfacaGLDbaaaeaacaWGKbGaamiDaaaacqGH9aqp % cqaH6oWAdaWadaqaaiaadIeadaWgaaWcbaGaaGOmaaqabaGccaWGpb % WaaSbaaSqaaiaaikdaaeqaaaGccaGLBbGaayzxaaWaamWaaeaacaWG % jbWaaWbaaSqabeaacqGHsislaaaakiaawUfacaGLDbaaaaa!4FEE! Rate = \frac{{ - d\left[ {{H_2}{O_2}} \right]}}{{dt}} = \kappa \left[ {{H_2}{O_2}} \right]\left[ {{I^ - }} \right]\)

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^– \(\to\) H₂O + IO^–

2. H₂O₂ + IO^– \(\to\) 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:

1. 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.

2. 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.

3. 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.

INTEXT QUESTION

For a reaction, A + B \(\to\) Product; the rate law is given by, \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOCaiabg2 % da9iabeQ7aRnaadmaabaGaamyqaaGaay5waiaaw2faamaaCaaaleqa % baGaaGymaiaac+cacaaIYaaaaOWaamWaaeaacaWGcbaacaGLBbGaay % zxaaWaaWbaaSqabeaacaaIYaaaaaaa!4260! r = \kappa {\left[ A \right]^{1/2}}{\left[ B \right]^2}\). What is the order of the reaction?

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 ?

Integrated Rate Equations

We have already noted that the concentration dependence of rate is called differential rate equation. It is not always convenient to determine the instantaneous rate, as it is measured by determination of slope of the tangent at point ‘t’ in concentration vs time plot (Fig. 4.1). 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.

Zero Order Reactions

Zero order reaction means that the rate of the reaction is proportional to zero power of the concentration of reactants. Consider the reaction, 

R \(\to\) P

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuaiaadg % gacaWG0bGaamyzaiabg2da9iabgkHiTmaalaaabaGaamizamaadmaa % baGaamOuaaGaay5waiaaw2faaaqaaiaadsgacaWG0baaaiabg2da9i % abeQ7aRnaadmaabaGaamOuaaGaay5waiaaw2faamaaCaaaleqabaGa % aGimaaaaaaa!4795! Rate = - \frac{{d\left[ R \right]}}{{dt}} = \kappa {\left[ R \right]^0}\)

As any quantity raised to power zero is unity

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuaiaadg % gacaWG0bGaamyzaiabg2da9iabgkHiTmaalaaabaGaamizamaadmaa % baGaamOuaaGaay5waiaaw2faaaqaaiaadsgacaWG0baaaiabg2da9i % abeQ7aRjabgEna0kaaigdaaaa!46B7! Rate = - \frac{{d\left[ R \right]}}{{dt}} = \kappa \times 1\)
d[R] = – k dt

Integrating both sides

[R]    = – k t + I                                                                          (4.5)

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 (4.5)

[R]₀ = –k × 0 + I

[R]₀  = I

Substituting the value of I in the equation (4.5)

[R]    = -kt + [R]₀                                                                       (4.6)

Comparing (4.6) with equation of a straight line, y = mx + c, if we plot [R] against t, we get a straight line (Fig. 4.3) with slope = –k and intercept equal to [R]₀.

Fig. 4.3: Variation in the concentration vs time plot for a zero order reaction

Further simplifying equation (4.6), we get the rate constant, k as

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqOUdSMaey % ypa0ZaaSaaaeaadaWadaqaaiaadkfaaiaawUfacaGLDbaadaWgaaWc % baGaaGimaaqabaGccqGHsisldaWadaqaaiaadkfaaiaawUfacaGLDb % aaaeaacaWG0baaaaaa!4126! \kappa = \frac{{{{\left[ R \right]}_0} - \left[ R \right]}}{t}\)   (4.7)

Zero order reactions are relatively uncommon but they occur under special conditions. Some enzyme catalysed reactions and reactions which occur on metal surfaces are a few examples of zero order reactions. The decomposition of gaseous ammonia on a hot platinum surface is a zero order reaction at high pressure.

Rate = \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqOUdS2aam % WaaeaacaWGobGaamisamaaBaaaleaacaaIZaaabeaaaOGaay5waiaa % w2faamaaCaaaleqabaGaaGimaaaakiabg2da9iabeQ7aRbaa!3FD6! \kappa {\left[ {N{H_3}} \right]^0} = \kappa \)

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.

