Chapter 4 : Chemical Kinetics

Chemical kinetics is that branch of chemistry which deals with the study of the rate of chemical reaction. In this chapter, you will be dealing with the average and instantaneous rate of reaction, the factors such as temperature, pressure, concentration and catalyst which affect the rate of reactions. Differences between elementary and complex reactions are also given. Differences between molecularity and order of a reaction are also given. Description of collision theory and derivation of integrated rate of equations for the zero and first order reactions along- with determination of its rate constants.

Exercise 1 ( Page No. : 116 )

For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

Exercise 1 ( Page No. : 116 )

In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L^-1 to 0.4 mol L^-1in 10 minutes. Calculate the rate during this interval?

Exercise 1 ( Page No. : 116 )

For a reaction, A + B → Product; the rate law is given by, r = k [A]^½ [B]². What is the order of the reaction?

The order of the reaction = 1/2 + 2

= 2.5

Exercise 1 ( Page No. : 116 )

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?

The order of reaction is defined as the sum of the powers of concentrations In the rate law.

The rate of second order reaction can be expressed as rate = k [A]²

The reaction X → Y follows second order kinetics.

Therefore, the rate equation for this reaction will be:

Rate = k [X]^{2 ________________________}(1)

Let [X] = a mol L^-1, then equation (1) can be written as:

Rate = k [a]²

= ka²

If the concentration of X is increased to three times, then [X] = 3a mol L^-1

Now, the rate equation will be:

Rate = k [3a]²

= 9 (ka²)

Hence, the rate of formation will increase by 9 times.

Exercise 1 ( Page No. : 116 )

A first order reaction has a rate constant 1.15 10^-3s^-1. How long will 5 g of this reactant take to reduce to 3 g?

From the question, we can write down the following information:

Initial amount = 5 g

Final concentration = 3 g

Rate constant = 1.15 10 ^{- 3}s^{- 1}

We know that for a 1^storder reaction,

= 444.38 s

= 444 s (approx)

Exercise 1 ( Page No. : 116 )

Time required to decompose SO₂Cl₂to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

We know that for a 1^storder reaction,

t½ = 0.693 / k

It is given that t_1/2= 60 min

k = 0.693 / t½

= 0.693 / 60

= 0.01155 min^-1

= 1.155 min^-1

k = 1.925 x 10^-2 s^-1

Exercise 1 ( Page No. : 116 )

What will be the effect of temperature on rate constant?

The rate constant of a reaction is nearly doubled with a 10° rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by Arrhenius equation,

k = Ae^{- E_a / RT}

Where,

A is the Arrhenius factor or the frequency factor

T is the temperature

R is the gas constant

E_ais the activation energy

Exercise 1 ( Page No. : 116 )

The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate E_a.

It is given that T₁ = 298 K

∴T2 = (298 + 10) K

= 308 K

We also know that the rate of the reaction doubles when temperature is increased by 10°.

Therefore, let us take the value of k₁ = k and that of k₂ = 2k

Also, R = 8.314 J K ^{- 1} mol ^{- 1}

Now, substituting these values in the equation:

= 52897.78 J mol^{- 1}

= 52.9 kJ mol^{- 1}

Exercise 1 ( Page No. : 116 )

The activation energy for the reaction 2HI_(g) → H₂ + I_2(g) is 209.5 kJ mol^-1 at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

In the given case:

E_a = 209.5 kJ mol^{- 1} = 209500 J mol^{- 1}

T = 581 K

R = 8.314 JK^{- 1} mol^{- 1}

Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as

Exercise 2 ( Page No. : 120 )

From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

(i) 3 NO(g) → N₂O(g) Rate = k[NO]²

(ii) H₂O₂ (aq) + 3 I^- (aq) + 2 H⁺→ 2 H₂O (l) + I₃^- 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]

(i) Given rate = k [NO]²

Therefore, order of the reaction = 2

Dimension of k = Rate / [NO]²

= mol L^-1 s^-1 / (mol L^-1)²

= mol L^-1 s^-1 / mol² L^-2

= L mol^-1 s^-1

(ii) Given rate = k [H₂O₂][I^-]

