CHEMICAL KINETICS

Chemical Kinetics helps us to understand how chemical reactions occur.
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 ΔG < 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.
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:
(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,
Δt = t2 – t1
Δ[R] = [R]2 – [R]1
Δ [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 / Time taken       R/t ……………(4.1)
Rate of appearance of P
=Increase in concentration of P / Time taken        P/t ………………(4.2)
Since, Δ[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, rav.
Average rate depends upon the change in concentration of reactants or products and the time taken for that change to occur (Fig. 4.1).
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 4.1
From the concentrations of C4H9Cl (butyl chloride) at different times given 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
Solution
We can determine the difference in concentration over different intervals of time and thus determine the average rate by dividing Δ[R] by Δt(Table 4.1).
Table 4.1: Average rates of hydrolysis of butyl chloride
[C4H9CL]T1/ MOL L-1[C4H9CL]T2/ MOL L-1T1/ST2/SRAV×104/MOL L -1S-1
= {[C4H9CL]T2 – [C4H9CL]T1 / (T2 – T1)} × 104
0.1000.09050501.90
0.09050.0820501001.70
0.08200.07411001501.58
0.07410.06211502001.40
0.06710.05492003001.22
0.05490.04393004001.10
0.04390.03354005001.04
0.02100.0177008000.4
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 Δt approaches zero). Hence, mathematically for an infinitesimally small dt instantaneous rate is given by
rav R/t   P/t……………………….(4.3)
As   Δt → 0   or   rinst =−d [R ]/dt= d [P ]/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).
The slope of this tangent gives the instantaneous rate.
So, rinst at 600 s = – 0.0165 – 0.037/800 – 400 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
Rate of reaction = – Hg/t    −Cl2/t   HgCl2/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 Δ[HI] is divided by 2. The rate of this reaction is given by
Rate of reaction   1/2    HI/t    H2/t   I2/t
Similarly, for the reaction
5 Br- (aq) + BrO3– (aq) + 6 H+ (aq) → 3 Br2 (aq) + 3 H2O (l)
Rate   1/5  Br/t  BrO3/t  1/6  H/t  1/3  H2O/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 4.2 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–1and 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?
Solution
Average Rate  1/2  N2O5/t   1/2   2.08×2.33 mol L1/184 min
= 6.79 × 10 mol L /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 1/4    NO2/t
NO2/t    6.79 × 10–4 × 4 mol L–1 min–1 = 2.72 × 10–3 mol L–1min–1
Intext Questions
4.1 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.
4.2 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?
4.2 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.
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.
4.2.2 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 → cC + dD
where a, b, c and d are the stoichiometric coefficients of reactants and products.
The rate expression for this reaction is
Rate ∝ [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)
d R/dt k A x By……………………(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) → 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
EXPERIMENTINITIAL [NO] / MOL L-1INITIAL [O2] / MOL L-1INITIAL RATE OF FORMATION OF NO2 / MOL L-1S-1
10.300.300.096
20.600.300.384
30.300.600.192
40.600.600.768
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 O2is doubled the rate also gets doubled indicating that rate depends on concentration of O2to the first power. Hence, the rate equation for this reaction will be
Rate = k[NO] [O2]
The differential form of this rate expression is given as
d R/dt k NO2 O2
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]°
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.
4.2.3 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 4.3 Calculate the overall order of a reaction which
solution
has the rate expression
(a) Rate = k [A]1/2 [B]3/2
(b) Rate = k [A]3/2 [B]–1
(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 → cC + dD
Rate = k [A]x [B]y
Where x + y = n = order of the reaction
k =Rate/ [A]x [B]y
= concentration / time × 1/ concentrationn where [A] [B]
Taking SI units of concentration, mol L and time, s, the units of k for different reaction order are listed in Table 4.3
Table 4.3: Units of rate constant
REACTIONORDERUNITS OF RATE CONSTANT
Zero order reaction0mol L-1/s × 1/(mol L-1)0 = mol L-1S-1
first order reaction1mol L-1/s × 1/(mol L-1)1 = S-1
Second order reaction2mol L-1/s × 1/(mol L-1)2 = mol-1 L-1s-1
Example 4.4
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
(i) 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.
(ii) The unit of a first order rate constant is s–1 therefore k = 3 × 10–4 s–1 represents a first order reaction.
4.2.4 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 → 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, the molecularity greater than three is not observed.
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.
The rate equation for this reaction is found to be
Rate= −d [H2 O2 ]/dt
= k [H2 O2 ][I− ]
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.
(iii) For complex reaction, order is given by the slowest step and generally, molecularity of the slowest step is same as the order of the overall reaction.
Intext Questions
4.3 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?
4.4 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 ?
4.3 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.
4.3.1 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 →P
Rate = −d [R ]/dt = k [R ]°
As any quantity raised to power zero is unity
Rate = − d [R ]/dt =k ×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.
Further simplifying equation (4.6), we get the rate constant, k as
k = (R0 × R) / 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 = 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.
