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NCERT Solutions · Class 12 Chemistry · NCERT Chemistry Part I and II Part I · Chapter 3

Chapter 3: Chemical Kinetics (Physical Chemistry)

Step-by-step solutions to all 9 intext questions and 30 exercises of Chapter 3, Chemical Kinetics (NCERT Class 12 Chemistry Part I, 2026-27): average rate, rate law and order, units of rate constants, initial rate method, integrated rate equations for zero and first order reactions, half-life, radioactive decay and carbon dating, pressure data, and the Arrhenius equation with activation energy. All 39 questions are answered, with the key answer highlighted.

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First order: $k = \frac{2.303}{t} lo g \frac{[ R ] _{0}}{[ R ]}$ , $t_{1/2} = \frac{0.693}{k}$ . Zero order: $[R] = [R]_{0} - k t$ , rate $= k$ . Units of $k$ for order $n$ : (mol L⁻¹) $^{1 - n}$ s⁻¹. Gas-phase first order (A → B + C): $p_{A} = 2 p_{0} - p_{t}$ . Arrhenius: $k = A e^{- E_{a} / R T}$ , $lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})$ , $R = 8.314$ J K⁻¹ mol⁻¹.

Intext questions

3.1

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.

Solution

$r_{av} = - \frac{Δ [ R ]}{Δ t} = - \frac{0.02 - 0.03}{25} = 4 \times 1 0^{- 4} M min^{- 1}$

In seconds: $\frac{4 \times 1 0 ^{- 4}}{60} \approx 6.67 \times 1 0^{- 6}$ M s⁻¹.

$4 \times 1 0^{- 4}$ M min⁻¹ = $6.67 \times 1 0^{- 6}$ M s⁻¹.

3.2

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

Solution

$Rate = - \frac{1}{2} \frac{Δ [ A ]}{Δ t} = - \frac{1}{2} \times \frac{0.4 - 0.5}{10} = 0.005 mol L^{- 1} min^{- 1}$

(The rate of disappearance of A itself is 0.01 mol L⁻¹ min⁻¹.)

0.005 mol L⁻¹ min⁻¹.

3.3

For a reaction, A + B → Product; the rate law is given by,

r = k [A]^{1/2} [B]^{2}

. What is the order of the reaction?

Solution

Order $= \frac{1}{2} + 2 = 2.5$ .

2.5

3.4

The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?

Solution

Rate $= k [X]^{2}$ . With $[X]$ tripled, rate $= k (3 [X])^{2} = 9 k [X]^{2}$ : it becomes 9 times.

The rate increases 9 times.

3.5

A first order reaction has a rate constant

1.15 \times 1 0^{- 3}

s⁻¹. How long will 5 g of this reactant take to reduce to 3 g?

Solution

$t = \frac{2.303}{k} lo g \frac{[ R ] _{0}}{[ R ]} = \frac{2.303}{1.15 \times 1 0 ^{- 3}} lo g \frac{5}{3} = 2003 \times 0.2218 \approx 444 s$

About 444 s.

3.6

Time required to decompose

SO X_{2} Cl X_{2}

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

Solution

$k = \frac{0.693}{t _{1/2}} = \frac{0.693}{60 min} = 0.01155 min^{- 1} = 1.925 \times 1 0^{- 4} s^{- 1}$

$k = 1.925 \times 1 0^{- 4}$ s⁻¹.

3.7

What will be the effect of temperature on rate constant?

Solution

The rate constant increases with temperature; for many reactions it nearly doubles for a 10 K rise. Higher temperature gives more molecules energy above the activation energy. Quantitatively, by the Arrhenius equation $k = A e^{- E_{a} / R T}$ .

k increases with temperature (roughly doubles per 10 K), following $k = A e^{- E_{a} / R T}$ .

3.8

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

E_{a}

Solution

$lo g 2 = \frac{E _{a}}{2.303 \times 8.314} (\frac{308 - 298}{298 \times 308})$

$E_{a} = \frac{0.3010 \times 2.303 \times 8.314 \times 298 \times 308}{10} \approx 52.9 kJ mol^{- 1}$

About 52.9 kJ mol⁻¹.

3.9

The activation energy for the reaction

2 HI (g) H X_{2} + I X_{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.

Solution

$x = e^{- E_{a} / R T} = exp (- \frac{209500}{8.314 \times 581}) = e^{- 43.37} \approx 1.47 \times 1 0^{- 19}$

About $1.47 \times 1 0^{- 19}$ .