Therefore, order of the reaction = 2

Dimension of k = Rate / [H₂O₂][I^-]

= mol L^-1 s^-1/ (mol L^-1) (mol L^-1)

= L mol^-1 s^-1

(iii) Given rate =k [CH₃CHO]^3/2

Therefore, order of reaction = 3/2

Dimension of k = Rate / [CH₃CHO]^3/2

= mol L^-1 s^-1/ (mol L^-1)^3/2

= mol L^-1 s^-1/ mol^3/2 L^-3/2

= L^½ mol^-½s^-1

(iv) Given rate = k [C₂H₅Cl]

Therefore,order of the reaction = 1

Dimension of k = Rate / [C₂H₅Cl]

= mol L^-1 s^-1/ mol L^-1

= s^-1

Exercise 2 ( Page No. : 120 )

For the reaction:

2A + B → A₂B

the rate = k[A][B]²with k= 2.0 x 10^-6mol^-2L²s^-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.

The initial rate of the reaction is

Rate = k [A][B]²

= (2.0 × 10^{- 6}mol^{- 2}L²s^{- 1}) (0.1 mol L^{- 1}) (0.2 mol L^{- 1})²

= 8.0 × 10^{- 9}mol^{- 2}L²s^{- 1}

When [A] is reduced from 0.1 mol L^{- 1}to 0.06 mol^{- 1}, the concentration of A reacted = (0.1 - 0.06) mol L^{- 1} = 0.04 mol L^{- 1}

Therefore, concentration of B reacted= 1/2 x 0.04 mol L^-1 = 0.02 mol L^{- 1}

Then, concentration of B available, [B] = (0.2 - 0.02) mol L^{- 1}

= 0.18 mol L^{- 1}

After [A] is reduced to 0.06 mol L^{- 1}, the rate of the reaction is given by,

Rate = k [A][B]²

= (2.0 × 10^{- 6}mol^{- 2}L²s^{- 1}) (0.06 mol L^{- 1}) (0.18 mol L^{- 1})²

= 3.89 mol L^{- 1}s^{- 1}

Exercise 2 ( Page No. : 120 )

The decomposition of NH₃on platinum surface is zero order reaction. What are the rates of production of N₂and H₂if k = 2.5 x 10^-4mol^-1L s^-1?

The decomposition of NH₃on platinum surface is represented by the following equation.

= 7.5 × 10^{- 4}mol L^{- 1}s^{- 1}

Exercise 2 ( Page No. : 120 )

The decomposition of dimethyl ether leads to the formation of CH₄, H₂and 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 (P_CH3OCH3)^3/2

If the pressure is measured in bar andtime in minutes, then what are the units of rate and rate constants?

If pressure is measured in bar and time in minutes, then

Unit of rate = bar min^{- 1}

Rate = k (P_CH3OCH3)^3/2

⇒ k = Rate / (P_CH3OCH3)^3/2

Therefore, unit of rate constants (k) = bar min^-1/ bar^3/2

= bar^-½ min^{- 1}

Exercise 2 ( Page No. : 120 )

Mention the factors that affect the rate of a chemical reaction.

The factors that affect the rate of a reaction areas follows.

(i) Concentration of reactants (pressure in case of gases)

(ii) Temperature

(iii) Presence of a catalyst

Exercise 2 ( Page No. : 120 )

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?

Let the concentration of the reactant be [A] = a

Rate of reaction, R = k [A]² = ka²

(i) If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be

R = k(2a)²

= 4ka²

= 4 R

Therefore, the rate of the reaction would increase by 4 times.

(ii) If the concentration of the reactant is reduced to half, i.e. [A] = 1/2 a, then the rate of the reaction would be

R = k(1/2a)²

= 1/4 Ka²

= 1/4 R

Exercise 2 ( Page No. : 120 )

What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?

The rate constant is nearly doubled with a rise in temperature by 10° for a chemical reaction.

The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation,

K = Ae^{-E_a/ RT}

where, kis the rate constant,

A is the Arrhenius factor or the frequency factor,

R is the gas constant,

T is the temperature, and

E_a is the energy of activation for the reaction

Exercise 2 ( Page No. : 120 )

In a pseudo first order hydrolysis of ester in water, the following results were obtained:

t/s	0	30	60	90
[Ester]mol L^{- 1}	0.55	0.31	0.17	0.085

(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.