4.3.2 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 →P
Rate dR/dt k R
or d R/ R -kdt
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
ln R/R0 kt
or k = 1/t [R ]0 / [R] …………………(4.10)
At time t1 from equation (4.8)
*ln[R]1 = – kt1 + *ln[R]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)
ln R1 / R2 k t2 t1
k 1/ t2 t1 ln R1/R2………………………(4.13)
Equation (4.9) can also be written as
ln R / R0 kt
Taking antilog of both sides
[R] = [R]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
The first order rate equation (4.10) can also be written in the form
k 2.303 / t log R0/ R…………………………(4.15)
*log R0/ R kt/2.303………………………(4.15)
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) → C2H6(g)
Rate = k [C2H4]
All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics.
22688Ra → 2He + 22286Rn
Rate = k [Ra]
Decomposition of N2O5 and N2O are some more examples of first order reactions.
Example 4.5 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.
Solution
For a first order reaction
Log [R1] / [R2] = k(t2 − t1) / 2.303
k = 2.303 / (t2 − t1) log [R1] / [R2]
=2.303 / (t2 − t2) log 1.24 × 10-2 mol L−1 / 0.20 × 10-2 mol L−1
=2.303 / 60 log 6.2 min−1
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)
pA , pB , 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 0 atm       0 atm
At t = t       (pi−x)   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
k =( 2.303 / t ) ( log pi / pA)……………………….(4.16)
= ( 2.303 / t) log (pi /2 pi − pt )
Example 4.6
The following data were obtained during the first order thermal decomposition of N2O5 (g) at constant volume:
2N2O5 g    2N2O4g    O2
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.
2N2O5g       2N2O4 O2g
Start t = 0 0.5 atm          0 atm       0 atm
At time t  (0.5 − 2x) atm     2x atm       x atm
pt = p N2O5 + pN2O4 + pO2
= (0.5 – 2x) + 2x + x = 0.5 + x
x        pt 0.5
pN2O5 = 0.5 – 2x
= 0.5 – 2 (pt – 0.5) = 1.5 – 2pt
At t = 100 s; pt = 0.512 atm
pN2O5 = 1.5 – 2 × 0.512 = 0.476 atm
Using equation (4.16)
k=2.303/t log pi/ pA = (2.303 /100s) log 0.5 / 0.476atm
=2.303/100 s× 0.0216 = 4.98 × 10−4 s −1
4.3.3 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 t1/2.
For a zero order reaction, rate constant is given by equation 4.7.
k R0 R / t
At t t1/2, R    1/2R0
The rate constant at t1/ 2 becomes
k R0 1/2R0 / t1/2
t1/2 R0/2k
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,
k 2.303/t log R0/ R………………………………(4.15)
at t1/2 R    R0/2…………………(4.16)
So, the above equation becomes
k 2.303/t1/2 log R0/ R/2
or
t1/2 = 2.303/k log 2
t1/2 2.303/k 0.301
t1/2 0.693 / k………………………….(4.17)
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 ∝ [R]0. For first order reaction t1/2 is independent of [R]0.
Example 4.7
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
t1/2 = 0.693 / k
t1/2 = 0.693 / 5.5 ×10 –14 s –1
= 1.26 × 1014s
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.
Example 4.8
When reaction is completed 99.9%, [R]n = [R]0 – 0.999[R]0
Solution
k = 2.303 /t log [R]0/[R]
= 2.303 / t log [R]0/[R]0 − 0.999[R]0 = 2.303/t log 103
t = 6.909 / k
For half-life of the reaction
t1/2 = 0.693/k
t / t1/2 = 6.909 / k × k / 0.693 =10
Table 4.4: Integrated Rate Laws for the Reactions of Zero and First Order
ORDERREACTION TYPEDIFFERENTIAL RATE LAWINTEGRATED RATE LAWSTRAIGHT LINE PLOTHLF LIFEUNITS OF K
0R → Pd[R]/dt = -kkt = [R]0-[R][R] vs t[R]0/ 2kconc time -1 or mol L-1 s-1
0R → Pd[R]/dt = -k[R][R] = [R]0e-kt or kt = ln{[R]0 / [R]}ln[R] vs tln 2/ktime-1 or s-1
4.4 Pseudo First Order Reaction
The order of a reaction is sometimes altered by conditions. Consider a chemical reaction between two substances when one reactant is present in large excess. During the hydrolysis of 0.01 mol of ethyl acetate with 10 mol of water, amounts of the various constituents 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 9.9 mol      0.01 mol    0.01 mol
The concentration of water does not get altered much during the course of the reaction. So, in the rate equation
Rate = k′ [CH3COOC2H5] [H2O]
the term [H2O] can be taken as constant. The equation, thus, becomes
Rate = k [CH3COOC2H5]
where k = k′ [H2O]
and 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 = k [C12H22O11]
Example 4.9 Hydrolysis of methyl acetate in aqueous solution has been studied by titrating the liberated acetic acid against sodium hydroxide. The concentration of the ester at different times is given below.
T/MIN0306090
C/MOL L-10.85000.80040.75380.7096
Show that it follows a pseudo first order reaction, as the concentration of water remains nearly constant (55 mol L-1), during the course of the reaction. What is the value of k′ in this equation? Rate = k′ [CH3COOCH3][H2O]
Solution For pseudo first order reaction, the reaction should be first order with respect to ester when [H2O] is constant. The rate constant k for pseudo first order reaction is
k 2.303 / t log C0/C          where k k’ H2O
From the above data we note
T/MINC/MOL L-1K’/MIN-1
00.8500-
300.80042004 ×10-3
600.75382002 ×10-3
900.70962005 ×10-3
It can be seen that k ́ [H2O] is constant and equal to 2.004 × 10-3 min–1 and hence, it is pseudo first order reaction. We can now determine k from
k ́ [H2O] = 2.004 × 10–3 min–1
k ́ [55 mol L–1] = 2.004 × 10–3 min–1
k ́ = 3.64 × 10–5 mol–1 L min–1
Intext Questions
4.5 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?