First ORDER Reactions

In this class of reactions, the rate of the reaction is proportional to the first power of the concentration of the reactant R. For example,

R \(\to\) P 

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGsb % GaamyyaiaadshacaWGLbGaeyypa0ZaaSaaaeaacaWGKbWaamWaaeaa % caWGsbaacaGLBbGaayzxaaaabaGaamizaiaadshaaaGaeyypa0Jaeq % OUdS2aamWaaeaacaWGsbaacaGLBbGaayzxaaaabaGaam4Baiaadkha % caaMc8UaaGPaVlaaykW7daWcaaqaaiaadsgadaWadaqaaiaadkfaai % aawUfacaGLDbaaaeaadaWadaqaaiaadkfaaiaawUfacaGLDbaaaaGa % eyypa0JaeyOeI0IaeqOUdSMaamizaiaadshacaaMc8UaaGPaVdaaaa!5B7C! \begin{array}{l} Rate = \frac{{d\left[ R \right]}}{{dt}} = \kappa \left[ R \right]\\ or\,\,\,\frac{{d\left[ R \right]}}{{\left[ R \right]}} = - \kappa dt\,\, \end{array}\)

Integrating this equation, we get

ln [R] = – kt + I                                                                          (4.8)

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 (4.8) can be written as

ln [R]₀ = –k × 0 + I

ln [R]₀= I

Substituting the value of I in equation (4.8)

ln[R] = -kt + ln[R]₀   (4.9)   

Rearranging this equation

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGjb % GaamOBaiaaykW7caaMc8+aaSaaaeaadaWadaqaaiaadkfaaiaawUfa % caGLDbaaaeaadaWadaqaaiaadkfaaiaawUfacaGLDbaadaWgaaWcba % GaaGimaaqabaaaaOGaeyypa0JaeyOeI0IaeqOUdSMaamiDaaqaaiaa % d+gacaWGYbGaaGPaVlaaykW7cqaH6oWAcqGH9aqpdaWcaaqaaiaaig % daaeaacaWG0baaaiaaykW7caaMc8UaciiBaiaac6gacaaMc8UaaGPa % VpaalaaabaWaamWaaeaacaWGsbaacaGLBbGaayzxaaWaaSbaaSqaai % aaicdaaeqaaaGcbaWaamWaaeaacaWGsbaacaGLBbGaayzxaaaaaiaa % ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG % PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7daqa % daqaaiaaisdacaGGUaGaaGymaiaaicdaaiaawIcacaGLPaaaaaaa!7B41! \begin{array}{l} In\,\,\frac{{\left[ R \right]}}{{{{\left[ R \right]}_0}}} = - \kappa t\\ or\,\,\kappa = \frac{1}{t}\,\,\ln \,\,\frac{{{{\left[ R \right]}_0}}}{{\left[ R \right]}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {4.10} \right) \end{array}\)

At time t₁ from equation (4.8)

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaWbaaSqabe % aacqGHxiIkaaGcciGGSbGaaiOBamaadmaabaGaamOuaaGaay5waiaa % w2faamaaBaaaleaacaaIXaaabeaakiabg2da9iabgkHiTiabeQ7aRj % aadshadaWgaaWcbaGaaGymaaqabaGccqGHRaWkdaahaaWcbeqaaiab % gEHiQaaakiGacYgacaGGUbWaamWaaeaacaWGsbaacaGLBbGaayzxaa % WaaSbaaSqaaiaaicdaaeqaaaaa!49E4! ^ * \ln {\left[ R \right]_1} = - \kappa {t_1}{ + ^ * }\ln {\left[ R \right]_0}\)        (4.11)

At time t₂

ln[R]₂ = – kt₂ + ln[R]₀                 (4.12)

where, [R]₁ and [R]₂ are the concentrations of the reactants at time t₁ and t₂ respectively.

Subtracting (4.12) from (4.11)

ln[R]₁– ln[R]₂ = – kt₁ – (–kt₂)

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaciGGSb % GaaiOBamaalaaabaWaamWaaeaacaWGsbaacaGLBbGaayzxaaWaaSba % aSqaaiaaigdaaeqaaaGcbaWaamWaaeaacaWGsbaacaGLBbGaayzxaa % WaaSbaaSqaaiaaikdaaeqaaaaakiabg2da9iaadUgadaqadaqaaiaa % dshadaWgaaWcbaGaaGOmaaqabaGccqGHsislcaWG0bWaaSbaaSqaai % aaigdaaeqaaaGccaGLOaGaayzkaaaabaGaam4Aaiabg2da9maalaaa % baGaaGymaaqaamaabmaabaGaamiDamaaBaaaleaacaaIYaaabeaaki % abgkHiTiaadshadaWgaaWcbaGaaGymaaqabaaakiaawIcacaGLPaaa % aaGaciiBaiaac6gadaWcaaqaamaadmaabaGaamOuaaGaay5waiaaw2 % faamaaBaaaleaacaaIXaaabeaaaOqaamaadmaabaGaamOuaaGaay5w % aiaaw2faamaaBaaaleaacaaIYaaabeaaaaGccaaMc8UaaGPaVlaayk % W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa % VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 % +aaeWaaeaacaaI0aGaaiOlaiaaigdacaaIZaaacaGLOaGaayzkaaaa % aaa!7BDE! \begin{array}{l} \ln \frac{{{{\left[ R \right]}_1}}}{{{{\left[ R \right]}_2}}} = k\left( {{t_2} - {t_1}} \right)\\ k = \frac{1}{{\left( {{t_2} - {t_1}} \right)}}\ln \frac{{{{\left[ R \right]}_1}}}{{{{\left[ R \right]}_2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {4.13} \right) \end{array}\)