(i) Average rate of reaction between the time interval, 30 to 60 seconds, = d[ester] / dt

= (0.31-0.17) / (60-30)

= 0.14 / 30

= 4.67 × 10 - 3mol L^{- 1}s ^{- 1}

(ii) For a pseudo first order reaction,

k = 2.303/ t log [R]_º / [R]

For t= 30 s, k₁ = 2.303/ 30 log 0.55 / 0.31

= 1.911 × 10 ^{- 2}s^{- 1}

For t = 60 s, k₂ = 2.303/ 60 log 0.55 / 0.17

= 1.957 × 10^{- 2}s ^{- 1}

For t= 90 s, k₃ = 2.303/ 90 log 0.55 / 0.085

= 2.075 × 10 ^{- 2}s ^{- 1}

Then, average rate constant, k = k₁ + k₂+ k₃/ 3

= 1.911 × 10 ^{- 2}+ 1.957 × 10^{- 2} + 2.075 × 10 ^{- 2} / 3

= 1.981 x 10^-2s ^{- 1}

Exercise 2 ( Page No. : 120 )

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?

Exercise 2 ( Page No. : 120 )

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^{- 1}	0.20	0.20	0.40
B/ mol L^{- 1}	0.30	0.10	0.05
r₀/ mol L^{- 1} s^{- 1}	5.07 × 10^{- 5}	5.07 × 10^{- 5}	1.43 × 10^{- 4}

What is the order of the reaction with respect to A and B?

Let the order of the reaction with respect to A be xand with respect to B be y.

Therefore,

= 1.496

= 1.5 (approximately)

Hence, the order of the reaction with respect to A is 1.5 and with respect to B is zero.

Exercise 2 ( Page No. : 120 )

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^{- 1}	Initial rate of formation of D/mol L^{- 1} min^{- 1}
I	0.1	0.1	6.0 × 10^{- 3}
II	0.3	0.2	7.2 × 10^{- 2}
III	0.3	0.4	2.88 × 10^{- 1}
IV	0.4	0.1	2.40 × 10^{- 2}

Determine the rate law and the rate constant for the reaction.

Exercise 2 ( Page No. : 120 )

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^{- 1}	Initial rate/mol L^{- 1} min^{- 1}
I	0.1	0.1	2.0 × 10^{- 2}
II	--	0.2	4.0 × 10^{- 2}
III	0.4	0.4	--
IV	--	0.2	2.0 × 10^{- 2}

The given reaction is of the first order with respect to A and of zero order with respect to B.

Therefore, the rate of the reaction is given by,

Rate = k [A]¹[B]⁰

⇒ Rate = k [A]

From experiment I, we obtain

2.0 x 10^-2mol L^-1min^-1= k (0.1 mol L^-1)

⇒ k = 0.2 min^-1

From experiment II, we obtain

4.0 x 10^-2mol L^-1min^-1= 0.2 min^-1[A]

⇒ [A] = 0.2 mol L^-1

From experiment III, we obtain

Rate = 0.2 min^-1 x 0.4 mol L^-1

= 0.08 mol L^-1min^-1

From experiment IV, we obtain

2.0 x 10^-2mol L^-1min^-1= 0.2 min^-1[A]

⇒ [A] = 0.1 mol L^-1

Exercise 2 ( Page No. : 120 )

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

(i) Half life, t_½ = 0.693 / k

= 0.693 / 200 s^-1

= 3.47 ×10 ^-3 s (approximately)

(ii) Half life, t_½ = 0.693 / k

= 0.693 / 2 min^-1

= 0.35 min (approximately)

(iii) Half life,t_½ = 0.693 / k

= 0.693 / 4 years^-1

= 0.173 years (approximately)

Exercise 2 ( Page No. : 120 )

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.

Here,

K = 0.693 / t_½

= 0.693 / 5730 years^-1

It is known that,

= 1845 years (approximately)

Hence, the age of the sample is 1845 years.