4.6 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.
4.5 Temperature Depndence 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°C, 5 h at 25°C and 10 days at 0°C. You also know that in a mixture of potassium 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 (4.18). It was first proposed by Dutch chemist, J.H. van’t Hoff but Swedish chemist, Arrhenius provided its physical justification and interpretation .
k = 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
H2 (g) + I2 ( g ) → 2HI ( g )
According to Arrhenius, this reaction can take place only when a molecule of hydrogen and a molecule of iodine collide Intermediate to form an unstable intermediate (Fig. 4.6). 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 (Ea). Fig. 4.7 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 heat of the reaction depends upon the nature of reactants and products.
All the molecules in the reacting species do not have the same kinetic energy. Since it is difficult to predict the 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. 4.8). 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 of molecules with energies higher or lower than this value. When the temperature is raised, the maximum of the curve moves to the higher energy value (Fig. 4.9) 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 Maxwell Boltzmann distribution curve (Fig. 4.9).
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 (4.18) 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 (4.18)
ln k = –Ea/RT+ ln A………………………(4.19)
The plot of ln k vs 1/T gives a straight line according to the equation (4.19) as shown in Fig. 4.10.
Thus, it has been found from Arrhenius equation (4.18) 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.
In Fig. 4.10, slope = –Ea / R and intercept = ln A. So we can calculate Ea and A using these values.
At temperature T1, equation (4.19) is
ln k1 = – Ea/RT + ln A……………………..(4.20)
At temperature T2, equation (4.19) is
ln k2 = – Ea/RT2 + ln A…………………………(4.21)
(since A is constant for a given reaction) k1 and k2 are the values of rate constants at temperatures T1 and T2 respectively.
Subtracting equation (4.20) from (4.21), we obtain
ln k2 – ln k1 = Ea/RT1 – Ea/RT2
ln k2/k1 = Ea/R [1/T1 - 1/T2]
log k2/k1 = Ea/2.303R [1/T1 - 1/T2............................(4.22)
log k2/k1 = Ea/2.303 R [ (T2 - T1) / T1T2]
Example 4.10 The rate constants of a reaction at 500K and 700K are 0.02s-1 and 0.07s−1respectively. Calculate the values of Ea and A.
Solution
log k2/k1 = Ea/2.303 R [(T2 − T1) / T1T2]
log 0.07/0.02 =(Ea/ 2.303 × 8.314 JK −1mol −1)[(700 − 500)/ (700× 500)] 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
0.02 = Ae-18230.8/8.314 × 500
A = 0.02/0.012 = 1.61
Example 4.11 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.
Solution
We know that log k2 – log k1 = Ea/2.303R[1/T1 − 1/T2]
log k2 = log k1 + Ea/2.303R[1/T1 − 1/T2]
= log (1.60 × 10−5) + 209000 J mol L−1/2.303 × 8.314 J mol L−1K−1 [1/600K − 1/700K]
log k2 = – 4.796 + 2.599 = – 2.197
k2 = 6.36 × 10–3 s–1
4.5.1 Effect of Catalyst
A catalyst is a substance which alters the rate of a reaction without itself undergoing any permanent chemical change. For example, MnO2 catalyses the following reaction so as to increase its rate considerably.
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. 4.11.
It is clear from Arrhenius equation (4.18) 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 not alter Gibbs energy, ΔG 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.
4.6 Collision Theory of Chemical Reaction
Though Arrhenius equation is applicable under a wide range of circumstances, collision theory, which was developed by Max Trautz and William Lewis in 1916 -18, provides a greater insight into the energetic and mechanistic aspects of reactions. It is based on kinetic theory of gases. According to this theory, the reactant molecules are 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
Rate = ZABe -Ea / RT…………………….(4.23)
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 (4.23) with Arrhenius equation, we can say that A is related to collision frequency.
Equation (4.23) 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. 4.12. The proper orientation of reactant molecules lead to bond formation whereas improper orientation makes them simply bounce back and no products are formed.
To account for effective collisions, 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.,
Rate = PZAB e-Ea/ RT
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.
Intext Question
4.7 What will be the effect of temperature on rate constant ?
4.8 The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate Ea. 4.9 The activation energy for the reaction
2 HI(g) → H2 + I2(g)
is 209.5 kJ mol–1 is 209.5kJ mol -1 at 581K.Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
Summary
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 = PZ AB e−Ea / RT .
Exercises
4.1 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+ → H2O(l) + I−3 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]
4.2 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.
4.3 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?
4.4 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.,
Rate = k(PCH3OCH3)3/2
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?4.5 Mention the factors that affect the rate of a chemical reaction.
4.6 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 ?