Equation (4.9) can also be written as

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6 % gadaWcaaqaamaadmaabaGaamOuaaGaay5waiaaw2faaaqaamaadmaa % baGaamOuaaGaay5waiaaw2faamaaBaaaleaacaaIWaaabeaaaaGccq % GH9aqpcqGHsislcaWGRbGaamiDaaaa!4248! \ln \frac{{\left[ R \right]}}{{{{\left[ R \right]}_0}}} = - kt\)

Taking antilog of both sides

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamWaaeaaca % WGsbaacaGLBbGaayzxaaGaeyypa0ZaamWaaeaacaWGsbaacaGLBbGa % ayzxaaWaaSbaaSqaaiaaicdaaeqaaOGaamyzamaaCaaaleqabaGaey % OeI0Iaam4Aaiaadshaaaaaaa!416B! \left[ R \right] = {\left[ R \right]_0}{e^{ - kt}}\)           (4.14)

Comparing equation (4.9) with y = mx + c, if we plot ln [R] against t (Fig. 4.4) we get a straight line with slope = –k and intercept equal to ln [R]₀

Fig. 4.4: A plot between ln[R] and t for a first order reaction

The first order rate equation (4.10) can also be written in the form

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGRb % Gaeyypa0ZaaSaaaeaacaaIYaGaaiOlaiaaiodacaaIWaGaaG4maaqa % aiaadshaaaGaciiBaiaac+gacaGGNbWaaSaaaeaadaWadaqaaiaadk % faaiaawUfacaGLDbaadaWgaaWcbaGaaGimaaqabaaakeaadaWadaqa % aiaadkfaaiaawUfacaGLDbaaaaGaaGPaVlaaykW7caaMc8UaaGPaVl % aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua % aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8+aaeWaaeaacaaI0a % GaaiOlaiaaigdacaaI1aaacaGLOaGaayzkaaaabaWaaWbaaSqabeaa % cqGHxiIkaaGcciGGSbGaai4BaiaacEgadaWcaaqaamaadmaabaGaam % OuaaGaay5waiaaw2faamaaBaaaleaacaaIWaaabeaaaOqaamaadmaa % baGaamOuaaGaay5waiaaw2faaaaacaaMc8Uaeyypa0ZaaSaaaeaaca % WGRbGaamyEaaqaaiaaikdacaGGUaGaaG4maiaaicdacaaIZaaaaaaa % aa!78F2! \begin{array}{l} k = \frac{{2.303}}{t}\log \frac{{{{\left[ R \right]}_0}}}{{\left[ R \right]}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {4.15} \right)\\ ^ * \log \frac{{{{\left[ R \right]}_0}}}{{\left[ R \right]}}\, = \frac{{ky}}{{2.303}} \end{array}\)
If we plot a graph between log [R]₀/[R] vs t, (Fig. 4.5), the slope = k/2.303

Hydrogenation of ethene is an example of first order reaction.

C₂H₄(g) + H₂ (g) \(\to\) C₂H₆(g)

Rate = k [C₂H₄]

All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics.

Fig. 4.5: Plot of log [R]₀/[R] vs time for a first order reaction

Decomposition of N₂O₅ and N₂O are some more examples of first order reactions.

EXAMPLE 5

The initial concentration of N₂O₅ in the following first order reaction N₂O₅ (g) \(\to\)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 mol L . Calculate the rate constant of the reaction at 318 K.