Exercise 2 ( Page No. : 120 )

The experimental data for decomposition of N₂O₅

[2N₂O₅ → 4NO₂ + O₂]

in gas phase at 318K are given below:

t/s	0	400	800	1200	1600	2000	2400	2800	3200
10² × [N₂O₅] mol L^-1	1.63	1.36	1.14	0.93	0.78	0.64	0.53	0.43	0.35

(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).

(i)

(ii) Time corresponding to the concentration, 1630x10² / 2 mol L^-1 = 81.5 mol L^-1 is the half life. From the graph, the half life is obtained as 1450 s.

(iii)

t/s	10² × [N₂O₅] mol L^-1	Log [N₂O₅]
0	1.63	-1.79
400	1.36	-1.87
800	1.14	-1.94
1200	0.93	-2.03
1600	0.78	-2.11
2000	0.64	-2.19
2400	0.53	-2.28
2800	0.43	-2.37
3200	0.35	-2.46

(iv) The given reaction is of the first order as the plot, log[N₂O₅] v/s t, is a straight line. Therefore, the rate law of the reaction is

Rate = k [N₂O₅]

(v) From the plot, log[N₂O₅]

v/s t, we obtain

Slope = -2.46 -(1.79) / 3200-0

= -0.67 / 3200

Again, slope of the line of the plot log[N₂O₅] v/s t is given by

- k / 2.303

.Therefore, we obtain,

- k / 2.303 = - 0.67 / 3200

⇒ k = 4.82 x 10^-4 s^-1

(vi) half life is given by,

t_½ = 0.693 / k

= 0.639 / 4.82x10^-4s

=1.438 x 10³s

Or we can say

1438 S

Which is very near to what we obtain from graph.

Exercise 2 ( Page No. : 120 )

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 1/16^th value?

Exercise 2 ( Page No. : 120 )

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.

Therefore, 0.7814 μg of ⁹⁰Sr will remain after 10 years.

Again,

Therefore, 0.2278 μg of ⁹⁰Sr will remain after 60 Years

Exercise 2 ( Page No. : 120 )

For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

For a first order reaction, the time required for 99% completionis

t₁= 2.303/k Log 100/100-99

= 2.303/k Log 100

= 2x 2.303/k

For a first order reaction, the time required for 90% completion is

t₂= 2.303/k Log 100/100-90

= 2.303/k Log 10

= 2.303/k

Therefore, t₁= 2t₂

Hence, the time required for 99% completion of a first order reaction is twice the time required for the completion of 90% of the reaction.

Exercise 2 ( Page No. : 120 )

A first order reaction takes 40 min for 30% decomposition. Calculate t_1/2.

For a first order reaction,

t= 2.303/k Log [R] _º/ [R]

k = 2.303/40min Log 100/ 100-30

= 2.303/40min Log 10/ 7

= 8.918 x 10^-3 min^-1

Therefore, t_1/2 of the decomposition reaction is

t_1/2= 0.693/k

= 0.693 / 8.918 x 10^-3 min

= 77.7 min (approximately)

Exercise 2 ( Page No. : 120 )

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

The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.

After time, t, total pressure, P_t = (P_º - p) + p + p

⇒ P_t = (P_º + p)

⇒ p = P_t - P_º

therefore, P_º - p = P_º- P_t - P_º

= 2 P_º- P_t

For a first order reaction,

k = 2.303/t Log P_º/ P_º- p

= 2.303/t Log P_º/ 2 P_º- P_t

When t = 360 s, k = 2.303 / 360s log 35.0 / 2x35.0 - 54.0

= 2.175 × 10 ^{- 3} s ^{- 1}

When t = 720 s, k = 2.303 / 720s log 35.0 / 2x35.0 - 63.0

= 2.235 × 10 ^{- 3} s ^{- 1}

Hence, the average value of rate constant is

k = (2.175 × 10 ^{- 3}+ 2.235 × 10 ^{- 3} ) / 2 s ^{- 1}

= 2.21 × 10 ^{- 3}s^{- 1}

Exercise 2 ( Page No. : 120 )

The following data were obtained during the first order thermal decomposition of SO₂Cl₂at a constant volume.