4.7 What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?
4.8 In a pseudo first order hydrolysis of ester in water, the following results were obtained:
T/S0306090
[ESTER]/MOL L-10.550.310.170.085
(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
4.9 A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?
4.10 In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below:
A/MOL L-10.200.200.40
B/MOL L-10.300.100.05
R0/MOL L-1S-15.07 × 1055.07 × 10-51.43 × 10-4
What is the order of the reaction with respect to A and B?
4.11 The following results have been obtained during the kinetic studies of the reaction:
2A + B →C + D
EXPERIMENT[A]/MOL L-1[B]/MOL L-1INITIAL RATE OF FORMATION OF D/ MOL L-1 MIN-1
I0.10.16.0 × 10-3
II0.30.27.2 × 10-2
III0.30.42.88 × 10-2
IV0.40.12.40 × 10-2
Determine the rate law and the rate constant for the reaction.
4.12 The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
EXPERIMENT[A]/MOL L-1[B]/MOL L-1INITIAL RATE OF FORMATION OF D/ MOL L-1 MIN-1
I0.10.12.0 × 10-2
II-0.24.0 × 10-2
III0.40.4-
IV-0.22.0 × 10-2
4.13 Calculate the half-life of a first order reaction from their rate constants given below:
(i) 200 s–1 (ii) 2 min–1 (iii) 4 years–1
4.14 The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.
4.15 The experimental data for decomposition of N2O5 [2N2O5 → 4NO2 + O2]in gas phase at 318K are given below:
t/s0400800120016002000240028003200
102 × [N2O5]/ mol L-11.631.361.140.930.780.640.530.430.35
(i) Plot [N2O5] against t.
(ii) Find the half-life period for the reaction.
(iii) Draw a graph between log[N2O5] and t.
(iv) What is the rate law ?
(v) Calculate the rate constant.
(vi) Calculate the half-life period from k and compare it with (ii).
4.16 The rate constant for a first order reaction is 60 s–1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?
4.17 During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1μg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.
4.18 For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
4.19 A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.
4.20 For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.
t(sec) p(mm of Hg) 0 35.0 360 54.0 720 63.0
calculate the rate constant.
4.21 The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume. SO2Cl2 ( g ) → SO2 ( g ) + Cl2( g )
Experiment100.5
21000.6
Calculate the rate of the reaction when total pressure is 0.65 atm.
4.22 The rate constant for the decomposition of N2O5at various temperatures is given below:
T/°C 0 20 40 60 80 105×k/s-1 0.0787 1.70 25.7 178 2140
Draw a graph between ln k and 1/T and calculate the values of A and Ea. Predict the rate constant at 30° and 50° C.
4.23 The rate constant for the decomposition of hydrocarbons is 2.418 × 10–5s–1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.
4.24 Consider a certain reaction A → Products with k = 2.0 × 10 –2s–1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L–1.
4.25 Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours ?
4.26 The decomposition of hydrocarbon follows the equation k = (4.5 × 1011s–1) e-28000K/TCalculate Ea.
4.27 The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 – 1.25 × 104K/T
Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?
4.28 The decomposition of A into product has value of k as 4.5 × 103 s–1 at 10° C and energy of activation 60 kJ mol–1. At what temperature would k be 1.5 × 104s–1?
4.29 The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 1010 s–1. Calculate k at 318K and Ea.
4.30 The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Answers to Some Intext Questions
4.1 rav = 6.66 × 10-6 Ms–1
4.2 Rate of reaction = rate of diappearance of A = 0.005 mol litre-1min–1
4.3 Order of the reaction is 2.5
4.4 X →Y
Rate = k[X]2
The rate will increase 9 times
4.5 t = 444 s
4.6 1.925 × 10–4 s–1
4.8 Ea = 26.43 kJ mol–1
4.9 1.462 × 10–19
I. Multiple Choice Questions (Type-I)
1. The role of a catalyst is to change ______________.
(i) gibbs energy of reaction.
(ii) enthalpy of reaction.
(iii) activation energy of reaction.
(iv) equilibrium constant.
2. In the presence of a catalyst, the heat evolved or absorbed during the reaction ___________.
(i) increases.
(ii) decreases.
(iii) remains unchanged.
(iv) may increase or decrease.
3. Activation energy of a chemical reaction can be determined by _____________.
(i) determining the rate constant at standard temperature.
(ii) determining the rate constants at two temperatures.
(iii) determining probability of collision.
(iv) using catalyst.
4. Consider Fig. 4.1 and mark the correct option.
(i) Activation energy of forward reaction is E1 + E2 and product is less stable than reactant.
(ii) Activation energy of forward reaction is E1+E2 and product is more stable than reactant.
(iii) Activation energy of both forward and backward reaction is E1+E2 and reactant is more stable than product.
(iv) Activation energy of backward reaction is E1 and product is more stable than reactant.
5. Consider a first order gas phase decomposition reaction given below :
A(g) → B(g) + C(g)
The initial pressure of the system before decomposition of A was pi. After lapse of time ‘t’, total pressure of the system increased by x units and became ‘pt’ The rate constant k for the reaction is given as _________.
(i) k = 2.303/t log (pi/pi-x)
(ii) k = 2.303/t log (pi/2pi-pt)
(iii) k = 2.303/t log (pi/2pi+pt)
(iv) k = 2.303/t log (pi/2pi+x)
6. According to Arrhenius equation rate constant k is equal to A e–Ea /RT . Which of the following options represents the graph of ln k vs 1/T ?