SOLUTION

For a first order reaction

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaciGGSb % Gaai4BaiaacEgadaWcaaqaamaadmaabaGaamOuaaGaay5waiaaw2fa % amaaBaaaleaacaaIXaaabeaaaOqaamaadmaabaGaamOuaaGaay5wai % aaw2faamaaBaaaleaacaaIYaaabeaaaaGccqGHsisldaWcaaqaaiab % eQ7aRnaabmaabaGaamiDamaaBaaaleaacaaIYaaabeaakiabgkHiTi % aadshadaWgaaWcbaGaaGymaaqabaaakiaawIcacaGLPaaaaeaacaaI % YaGaaiOlaiaaiodacaaIWaGaaG4maaaaaeaacaWGRbGaeyypa0ZaaS % aaaeaacaaIYaGaaiOlaiaaiodacaaIWaGaaG4maaqaamaabmaabaGa % amiDamaaBaaaleaacaaIYaaabeaakiabgkHiTiaadshadaWgaaWcba % GaaGymaaqabaaakiaawIcacaGLPaaaaaGaciiBaiaac+gacaGGNbWa % aSaaaeaadaWadaqaaiaadkfaaiaawUfacaGLDbaadaWgaaWcbaGaaG % ymaaqabaaakeaadaWadaqaaiaadkfaaiaawUfacaGLDbaadaWgaaWc % baGaaGOmaaqabaaaaaGcbaGaeyypa0ZaaSaaaeaacaaIYaGaaiOlai % aaiodacaaIWaGaaG4maaqaamaabmaabaGaamiDamaaBaaaleaacaaI % YaaabeaakiabgkHiTiaadshadaWgaaWcbaGaaGymaaqabaaakiaawI % cacaGLPaaaaaGaciiBaiaac+gacaGGNbWaaSaaaeaacaaIXaGaaiOl % aiaaikdacaaI0aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgk % HiTiaaikdaaaGccaWGTbGaam4BaiaadYgacaWGmbWaaWbaaSqabeaa % cqGHsislcaaIXaaaaaGcbaGaaGimaiaac6cacaaIYaGaaGimaiabgE % na0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaIYaaaaOGaamyB % aiaad+gacaWGSbGaamitamaaCaaaleqabaGaeyOeI0IaaGymaaaaaa % aakeaacqGH9aqpdaWcaaqaaiaaikdacaGGUaGaaG4maiaaicdacaaI % ZaaabaGaaGOnaiaaicdaaaGaciiBaiaac+gacaGGNbGaaGOnaiaac6 % cacaaIYaGaciyBaiaacMgacaGGUbWaaWbaaSqabeaacqGHsislcaaI % XaaaaaGcbaGaam4Aaiabg2da9iaaicdacaGGUaGaaGimaiaaiodaca % aIWaGaaGinaiaaykW7caaMc8UaciyBaiaacMgacaGGUbWaaWbaaSqa % beaacqGHsislcaaIXaaaaaaaaa!AABA! \begin{array}{l} \log \frac{{{{\left[ R \right]}_1}}}{{{{\left[ R \right]}_2}}} - \frac{{\kappa \left( {{t_2} - {t_1}} \right)}}{{2.303}}\\ k = \frac{{2.303}}{{\left( {{t_2} - {t_1}} \right)}}\log \frac{{{{\left[ R \right]}_1}}}{{{{\left[ R \right]}_2}}}\\ = \frac{{2.303}}{{\left( {{t_2} - {t_1}} \right)}}\log \frac{{1.24 \times {{10}^{ - 2}}mol{L^{ - 1}}}}{{0.20 \times {{10}^{ - 2}}mol{L^{ - 1}}}}\\ = \frac{{2.303}}{{60}}\log 6.2{\min ^{ - 1}}\\ k = 0.0304\,\,{\min ^{ - 1}} \end{array}\)

Let us consider a typical first order gas phase reaction

A(g) \(\to\) B(g) + C(g)

Let p_i 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 p_t  = p_A + p_B + p_C (pressure units) p_A, p_B and p_C 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)    \(\to\)  B(g)       +          C(g)

At t = 0              p_i atm            0 atm                 0 atm 

At time t           (p_i-x) atm       x atm                 x atm

where, p_i is the initial pressure at time t = 0

p_t = (p_i – x) + x + x = p_i + x x = (p_t - p_i)

where, p_A = p_i – x = p_i – (p_t – p_i)

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4Aaiabg2 % da9maabmaabaWaaSaaaeaacaaIYaGaaiOlaiaaiodacaaIWaGaaG4m % aaqaaiaadshaaaaacaGLOaGaayzkaaWaaeWaaeaaciGGSbGaai4Bai % aacEgadaWcaaqaaiaadchadaWgaaWcbaGaaGymaaqabaaakeaacaWG % WbWaaSbaaSqaaiaadgeaaeqaaaaaaOGaayjkaiaawMcaaaaa!4660! k = \left( {\frac{{2.303}}{t}} \right)\left( {\log \frac{{{p_1}}}{{{p_A}}}} \right)\)      (4.16)

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIYaGaaiOlaiaaiodacaaIWaGaaG4maaqaaiaadshaaaGaciiBaiaa % c+gacaGGNbWaaSaaaeaacaWGWbWaaSbaaSqaaiaadMgaaeqaaaGcba % WaaeWaaeaacaaIYaGaamiCamaaBaaaleaacaWGPbaabeaakiabgkHi % TiaadchadaWgaaWcbaGaamiDaaqabaaakiaawIcacaGLPaaaaaaaaa!4709! \frac{{2.303}}{t}\log \frac{{{p_i}}}{{\left( {2{p_i} - {p_t}} \right)}}\)

EXAMPLE 6

The following data were obtained during the first order thermal decomposition of N₂O₅ (g) at constant volume:

2N₂O₅ (g)  \(\to\) 2N ₂O₄ (g) + O₂ (g)

S.No.                              Time/s             Total Pressure/(atm)

1.                                        0                                 0.5

2.                                       100                             0.512

Calculate the rate constant.