Exercises

3.1

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

3 NO (g) N X_{2} O (g)

, Rate

= k [NO]^{2}

(ii)

H X_{2} O X_{2} (aq) + 3 I X^{-} (aq) + 2 H X^{+} 2 H X_{2} O (l) + I X_{3} X^{-}

, Rate

= k [H X_{2} O X_{2}] [I X^{-}]

(iii)

CH X_{3} CHO (g) CH X_{4} (g) + CO (g)

, Rate

= k [CH X_{3} CHO]^{3/2}

(iv)

C X_{2} H X_{5} Cl (g) C X_{2} H X_{4} (g) + HCl (g)

, Rate

= k [C X_{2} H X_{5} Cl]

Solution

Units of $k$ = (mol L⁻¹) $^{1 - n}$ s⁻¹:

(i) order 2; $k$ in L mol⁻¹ s⁻¹
(ii) order 2; $k$ in L mol⁻¹ s⁻¹
(iii) order 3/2; $k$ in L $^{1/2}$ mol $^{- 1/2}$ s⁻¹
(iv) order 1; $k$ in s⁻¹

(i) 2, L mol⁻¹ s⁻¹; (ii) 2, L mol⁻¹ s⁻¹; (iii) 3/2, L^½ mol^−½ s⁻¹; (iv) 1, s⁻¹.

3.2

For the reaction: 2A + B → A₂B, the rate =

k [A] [B]^{2}

with

k = 2.0 \times 1 0^{- 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.

Solution

Initial: rate $= 2.0 \times 1 0^{- 6} \times 0.1 \times (0.2)^{2} = 8.0 \times 1 0^{- 9}$ mol L⁻¹ s⁻¹.

Later: A used up $= 0.1 - 0.06 = 0.04$ mol L⁻¹, so B used up $= 0.02$ mol L⁻¹ (2A react with 1B) and $[B] = 0.18$ mol L⁻¹.

Rate $= 2.0 \times 1 0^{- 6} \times 0.06 \times (0.18)^{2} \approx 3.89 \times 1 0^{- 9}$ mol L⁻¹ s⁻¹.

Initial $8.0 \times 1 0^{- 9}$ ; later $3.89 \times 1 0^{- 9}$ mol L⁻¹ s⁻¹.

3.3

The decomposition of

NH X_{3}

on platinum surface is zero order reaction. What are the rates of production of

N X_{2}

and

H X_{2}

k = 2.5 \times 1 0^{- 4}

mol L⁻¹ s⁻¹?

Solution

$2 NH X_{3} N X_{2} + 3 H X_{2}$ ; for zero order, rate $= k = 2.5 \times 1 0^{- 4}$ mol L⁻¹ s⁻¹, where rate $= \frac{d [ N X _{2} ]}{d t} = \frac{1}{3} \frac{d [ H X _{2} ]}{d t}$ .

$\frac{d [ N X _{2} ]}{d t} = 2.5 \times 1 0^{- 4}$ mol L⁻¹ s⁻¹
$\frac{d [ H X _{2} ]}{d t} = 3 \times 2.5 \times 1 0^{- 4} = 7.5 \times 1 0^{- 4}$ mol L⁻¹ s⁻¹

(For a zero order reaction the unit of $k$ is mol L⁻¹ s⁻¹; the book's "mol⁻¹ L s⁻¹" is a misprint.)

N₂: $2.5 \times 1 0^{- 4}$ and H₂: $7.5 \times 1 0^{- 4}$ mol L⁻¹ s⁻¹.

3.4

The decomposition of dimethyl ether leads to the formation of

CH X_{4}

H X_{2}

and CO and the reaction rate is given by Rate

= k [CH X_{3} OCH X_{3}]^{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_{CH X_{3} OCH X_{3}})^{3/2}

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

Solution

Rate: bar min⁻¹
$k = \frac{rate}{p ^{3/2}} = \frac{bar min ^{- 1}}{bar ^{3/2}}$ = bar $^{- 1/2}$ min⁻¹

Rate in bar min⁻¹; k in bar^−½ min⁻¹.

3.5

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

Solution

Concentration of reactants (pressure, for gases)
Temperature
Presence of a catalyst
Nature of the reactants
Surface area of solid reactants
Exposure to radiation (light), for photochemical reactions

Concentration (pressure), temperature, catalyst, nature of reactants, surface area, and radiation.

3.6

A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half?

Solution

Rate $= k [A]^{2}$ .

(i) $(2)^{2} = 4$ : the rate becomes 4 times.
(ii) $(\frac{1}{2})^{2} = \frac{1}{4}$ : the rate becomes one-fourth.

(i) 4 times. (ii) One-fourth.

3.7

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

Solution

The rate constant increases with temperature; for a 10 K rise it roughly doubles. This is described by the Arrhenius equation:

$k = A e^{- E_{a} / R T}$

where $A$ is the frequency factor and $E_{a}$ the activation energy. In log form, $lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})$ . A plot of $ln k$ against $1/ T$ is a straight line of slope $- \frac{E _{a}}{R}$ .

k increases with T, as given by $k = A e^{- E_{a} / R T}$ .