7. Consider the Arrhenius equation given below and mark the correct option. k = A e–Ea /RT
(i) Rate constant increases exponentially with increasing activation energy and decreasing temperature.
(ii) Rate constant decreases exponentially with increasing activation energy and decreasing temperature.
(iii) Rate constant increases exponentially with decreasing activation energy and decreasing temperature.
(iv) Rate constant increases exponentially with decreasing activation energy and increasing temperature.
8. A graph of volume of hydrogen released vs time for the reaction between zinc and dil.HCl is given in Fig. 4.2. On the basis of this mark the correct option.
(i) Average rate upto 40s is (V3 − V2)/40
(ii) Average rate upto 40 seconds is (V3 − V2)/(40 – 30)
(iii) Average rate upto 40 seconds is V3/40
(iv) Average rate upto 40 seconds is (V3 − V1)/(40 − 20)
9. Which of the following statements is not correct about order of a reaction.
(i) The order of a reaction can be a fractional number.
(ii) Order of a reaction is experimentally determined quantity.
(iii) The order of a reaction is always equal to the sum of the stoichiometric coefficients of reactants in the balanced chemical equation for a reaction.
(iv) The order of a reaction is the sum of the powers of molar concentration of the reactants in the rate law expression.
10. Consider the graph given in Fig. 4.2. Which of the following options does not show instantaneous rate of reaction at 40th second?
(i) V5 − V2 / 50 − 30
(ii) V4 − V2 / 50 − 30
(iii) V3 − V2 / 40 − 30
(iv) V3 − V1 / 40 − 20
12. Which of the following expressions is correct for the rate of reaction given below?
5Br–(aq) + BrO3–(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l)
(i)Δ[Br–]/Δt = 5 Δ[H+]/Δt
(ii)Δ[Br–]/Δt = 6/5 Δ[H+]/Δt
(iii)Δ[Br–]/Δt = 5/6 Δ[H+]/Δt
(iv)Δ[Br–]/Δt = 6 Δ[H+]/Δt
13. Which of the following graphs represents exothermic reaction?
(i) (a) only
(ii) (b) only
(iii) (c) only
(iv) (a) and (b)
14. Rate law for the reaction A + 2B → C is found to be Rate = k [A][B] Concentration of reactant ‘B’ is doubled, keeping the concentration of ‘A’ constant, the value of rate constant will be______.
(i) the same
(ii) doubled
(iii) quadrupled
(iv) halved
15. Which of the following statements is incorrect about the collison theory of chemical reaction?
(i) It considers reacting molecules or atoms to be hard spheres and ignores their structural features.
(ii) Number of effective collisions determines the rate of reaction.
(iii) Collision of atoms or molecules possessing sufficient threshold energy results into the product formation.
(iv) Molecules should collide with sufficient threshold energy and proper orientation for the collision to be effective.
16. A first order reaction is 50% completed in 1.26 x 1014 s. How much time would it take for 100% completion?
(i) 1.26 x 1015 s
(ii) 2.52 x 1014 s
(iii) 2.52 x 1028 s
(iv) infinite
17. Compounds ‘A’ and ‘B’ react according to the following chemical equation. A (g) + 2 B (g) → 2C (g)Concentration of either ‘A’ or ‘B’ were changed keeping the concentrations of one of the reactants constant and rates were measured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate equations for this reaction.
ExperimentInitial concentration of [A]/mol L–1Initial concentration of [B]/mol L–1Initial rate of formation of [C]/mol L–1 s–1
1.0.300.300.10
2.0.300.600.40
3.0.600.300.20
(i) Rate = k [A]2 [B]
(ii) Rate = k [A] [B]2
(iii) Rate = k [A] [B]
(iv) Rate = k [A]2 [B]0
18. Which of the following statement is not correct for the catalyst?
(i) It catalyses the forward and backward reaction to the same extent.
(ii) It alters ΔG of the reaction.
(iii) It is a substance that does not change the equilibrium constant of a reaction.
(iv) It provides an alternate mechanism by reducing activation energy between reactants and products.
19. The value of rate constant of a pseudo first order reaction ____________.
(i) depends on the concentration of reactants present in small amount.
(ii) depends on the concentration of reactants present in excess.
(iii) is independent of the concentration of reactants.
(iv) depends only on temperature.
20. Consider the reaction . The concentration of both the reactants and the products varies exponentially with time. Which of the following figures correctly describes the change in concentration of reactants and products with time?

Lab Manual
CHEMICAL KINETICS
Rate of reaction can be measured either in terms of decrease in concentration of any one of the reactants or increase in concentration of any one of the products with time. For a hypothetical reaction, A → B
Rate of reaction = -[A]/T = [B]/T
Factors such as concentration, temperature and catalyst affect the rate of a reaction. In this unit you will learn the technique of determining the rate of a reaction and technique of studying the effect of concentration and temperature on the reaction rate.
EXPERIMENT 2.1
Aim
To study the effect of concentration and temperature variation respectively on the rate of reaction between sodium thiosulphate and hydrochloric acid.