SOLUTION

Let the pressure of N₂O₅(g) decrease by 2x atm. As two moles of 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.

                              2N₂O₅ (g)   \(\to\)     2N₂O₄ (g)    +     O₂ (g)

Start t = 0           0.5 m                   0 atm           0 atm

At time t          (0.5 – 2x) atm       2x atm           x atm

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGWb % WaaSbaaSqaaiaadshaaeqaaOGaeyypa0JaamiCamaaBaaaleaacaWG % obWaaSbaaWqaaiaaikdaaeqaaSGaam4tamaaBaaameaacaaI1aaabe % aaaSqabaGccqGHRaWkcaWGWbWaaSbaaSqaaiaad6eadaWgaaadbaGa % aGOmaaqabaWccaWGpbWaaSbaaWqaaiaaisdaaeqaaaWcbeaakiabgU % caRiaadchadaWgaaWcbaGaam4tamaaBaaameaacaaIYaaabeaaaSqa % baaakeaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 % Uaeyypa0JaaiikaiaaicdacaGGUaGaaGynaiabgkHiTiaaikdacaWG % 4bGaaiykaiabgUcaRiaaikdacaWG4bGaey4kaSIaamiEaiabg2da9i % aaicdacaGGUaGaaGynaiabgUcaRiaadIhaaeaacaWG4bGaeyypa0Ja % amiCamaaBaaaleaacaWG0baabeaakiabgkHiTiaaicdacaGGUaGaaG % ynaaqaaiaadchadaWgaaWcbaGaamOtamaaBaaameaacaaIYaaabeaa % liaad+eadaWgaaadbaGaaGynaaqabaaaleqaaOGaeyypa0JaaGimai % aac6cacaaI1aGaeyOeI0IaaGOmaiaadIhaaeaacaaMc8UaaGPaVlaa % ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG % PaVlabg2da9iaaicdacaGGUaGaaGynaiabgkHiTiaaikdadaqadaqa % aiaadchadaWgaaWcbaGaamiDaaqabaGccqGHsislcaaIWaGaaiOlai % aaiwdaaiaawIcacaGLPaaacqGH9aqpcaaIXaGaaiOlaiaaiwdacqGH % sislcaaIYaGaamiCamaaBaaaleaacaWG0baabeaaaOqaaiaadgeaca % WG0bGaaGPaVlaaykW7caaMc8UaaGPaVlaadshacqGH9aqpcaaIXaGa % aGimaiaaicdacaWGZbGaai4oaiaadchadaWgaaWcbaGaamiDaaqaba % GccqGH9aqpcaaIWaGaaiOlaiaaiwdacaaIXaGaaGOmaiaaykW7caaM % c8UaamyyaiaadshacaWGTbaabaGaamiCamaaBaaaleaacaWGobWaaS % baaWqaaiaaikdaaeqaaSGaam4tamaaBaaameaacaaI1aaabeaaaSqa % baGccqGH9aqpcaaIXaGaaiOlaiaaiwdacqGHsislcaaIYaGaey41aq % RaaGimaiaac6cacaaI1aGaaGymaiaaikdacqGH9aqpcaaIWaGaaiOl % aiaaisdacaaI3aGaaGOnaiaadggacaWG0bGaamyBaaaaaa!C9A3! \begin{array}{l} {p_t} = {p_{{N_2}{O_5}}} + {p_{{N_2}{O_4}}} + {p_{{O_2}}}\\ \,\,\,\,\,\,\, = (0.5 - 2x) + 2x + x = 0.5 + x\\ x = {p_t} - 0.5\\ {p_{{N_2}{O_5}}} = 0.5 - 2x\\ \,\,\,\,\,\,\,\,\,\,\, = 0.5 - 2\left( {{p_t} - 0.5} \right) = 1.5 - 2{p_t}\\ At\,\,\,\,t = 100s;{p_t} = 0.512\,\,atm\\ {p_{{N_2}{O_5}}} = 1.5 - 2 \times 0.512 = 0.476atm \end{array}\)

Using equation (4.16)