3.8

In a pseudo first order reaction in water, the following results were obtained: t/s: 0, 30, 60, 90; [A]/mol L⁻¹: 0.55, 0.31, 0.17, 0.085. Calculate the average rate of reaction between the time interval 30 to 60 seconds.

Solution

$r_{av} = - \frac{0.17 - 0.31}{60 - 30} = \frac{0.14}{30} \approx 4.67 \times 1 0^{- 3} mol L^{- 1} s^{- 1}$

About $4.67 \times 1 0^{- 3}$ mol L⁻¹ s⁻¹.

3.9

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?

Solution

(i) $- \frac{d [ R ]}{d t} = k [A] [B]^{2}$

(ii) $3^{2} = 9$ : the rate becomes 9 times.

(iii) $2 \times 2^{2} = 8$ : the rate becomes 8 times.

(i) Rate = k[A][B]². (ii) 9 times. (iii) 8 times.

3.10

In a reaction between A and B, the initial rate of reaction (

r_{0}

) 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;

r_{0}

/mol L⁻¹ s⁻¹:

5.07 \times 1 0^{- 5}

5.07 \times 1 0^{- 5}

1.43 \times 1 0^{- 4}

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

Solution

Let rate $= k [A]^{x} [B]^{y}$ .

Experiments 1 and 2 (same A, B changed from 0.30 to 0.10, rate unchanged): $y = 0$ .

Experiments 2 and 3 (B has no effect, A doubled): $\frac{1.43 \times 1 0 ^{- 4}}{5.07 \times 1 0 ^{- 5}} = 2.82 = 2^{x}$ , so $x = \frac{lo g 2.82}{lo g 2} \approx 1.5$ .

Order 1.5 in A and zero in B.

3.11

The following results have been obtained during the kinetic studies of the reaction: 2A + B → C + D. Experiment I: [A] = 0.1, [B] = 0.1, initial rate of formation of D =

6.0 \times 1 0^{- 3}

mol L⁻¹ min⁻¹; II: 0.3, 0.2,

7.2 \times 1 0^{- 2}

; III: 0.3, 0.4,

2.88 \times 1 0^{- 1}

; IV: 0.4, 0.1,

2.40 \times 1 0^{- 2}

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

Solution

Let rate $= k [A]^{x} [B]^{y}$ .

I and IV (B fixed, A ×4): rate × $\frac{2.40 \times 1 0 ^{- 2}}{6.0 \times 1 0 ^{- 3}} = 4$ , so $x = 1$ .
II and III (A fixed, B ×2): rate × $\frac{0.288}{0.072} = 4$ , so $y = 2$ .

Rate $= k [A] [B]^{2}$ . From I: $k = \frac{6.0 \times 1 0 ^{- 3}}{0.1 \times ( 0.1 ) ^{2}} = 6.0$ L² mol⁻² min⁻¹.

Rate = k[A][B]², k = 6.0 L² mol⁻² min⁻¹.

3.12

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 I: [A] 0.1, [B] 0.1, rate

2.0 \times 1 0^{- 2}

; II: [A] –, [B] 0.2, rate

4.0 \times 1 0^{- 2}

; III: [A] 0.4, [B] 0.4, rate –; IV: [A] –, [B] 0.2, rate

2.0 \times 1 0^{- 2}

(rates in mol L⁻¹ min⁻¹).

Solution

Rate $= k [A]$ . From I: $k = \frac{2.0 \times 1 0 ^{- 2}}{0.1} = 0.2$ min⁻¹.

II: $[A] = \frac{4.0 \times 1 0 ^{- 2}}{0.2} = 0.2$ mol L⁻¹
III: rate $= 0.2 \times 0.4 = 8.0 \times 1 0^{- 2}$ mol L⁻¹ min⁻¹
IV: $[A] = \frac{2.0 \times 1 0 ^{- 2}}{0.2} = 0.1$ mol L⁻¹

II: [A] = 0.2 M; III: rate = $8.0 \times 1 0^{- 2}$ mol L⁻¹ min⁻¹; IV: [A] = 0.1 M.

3.13

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.

Solution

$t_{1/2} = \frac{0.693}{k}$ :

(i) $\frac{0.693}{200} \approx 3.47 \times 1 0^{- 3}$ s
(ii) $\frac{0.693}{2} \approx 0.35$ min
(iii) $\frac{0.693}{4} \approx 0.173$ years

(i) $3.47 \times 1 0^{- 3}$ s. (ii) 0.35 min. (iii) 0.173 years.

3.14

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.