SO₂Cl₂₍g) → SO₂₍g) + Cl₂(g)

Experiment	Time/s^{- 1}	Total pressure/atm
1	0	0.5
2	100	0.6

Calculate the rate of the reaction when total pressure is 0.65 atm.

The thermal decomposition of SO₂Cl₂at a constant volume is represented by the following equation.

After time, t, total pressure,P_t = (P_º - p) + p + p

⇒ P_t = (P_º + p)

⇒ p = P_t - P_º

therefore, P_º - p = P_º- P_t - P_º

= 2 P_º- P_t

For a first order reaction,

k = 2.303/t Log P_º/ P_º- p

= 2.303/t Log P_º/ 2 P_º- P_t

When t= 100 s,

k = 2.303 / 100s log 0.5 / 2x0.5 - 0.6

= 2.231 × 10^{- 3}s^{- 1}

When Pt= 0.65 atm,

P₀+ p= 0.65

⇒ p= 0.65 - P₀

= 0.65 - 0.5

= 0.15 atm

Therefore, when the total pressure is 0.65 atm, pressure of SOCl₂ is

p_SOCL2 = P₀ - p

= 0.5 - 0.15

= 0.35 atm

Therefore, the rate of equation, when total pressure is 0.65 atm, is given by,

Rate = k(p_SOCL2)

= (2.23 × 10 ^{- 3}s ^{- 1}) (0.35 atm)

= 7.8 × 10 ^{- 4}atm s^{- 1}

Exercise 2 ( Page No. : 120 )

The rate constant for the decomposition of N₂O₅ at various temperatures is given below:

T/°C	0	20	40	60	80
10⁵ X 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 E_a.

Predict the rate constant at 30 º and 50 ºC.

From the given data, we obtain

T/°C	0	20	40	60	80
T/K	273	293	313	333	353
1/T / k^-1	3.66×10^{- 3}	3.41×10^{- 3}	3.19×10^{- 3}	3.0×10^{- 3}	2.83 ×10^{- 3}
10⁵ X K /S^-1	0.0787	1.70	25.7	178	2140
In k	-7.147	-4.075	-1.359	-0.577	3.063

Slope of the line,

In k= - 2.8

Therefore, k = 6.08x10^-2s^-1

Again when T = 50 + 273K = 323K,

1/T = 3.1 x 10^-3 K

In k = - 0.5

Therefore, k = 0.607 s^-1

Exercise 2 ( Page No. : 120 )

The rate constant for the decomposition of hydrocarbons is 2.418 x 10^-5 s^-1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

k= 2.418 × 10^-5 s^-1

T= 546 K

E_a= 179.9 kJ mol^{- 1} = 179.9 × 10³J mol^{- 1}

According to the Arrhenius equation,

= (0.3835 - 5) + 17.2082

= 12.5917

Therefore, A = antilog (12.5917)

= 3.9 × 10¹² s^{- 1}(approximately)

Exercise 2 ( Page No. : 120 )

Consider a certain reaction A → Products with k = 2.0 x 10^-2 s^-1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L^-1.

k= 2.0 × 110^-2 s^-1

T= 100 s

[A]_o= 1.0 moL ^{- 1}

Since the unit of kis s^{- 1}, the given reaction is a first order reaction.

Therefore, k = 2.303/t Log [A]_º/ [A]

⇒2.0 × 110^-2 s^-1= 2.303/100s Log 1.0/ [A]

⇒2.0 × 110^-2 s^-1= 2.303/100s ( - Log [A] )

⇒ - Log [A] = - (2.0 x 10^-2 x 100) / 2.303

⇒ [A] = antilog [- (2.0 x 10^-2 x 100) / 2.303]

= 0.135 mol L^{- 1}(approximately)

Hence, the remaining concentration of A is 0.135 mol L ^{- 1}.

Exercise 2 ( Page No. : 120 )

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t_1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?

For a first order reaction,

k = 2.303/t Log [R]_º/ [R]

It is given that, t_1/2= 3.00 hours

Therefore, k = 0.693 / t_1/2

= 0.693 / 3 h^-1

= 0.231 h^{- 1}

Then, 0.231 h^{- 1}= 2.303 / 8h Log [R]_º/ [R]

Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.