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqaH6o % WAcqGH9aqpdaWcaaqaaiaaikdacaGGUaGaaG4maiaaicdacaaIZaaa % baGaamiDaaaaciGGSbGaai4BaiaacEgadaWcaaqaaiaadchadaWgaa % WcbaGaaGymaaqabaaakeaacaWGWbWaaSbaaSqaaiaadgeaaeqaaaaa % kiabg2da9maalaaabaGaaGOmaiaac6cacaaIZaGaaGimaiaaiodaae % aacaaIXaGaaGimaiaaicdacaWGZbaaaiGacYgacaGGVbGaai4zamaa % laaabaGaaGimaiaac6cacaaI1aGaaGPaVlaadggacaWG0bGaamyBaa % qaaiaaicdacaGGUaGaaGinaiaaiEdacaaI2aGaaGPaVlaadggacaWG % 0bGaamyBaaaaaeaacqGH9aqpdaWcaaqaaiaaikdacaGGUaGaaG4mai % aaicdacaaIZaaabaGaaGymaiaaicdacaaIWaGaam4CaaaacqGHxdaT % caaIWaGaaiOlaiaaicdacaaIYaGaaGymaiaaiAdacqGH9aqpcaaI0a % GaaiOlaiaaiMdacaaI4aGaey41aqRaaGymaiaaicdadaahaaWcbeqa % aiabgkHiTiaaisdaaaGccaWGZbWaaWbaaSqabeaacqGHsislcaaIXa % aaaaaaaa!77ED! \begin{array}{l} \kappa = \frac{{2.303}}{t}\log \frac{{{p_1}}}{{{p_A}}} = \frac{{2.303}}{{100s}}\log \frac{{0.5\,atm}}{{0.476\,atm}}\\ = \frac{{2.303}}{{100s}} \times 0.0216 = 4.98 \times {10^{ - 4}}{s^{ - 1}} \end{array}\)

Half-Life of a Reaction

The half-life of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration. It is represented as t_1/2.

For a zero order reaction, rate constant is given by equation 4.7.

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqaH6o % WAcqGH9aqpdaWcaaqaamaadmaabaGaamOuaaGaay5waiaaw2faamaa % BaaaleaacaaIWaaabeaakiabgkHiTmaadmaabaGaamOuaaGaay5wai % aaw2faaaqaaiaadshaaaaabaGaamyqaiaadshacaaMc8UaaGPaVlaa % dshacqGH9aqpcaWG0bWaaSbaaSqaaiaaigdacaGGVaGaaGOmaaqaba % GccaGGSaWaamWaaeaacaWGsbaacaGLBbGaayzxaaGaeyypa0ZaaSaa % aeaacaaIXaaabaGaaGOmaaaadaWadaqaaiaadkfaaiaawUfacaGLDb % aadaWgaaWcbaGaaGimaaqabaaaaaa!550F! \begin{array}{l} \kappa = \frac{{{{\left[ R \right]}_0} - \left[ R \right]}}{t}\\ At\,\,t = {t_{1/2}},\left[ R \right] = \frac{1}{2}{\left[ R \right]_0} \end{array}\)

The rate constant at t_1/2 becomes

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqaH6o % WAcqGH9aqpdaWcaaqaamaadmaabaGaamOuaaGaay5waiaaw2faamaa % BaaaleaacaaIWaaabeaakiabgkHiTiaaigdacaGGVaGaaGOmamaadm % aabaGaamOuaaGaay5waiaaw2faamaaBaaaleaacaaIWaaabeaaaOqa % aiaadshadaWgaaWcbaGaaGymaiaac+cacaaIYaaabeaaaaaakeaaca % WG0bWaaSbaaSqaaiaaigdacaGGVaGaaGOmaaqabaGccqGH9aqpdaWc % aaqaamaadmaabaGaamOuaaGaay5waiaaw2faamaaBaaaleaacaaIWa % aabeaaaOqaaiaaikdacqaH6oWAaaaaaaa!513D! \begin{array}{l} \kappa = \frac{{{{\left[ R \right]}_0} - 1/2{{\left[ R \right]}_0}}}{{{t_{1/2}}}}\\ {t_{1/2}} = \frac{{{{\left[ R \right]}_0}}}{{2\kappa }} \end{array}\)

It is clear that t_1/2 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,

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqaH6o % WAcqGH9aqpdaWcaaqaaiaaikdacaGGUaGaaG4maiaaicdacaaIZaaa % baGaamiDaaaaciGGSbGaai4BaiaacEgadaWcaaqaamaadmaabaGaam % OuaaGaay5waiaaw2faamaaBaaaleaacaaIWaaabeaaaOqaamaadmaa % baGaamOuaaGaay5waiaaw2faaaaacaaMc8UaaGPaVlaaykW7caaMc8 % UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 % caaMc8UaaGPaVlaaykW7caaMc8+aaeWaaeaacaaI0aGaaiOlaiaaig % dacaaI1aaacaGLOaGaayzkaaaabaGaamyyaiaadshacaaMc8UaaGPa % VlaaykW7caaMc8UaamiDamaaBaaaleaacaaIXaGaai4laiaaikdaae % qaaOWaamWaaeaacaWGsbaacaGLBbGaayzxaaGaeyypa0ZaaSaaaeaa % daWadaqaaiaadkfaaiaawUfacaGLDbaadaWgaaWcbaGaaGimaaqaba % aakeaacaaIYaaaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa % VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 % UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8+aaeWaaeaacaaI % 0aGaaiOlaiaaigdacaaI2aaacaGLOaGaayzkaaaaaaa!9AED! \begin{array}{l} \kappa = \frac{{2.303}}{t}\log \frac{{{{\left[ R \right]}_0}}}{{\left[ R \right]}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {4.15} \right)\\ at\,\,\,\,{t_{1/2}}\left[ R \right] = \frac{{{{\left[ R \right]}_0}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {4.16} \right) \end{array}\)