Theory
Sodium thiosulphate reacts with hydrochloric acid and produces a colloidal solution of sulphur, which makes the solution translucent. The reaction occurs as follows:
Na2S2O3 (aq) + 2HCl (aq) → 2NaCl (aq) + H2O(l) + SO2(g) + S(s)
Ionic form of the above reaction is written as:
S2O32– (aq) + 2H+ (aq) → H2O (l) + SO2 (g) + S(s)
The property of the colloidal solution of sulphur to make the system translucent is used to study the rate of precipitation of sulphur. The rate of precipitation of sulphur increases with an increase in the concentration of the reacting species or with an increase in the temperature of the system. With an increase in the concentration, the number of molecular collisions per unit time between the reacting species increase and consequently chances of product formation increase. This results in an increase in the rate of precipitation of sulphur. Similarly, on increasing the temperature, the kinetic energy of the reacting species increases, so the number of collisions that result in the formation of products increase leading to a faster rate of reaction.
Material Required
Procedure
A. The effect of concentration on the rate of reaction
(i) Take a trough and fill half of it with water. This will serve as constant temperature bath, maintained at room temperature.
(ii) Rinse and fill the burette with 1.0 M HCl solution.
(iii) Take a 100 mL beaker and make a mark ‘X’ in the centre of the outer surface of the bottom with the help of a glass marker pencil. Fill 50 mL of 0.1M sodium thiosulphate solution in it. Place the beaker in the trough. The mark ‘X’ will be visible to the naked eye on account of the transparent nature of the system. Allow the beaker to stand in the trough for a few minutes so that it attains the temperature of the bath.
(iv) Add 1.0 mL of 1.0 M HCl solution with the help of a burette. Start the stopwatch when half the HCl solution i.e. (0.5 mL) has been transferred. Swirl the beaker while adding HCl.
(v) Record the time required for the mark ‘X’ on the bottom of the beaker to become invisible (This is considered as a stage of completion of the reaction).
(vi) Repeat the experiment by adding 2 mL, 4 mL, 8 mL and 16 mL of 1.0 M hydrochloric acid solution to fresh sodium thiosulphate solution every time and record the time required for the disappearance of the mark ‘X’ in each case separately.
B. The effect of temperature on the rate of reaction
(i) Take 50 mL of 0.1M sodium thiosulphate solution in a 100 mL beaker, on the outer surface of the bottom of which a mark ‘X’ has been made. Keep the beaker in a thermostat maintained at 30°C. Add 5 mL of 1.0 M hydrochloric acid solution with swirling. Start the stopwatch immediately when half the amount (i.e. 2.5 mL) of hydrochloric acid has been transferred.
(ii) Record the time at which the mark ‘X’ becomes invisible.
(iii) Repeat the experiment at temperatures 40°C, 50°C, 60°C and 70°C using fresh sodium thiosulphate solution each time and record the time required for the disappearance of the mark ‘X’.
(iv) Record your observations in Tables 2.1 and 2.2.
(v) Plot two graphs, one for the volume of HCl added (which determines concentration of HCl) and the time taken for the mark to become invisible and the other between temperature and the time taken for the mark to become invisible. For plotting the graph, the variation in time is plotted on x-axis and the variation in volume or temperature is plotted on y-axis.
Note : If thermostat (i.e. constant temperature bath) is not available for studying the rate of the reaction. Ordinary water bath may also be used for maintaining constant temperature but in this case heating of the bath from outside might be required for the adjustment of temperature. Water in the bath should also be stirred continuously.
Table 2.1 : Effect of concentration of HCl on the rate of reaction between sodium thiosulphate and hydrochloric acid
Amount of Na2S2O3 solution used each time = 50 mL
Concentration of Na2S2O3 solution = 0.1M, Room temperature = °C
Concentration of the HCl solution used in the reaction mixture = 1.0 M
Sl. No.Volume of HC1 added in mLTime ‘t’ in seconds for the mark ‘X’ to become invisible
1.1.0
2.2.0
3.4.0
4.8.0
5.16.0
Table 2.2 : The effect of temperature on the rate of reaction between sodium thiosulphate and hydrochloric acid
Volume of sodium thiosulphate solution used each time = 50 mL
Volume of HCl used each time = 5 mL
Sl. No.Temperature of the reaction mixture/°CTime ‘t’ in seconds for the mark ‘X’ to become invisible
1.30
2.40
3.50
4.60
5.70
Result
Write your conclusions on the basis of data in Tables 2.1 and 2.2.
Precautions
(a) Start the stopwatch when half of the hydrochloric acid solution has been transferred to the reaction flask and stop the watch when the mark ‘X’ becomes invisible.
(b) If a constant temperature bath is not available to maintain the constant temperature, heat the water of the bath in which the beaker is kept from time to time with constant stirring, and remove the burner when the required temperature is attained.
(c) Select suitable scale for plotting the graph.
Discussion Questions
(i) The reaction under examination is as follows:
S2O32–(aq) + 2H+ (aq) → H2O (l) + SO2 (g) + S(s)
Write the conditions under which the rate law expression for this reaction can be written in the following manner.
Rate of precipitation of sulphur = k [S2O32–][H+]2
(ii) Suppose the above rate law expression for the precipitation of sulphur holds good, then on doubling the concentration of S2O32– ion and H+ ion, by how many times will the rate of the reaction increase?