Exercise 2 ( Page No. : 120 )

The decomposition of hydrocarbon follows the equation

k = (4.5 x 10¹¹ s^-1) e^{-28000 K/T}

Calculate E_a.

The given equation is

k = (4.5 x 10¹¹ s^-1) e^{-28000 K/T} (i)

Arrhenius equation is given by,

k= Ae^-E_a/RT(ii)

From equation (i) and (ii), we obtain

E_a/ RT = 28000K / T

⇒ E_a= R x 28000K

= 8.314 J K^{- 1}mol^{- 1}× 28000 K

= 232792 J mol ^{- 1}

= 232.792 kJ mol^{- 1}

Exercise 2 ( Page No. : 120 )

The rate constant for the first order decomposition of H₂O₂is given by the following equation:

log k = 14.34 - 1.25 x 10⁴K/T

Calculate E_afor this reaction and at what temperature will its half-period be 256 minutes?

Arrhenius equation is given by,

k= Ae ^-E_a/RT

⇒In k = In A - E_a/RT

⇒In k = Log A - E_a/RT

⇒ Log k = Log A - E_a/2.303RT (i)

The given equation is

Log k = 14.34 - 1.25 10⁴ K/T (ii)

From equation (i) and (ii), we obtain

E_a/2.303RT = 1.25 10⁴ K/T

⇒ E_a=1.25 × 10⁴K × 2.303 × R

= 1.25 × 10⁴K × 2.303 × 8.314 J K^{- 1}mol^{- 1}

= 239339.3 J mol^{- 1} (approximately)

= 239.34 kJ mol - 1

Also, when t_1/2= 256 minutes,

k = 0.693 / t_1/2

= 0.693 / 256

= 2.707 × 10^{- 3}min ^{- 1}

= 4.51 × 10^{- 5}s^{- 1}

It is also given that, log k= 14.34 - 1.25 × 10⁴K/T

= 668.95 K

= 669 K (approximately)

Exercise 2 ( Page No. : 120 )

The decomposition of A into product has value of k as 4.5 x 10³ s^-1 at 10°C and energy of activation 60 kJ mol^-1. At what temperature would k be 1.5 x 10⁴ s^-1?

From Arrhenius equation, we obtain

log k₂/k₁= E_a / 2.303 R (T₂ - T₁) / T₁T₂

Also, k₁ = 4.5 × 10³ s^{- 1}

T₁ = 273 + 10 = 283 K

k₂ = 1.5 × 10⁴ s^{- 1}

E_a = 60 kJ mol ^{- 1} = 6.0 × 10⁴ J mol ^{- 1}

Then,

= 297 K = 24°C

Hence, k would be 1.5 × 10⁴ s^{- 1} at 24°C.

Exercise 2 ( Page No. : 120 )

The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 x 10¹⁰ s^-1. Calculate k at 318 K and E_a.

For a first order reaction,

t = 2.303 / k log a / a - x

At 298 K,

t = 2.303 / k log 100 / 90

= 0.1054 / k

At 308 K,

t' = 2.303 / k' log 100 / 75

= 2.2877 / k'

According to the question,

t = t'

⇒ 0.1054 / k = 2.2877 / k'

⇒ k' / k = 2.7296

From Arrhenius equation,we obtain

To calculate k at 318 K,

It is given that, A = 4 x 10¹⁰ s^-1, T = 318K

Again, from Arrhenius equation, we obtain

Therefore, k = Antilog (-1.9855)

= 1.034 x 10^-2 s ^-1

Exercise 2 ( Page No. : 120 )

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.

From Arrhenius equation, we obtain

Hence, the required energy of activation is 52.86 kJ mol^{- 1}.

Chapter 4 : Chemical Kinetics

Exercise 1 ( Page No. : 116 )

Exercise 1 ( Page No. : 116 )

Exercise 1 ( Page No. : 116 )

Exercise 1 ( Page No. : 116 )

Exercise 1 ( Page No. : 116 )

Exercise 1 ( Page No. : 116 )

Exercise 1 ( Page No. : 116 )

Exercise 1 ( Page No. : 116 )

Exercise 1 ( Page No. : 116 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

Exercise 2 ( Page No. : 120 )

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