So, the above equation becomes 

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGRb % Gaeyypa0ZaaSaaaeaacaaIYaGaaiOlaiaaiodacaaIWaGaaG4maaqa % aiaadshadaWgaaWcbaGaaGymaiaac+cacaaIYaaabeaaaaGcciGGSb % Gaai4BaiaacEgadaWcaaqaamaadmaabaGaamOuaaGaay5waiaaw2fa % amaaBaaaleaacaaIWaaabeaaaOqaamaadmaabaGaamOuaaGaay5wai % aaw2faaiaac+cacaaIYaaaaaqaaiaad+gacaWGYbGaaGPaVlaaykW7 % caaMc8UaaGPaVlaadshadaWgaaWcbaGaaGymaiaac+cacaaIYaaabe % aakiabg2da9maalaaabaGaaGOmaiaac6cacaaIZaGaaGimaiaaioda % aeaacqaH6oWAaaGaciiBaiaac+gacaGGNbGaaGOmaaqaaiaadshada % WgaaWcbaGaaGymaiaac+cacaaIYaaabeaakiabg2da9maalaaabaGa % aGOmaiaac6cacaaIZaGaaGimaiaaiodaaeaacqaH6oWAaaGaey41aq % RaaGimaiaac6cacaaIZaGaaGimaiaaigdaaeaacaWG0bWaaSbaaSqa % aiaaigdacaGGVaGaaGOmaaqabaGccqGH9aqpdaWcaaqaaiaaicdaca % GGUaGaaGOnaiaaiMdacaaIZaaabaGaeqOUdSgaaaaaaa!787C! \begin{array}{l} k = \frac{{2.303}}{{{t_{1/2}}}}\log \frac{{{{\left[ R \right]}_0}}}{{\left[ R \right]/2}}\\ or\,\,\,\,{t_{1/2}} = \frac{{2.303}}{\kappa }\log 2\\ {t_{1/2}} = \frac{{2.303}}{\kappa } \times 0.301\\ {t_{1/2}} = \frac{{0.693}}{\kappa } \end{array}\)

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 t_1/2 \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqGHDisTaaa!3796! \propto \)  [R]₀. For first order reaction t_1/2 is independent of [R]₀.

EXAMPLE 7

A first order reaction is found to have a rate constant, k = 5.5 × 10^-14 s^-1. Find the half-life of the reaction

SOLUTION

Half-life for a first order reaction is 

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaqaaaaa % aaaaWdbiaadshadaWgaaWcbaGaaGymaiaac+cacaaIYaaabeaakiab % g2da9maalaaabaGaaGimaiaac6cacaaI2aGaaGyoaiaaiodaaeaacq % aH6oWAaaaabaGaamiDamaaBaaaleaacaaIXaGaai4laiaaikdaaeqa % aOGaeyypa0ZaaSaaaeaacaaIWaGaaiOlaiaaiAdacaaI5aGaaG4maa % qaaiaaiwdacaGGUaGaaGynaiabgEna0kaaigdacaaIWaWaaWbaaSqa % beaacqGHsislcaaIXaGaaGinaaaakiaadohadaahaaWcbeqaaiabgk % HiTiaaigdaaaaaaOGaeyypa0JaaGymaiaac6cacaaIYaGaaGOnaiab % gEna0kaaigdacaaIWaWaaWbaaSqabeaacaaIXaGaaG4maaaakiaado % haaaaa!5D57! \begin{array}{l} {t_{1/2}} = \frac{{0.693}}{\kappa }\\ {t_{1/2}} = \frac{{0.693}}{{5.5 \times {{10}^{ - 14}}{s^{ - 1}}}} = 1.26 \times {10^{13}}s \end{array}\)

EXAMPLE 8

Show that in a first order reaction, time required for completion of 99.9% is 10 times of half-life (t_1/2) of the reaction.