(iii) Comment on the statement that for a given reaction, rate of the reaction varies but the rate constant remains constant at a particular temperature.
(iv) How does the rate constant of a reaction vary with temperature?
(v) Devise an experiment to study the dependence of rate of precipitation of sulphur upon the nature of monobasic acid for the reaction given below :
S2O32–(aq) + 2H+(aq) → H2O (l) + SO2 (g) + S(s)
(vi) Why is the stop watch/stop clock started when half of the reactant is delivered into the beaker?
(vii) The structure of S2O32– ion is described as follows:
The two sulphur atoms are marked here as (1) and (2). Which of the sulphur atoms, according to you, is precipitated as colloidal sulphur? How can you verify your answer experimentally?
(viii) What is the difference between the order and the molecularity of a reaction?
(ix) The molecularity of a reaction can’t be zero but the order can be zero? Explain.
(x) Can the order of a reaction be a fractional quantity?
(xi) Suppose the above reaction follows third order kinetics, then in what units, will the rate of the reaction and the rate constant be expressed?
EXPERIMENT 2.2
Aim
To study the effect of variation in concentration of iodide ions on the rate of reaction of iodide ions with hydrogen peroxide at room temperature.
Theory 
The reaction between iodide ions and hydrogen peroxide occurs in the acidic medium and can be represented in the following manner:2I–(aq) + H2O2 (l) + 2H+(aq) → I2(g) + 2H2O (l)
In this reaction, hydrogen peroxide oxidises iodide ions (I–) to molecular iodine. If calculated amount of sodium thiosulphate is added in the presence of starch solution as an indicator to the above reaction mixture, the liberated iodine reacts with thiosulphate ions as fast as it is formed and is reduced back to iodide ions till all the hiosulphate ions are oxidised to tetrathionate ions.
I2(g) + 2S2O32– (aq) → S4O62–(aq) + 2I–(aq)
After the complete consumption of thiosulphate ions, the concentration of iodine liberated in the reaction of hydrogen peroxide with iodide ions increases rapidly to a point where iodine forms intense blue complex with starch. The time required to consume a fixed amount of the thiosulphate ions is reproducible. Since the time for the appearance of colour is noted, the reaction is some times called a clock reaction.
Material Required
Hazard Warning : Cotact of hydrogenperoxide with combustible material may cause fire.
Procedure
(i) Take 25 mL of 3% hydrogen peroxide, 25 mL of 2.5 M H2SO4 solution, 5 mL of freshly prepared starch solution and 195 mL distilled water into a 500 mL conical flask marked as A. Stir this solution well and place it in a water bath maintained at room temperature.
(ii) Take four 250 mL conical flasks and mark them as B, C, D and E.
(iii) Take the sodium thiosulphate solution, potassium iodide solution, and distilled water in the flasks B, C and D in a proportion given in the following steps and keep the flask E for carrying out the reaction.
(iv) Take 10 mL of 0.04 M sodium thiosulphate solution, 10 mL of 0.1 M potassium iodide solution and 80 mL of distilled water in the conical flask marked B. Shake the contents of the flask well and keep it in a water bath.
(v) Take 10 mL of 0.04 M sodium thiosulphate solution, 20 mL of 0.1M potassium iodide solution and 70 mL of distilled water in the conical flask marked C. Shake the resulting solution well and place it in the same water bath in which reaction mixture of step
(iv) is kept.
(vi) Take 10 mL of 0.04 M sodium thiosulphate solution, 30 mL of 0.1 M potassium iodide solution and 60 mL of distilled water in the conical flask marked D. Shake the solution well and keep this flask also in the above water bath.
(vii) Take conical flask E. Pour 25 mL solution from flask A into it after measuring it with the help of a measuring cylinder. Now add 25 mL of solution from flask B into this flask with constant stirring. Start the stop watch when half of the solution from flask B has been transferred. Keep the flask E in a water bath to maintain the constant temperature and record the time required for the appearance of blue colour.
(viii) In exactly the same manner, repeat the experiment with the solutions of flasks C and D separately by using once again 25 mL of the solution of these flasks and 25 mL of solution from flask A. Note the time required for the appearance of blue colour in each case.
(ix) Repeat the experiment with solutions of flasks B, C and D twice and calculate the average time for the appearance of blue colour.
(x) Record your observations as given in Table 2.3.
(xi) Compare the time required for the appearance of blue colour for all the three systems and make a generalisation about the variation in the rate of the reaction with concentration of iodide ions.
Table 2.3 : Study of reaction rate between iodide ions and hydrogen peroxide in acidic medium
Sl. No.Composition of the systemTime taken for appearance of the blue colourAverage Time
First readingSecond reading
1.25 mL solution from flask A + 25 mL solution from flask B
2.25 mL solution from flask A + 25 mL solution from flask C
3.25 mL solution from flask A + 25 mL solution from flask D
Result
Write your conclusions on the basis of the data recorded in Table 2.3.
Precautions
(a) Always keep the concentration of sodium thiosulphate solution less than that of potassium iodide solution.
(b) Always use freshly prepared starch solution.
(c) Use fresh samples of hydrogen peroxide and potassium iodide.
(d) Always use the same measuring cylinders for measuring solutions in two different sets of observations. If after measuring one solution, the cylinder is used for measuring another solution, clean it before using.
(e) Record the time immediately after the appearance of blue colour.
Discussion Questions
(i) Distinguish between the role of iodine and iodide ions in this experiment.
(ii) Calculate the oxidation number of sulphur in tetrathionate ion (S4O62– ). Can the oxidation number be a fractional number?
(iii) Why does iodine impart blue colour to starch?
(iv) Explore the possibility of using an oxidant other than H2O2 in this experiment.
(v) Why is the reaction given the name clock reaction?
(vi) Why should the concentration of sodium thiosulphate solution taken be always less than that of potassium iodide solution?
EXPERIMENT 2.3
Aim
To study the rate of reaction between potassium iodate (KIO3) and sodium sulphite (Na2SO3).TheoryThe reaction between KIO3 and Na2SO3 indirectly involves the formation of iodide ions, which are oxidised in acidic medium by IO3– ions to iodine. The overall reaction proceeds in the following two steps.
IO3– + 3SO32– → I– + 3SO42– (1)
5I– + 6H+ + IO3– → 3H2O + 3I2 (2)
The evolved iodine produces blue colour with the starch solution in a manner described in the previous experiment. This reaction like the earlier reaction is also known as ‘clock reaction’.
Material Required
Procedure
(i) Take a 250 mL conical flask and mark it as ‘A’. Transfer 25 mL of 6% potassium iodate solution, 25 mL of 2.0 M H2SO4 and 50 mL of distilled water into it and shake the content of the flask well. Keep the flask in a trough half filled with water. This serves as constant temperature bath.
(ii) Take five 250 mL conical flasks and mark these as B, C, D, E and F respectively. Take 6% sodium sulphite solution, starch solution and distilled water in flasks B, C, D and E in the proportion given in the following steps and keep flask F for carrying out the reaction.
(iii) In the conical flask marked ‘B’ take 20 mL of sodium sulphite solution, 5 mL of starch solution and 75 mL of distilled water. Shake the contents of the flask well and keep it in the water bath.
(iv) In the conical flask marked ‘C’, take 15 mL of sodium sulphite solution, 5 mL of starch solution and 80 mL of distilled water. Shake the resulting solution well and keep it in the water bath.
(v) In conical flask ‘D’, take 10 mL of sodium sulphite solution, 5 mL of starch solution and 85 mL of distilled water. Shake the solution well and place the flask in the water bath.
(vi) In conical flask ‘E’, take 5 mL of sodium sulphite solution, 5 mL of starch solution and 90 mL of distilled water. Shake the content of the flask well and keep it in the water bath.
(vii) Take conical flask ‘F’. In this flask pour 25 mL of the solution from the conical flask marked ‘A’ and add 25 mL of the solution from the conical flask marked ‘B’. Start the stop watch when half of the solution from flask B has been added. Mix these two solutions thoroughly by constant stirring and keep it in the water bath. Record the time required for the appearance of blue colour (you may use stop watch/wrist watch for noting the time).
(viii) In a similar manner, repeat the experiment with the solutions in flasks C, D and E respectively by using 25 mL of the solution as in the experiment with solution from flask B and record the time required for the appearance of blue colour in each case.(Once again care should be taken to repeat the experiment for each case twice so as to take the average time required for the appearance of blue colour in each set).
(ix) Record your observations as given in Table 2.4.
(x) From the tabulated results, find out the relationship between the time of appearance of blue colour and the variation in concentration of sodium sulphite.Note : • Total amount of solution in each flask is 100 mL • Same amount of indicator has been used.
Table 2.4 : Study of the reaction rate between potassium iodate (KIO3) and sodium sulphite (Na2SO3) in acidic medium
Sl.No. Composition of the system Time taken for appearance of the blue colour in seconds Average Times/sec. First reading Second reading 1. 25 mL solution from flask A + 25 mL solution from flask B 2. 25 mL solution from flask A + 25 mL solution from flask C 3. 25 mL solution from flask A + 25 mL solution from flask D 4. 25 mL solution from flask A + 25 mL solution from flask E
Result
Write your conclusions on the basis of data recorded in Table 2.4.
Precautions
(a) As sodium sulphite is likely to be easily oxidised in air, therefore, always use its fresh solution.
(b) Keep the concentration of potassium iodate solution higher than the concentration of sodium sulphite solution.
(c) Use a freshly prepared starch solution.
(d) Start the stop watch when half of the solution from conical flask B, C, D or E is added to the conical flask F containing 25 mL solution from flask A.
Discussion Questions
(i) How would the time for the appearance of blue colour vary if the temperature of the experiment in the above case is enhanced by 10°C ?
(ii) Mention the factors that affect the rate of reaction in the present study.
(iii) Which of the acids, hydrochloric or nitric, would be suitable to make the medium acidic in this experiment? Explain your answer with reasons.
(iv) Out of the reactions (1) and (2) given below:
IO3– + 3SO32– → I– + 3SO42– (1)
5I– + IO3– + 6H+ → 3H2O + 3I2 (2)
which could be the rate determining reaction? What is the molecularity of the rate determining reaction?
(v) Can AsO33– be used in place of SO32– in the above reaction? Support your answer with proper reasoning.
(vi) Why is the concentration of potassium iodate solution kept higher than the concentration of sodium sulphite solution?

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