When reaction is completed 99.9%, [R]_n = [R]₀ – 0.999[R]₀

SOLUTION

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaqaaaaa % aaaaWdbiabeQ7aRjabg2da9maalaaabaGaaGOmaiaac6cacaaIZaGa % aGimaiaaiodaaeaacaWG0baaaiGacYgacaGGVbGaai4zamaalaaaba % WaamWaaeaacaWGsbaacaGLBbGaayzxaaWaaSbaaSqaaiaaicdaaeqa % aaGcbaWaamWaaeaacaWGsbaacaGLBbGaayzxaaaaaaqaaiabg2da9m % aalaaabaGaaGOmaiaac6cacaaIZaGaaGimaiaaiodaaeaacaWG0baa % aiGacYgacaGGVbGaai4zamaalaaabaWaamWaaeaacaWGsbaacaGLBb % GaayzxaaWaaSbaaSqaaiaaicdaaeqaaaGcbaWaamWaaeaacaWGsbaa % caGLBbGaayzxaaWaaSbaaSqaaiaaicdaaeqaaOGaeyOeI0IaaGimai % aac6cacaaI5aGaaGyoaiaaiMdadaWadaqaaiaadkfaaiaawUfacaGL % DbaadaWgaaWcbaGaaGimaaqabaaaaOGaeyypa0ZaaSaaaeaacaaIYa % GaaiOlaiaaiodacaaIWaGaaG4maaqaaiaadshaaaGaciiBaiaac+ga % caGGNbGaaGymaiaaicdadaahaaWcbeqaaiaaiodaaaaakeaacaWG0b % Gaeyypa0JaaGOnaiaac6cacaaI5aGaaGimaiaaiMdacaGGVaGaeqOU % dSgaaaa!7241! \begin{array}{l} \kappa = \frac{{2.303}}{t}\log \frac{{{{\left[ R \right]}_0}}}{{\left[ R \right]}}\\ = \frac{{2.303}}{t}\log \frac{{{{\left[ R \right]}_0}}}{{{{\left[ R \right]}_0} - 0.999{{\left[ R \right]}_0}}} = \frac{{2.303}}{t}\log {10^3}\\ t = 6.909/\kappa \end{array}\)

For half-life of the reaction

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaqaaaaa % aaaaWdbiaadshadaWgaaWcbaGaaGymaiaac+cacaaIYaaabeaakiab % g2da9iaaicdacaGGUaGaaGOnaiaaiMdacaaIZaGaai4laiabeQ7aRb % qaamaalaaabaGaamiDaaqaaiaadshadaWgaaWcbaGaaGymaiaac+ca % caaIYaaabeaaaaGccqGH9aqpdaWcaaqaaiaaiAdacaGGUaGaaGyoai % aaicdacaaI5aaabaGaeqOUdSgaaiabgEna0oaalaaabaGaeqOUdSga % baGaaGimaiaac6cacaaI2aGaaGyoaiaaiodaaaGaeyypa0JaaGymai % aaicdaaaaa!5569! \begin{array}{l} {t_{1/2}} = 0.693/\kappa \\ \frac{t}{{{t_{1/2}}}} = \frac{{6.909}}{\kappa } \times \frac{\kappa }{{0.693}} = 10 \end{array}\)

Table 4.4 summarises the mathematical features of integrated laws of zero and first order reactions.

                 Table 4.4: Integrated Rate Laws for the Reactions of Zero and First Order

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.

 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.

Rate = \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqaH6oWAcqGH9aqpdaWadaqaaiaadoeadaWgaaWcbaGaaGymaiaa % ikdaaeqaaOGaamisamaaBaaaleaacaaIYaGaaGOmaaqabaGccaWGpb % WaaSbaaSqaaiaaigdacaaIXaaabeaaaOGaay5waiaaw2faaaaa!4230! \kappa = \left[ {{C_{12}}{H_{22}}{O_{11}}} \right]\)

      Intext Questions

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?

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.

TTemperature Dependence OF the Rate of a Reaction

Most of the chemical reactions are accelerated by increase in temperature. For example, in decomposition of N₂O₅, the time taken for half of the original amount of material to decompose is 12 min at 50^oC, 5 h at 25^oC and 10 days at 0^oC. You also know that in a mixture of potassium permanganate (KMnO₄) and oxalic acid (H₂C₂O₄), potassium permanganate gets decolourised faster at a higher temperature thanthat 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 (4.18). It was first proposed by Dutch chemist, J.H. van’t Hoff but Swedish chemist, Arrhenius provided its physical justification and interpretation.

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqaH6oWAcqGH9aqpcaWGbbGaamyzamaaCaaaleqabaGaeyOeI0Ia % amyraiaadggacaGGVaGaamOuaiaadsfaaaaaaa!3FAB! \kappa = A{e^{ - Ea/RT}}\)       (4.18)

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

According to Arrhenius, this reaction can take place only when a molecule of hydrogen and a molecule of iodine collide to form an unstable intermediate (Fig. 4.6).

Fig. 4.6: Formation of HI through the intermediate

It exists for a very short time and then breaks up to form two molecules of hydrogen iodide.

The energy required to form this intermediate, called activated complex (C), is known as activation energy (E_a). Fig. 4.7 is obtained

Fig. 4.7: Diagram showing plot of potential energy vs reaction coordinate

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.