Solution

$k = \frac{0.693}{5730} = 1.21 \times 1 0^{- 4}$ year⁻¹.

$t = \frac{2.303}{k} lo g \frac{100}{80} = \frac{2.303}{1.21 \times 1 0 ^{- 4}} \times 0.0969 \approx 1845 years$

About 1845 years.

3.15

The experimental data for decomposition of

N X_{2} O X_{5}

[

2 N X_{2} O X_{5} 4 NO X_{2} + O X_{2}

] in gas phase at 318 K are given below: t/s: 0, 400, 800, 1200, 1600, 2000, 2400, 2800, 3200;

1 0^{2} \times [N X_{2} O X_{5}]

/mol L⁻¹: 1.63, 1.36, 1.14, 0.93, 0.78, 0.64, 0.53, 0.43, 0.35. (i) Plot

[N X_{2} O X_{5}]

against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between

lo g [N X_{2} O X_{5}]

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).

Solution

(i) The plot of $[N X_{2} O X_{5}]$ against t is a curve that falls steeply at first and then more slowly (exponential decay).

(ii) Half of $1.63 \times 1 0^{- 2}$ is $0.815 \times 1 0^{- 2}$ mol L⁻¹. From the data this is reached between 1200 s (0.93) and 1600 s (0.78), at about 1440 s.

(iii) Values of $lo g (1 0^{2} [N X_{2} O X_{5}])$ : 0.212, 0.134, 0.057, −0.032, −0.108, −0.194, −0.276, −0.367, −0.456. Plotted against t these lie on a straight line with a negative slope.

(iv) The straight-line log plot shows first order: rate $= k [N X_{2} O X_{5}]$ .

(v) Slope $= - \frac{k}{2.303}$ . Using the first and last points:

$k = \frac{2.303}{3200} lo g \frac{1.63}{0.35} = \frac{2.303 \times 0.668}{3200} \approx 4.8 \times 1 0^{- 4} s^{- 1}$

(vi) $t_{1/2} = \frac{0.693}{4.8 \times 1 0 ^{- 4}} \approx 1440$ s, which agrees with the value read from the graph in (ii).

(ii) About 1440 s. (iv) Rate = k[N₂O₅] (first order). (v) k ≈ $4.8 \times 1 0^{- 4}$ s⁻¹. (vi) 1440 s, the same as (ii).

3.16

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?

Solution

$t = \frac{2.303}{60} lo g 16 = \frac{2.303 \times 1.204}{60} \approx 4.62 \times 1 0^{- 2} s$

(This is exactly 4 half-lives: $(\frac{1}{2})^{4} = \frac{1}{16}$ .)

About $4.62 \times 1 0^{- 2}$ s.

3.17

During nuclear explosion, one of the products is ⁹⁰Sr with half-life of 28.1 years. If 1 μg 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?

Solution

$k = \frac{0.693}{28.1} = 0.02466$ year⁻¹, and $N = N_{0} e^{- k t}$ .

After 10 years: $N = 1 \times e^{- 0.2466} \approx 0.781$ μg
After 60 years: $N = 1 \times e^{- 1.480} \approx 0.228$ μg

About 0.781 μg after 10 years and 0.228 μg after 60 years.

3.18

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

Solution

$t = \frac{2.303}{k} lo g \frac{[ R ] _{0}}{[ R ]}$ .

99% complete: $[R] = 0.01 [R]_{0}$ , so $t_{99} = \frac{2.303}{k} lo g 100 = \frac{2 \times 2.303}{k}$
90% complete: $[R] = 0.1 [R]_{0}$ , so $t_{90} = \frac{2.303}{k} lo g 10 = \frac{2.303}{k}$

Therefore $t_{99} = 2 t_{90}$ .

$t_{99} = \frac{2.303}{k} lo g 100 = 2 \times \frac{2.303}{k} lo g 10 = 2 t_{90}$ .

3.19

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

t_{1/2}

Solution

30% decomposed means $[R] = 0.70 [R]_{0}$ :

$k = \frac{2.303}{40} lo g \frac{1}{0.70} = \frac{2.303 \times 0.155}{40} \approx 8.92 \times 1 0^{- 3} min^{- 1}$

$t_{1/2} = \frac{0.693}{8.92 \times 1 0 ^{- 3}} \approx 77.7$ min.

About 77.7 min.

3.20

For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained: t (sec) 0, 360, 720; P (mm of Hg) 35.0, 54.0, 63.0. Calculate the rate constant.

Solution

$(CH X_{3}) X_{2} CHN = NCH (CH X_{3}) X_{2} (g) C X_{6} H X_{14} (g) + N X_{2} (g)$ : 1 mol of gas gives 2 mol, so $p_{A} = 2 p_{0} - p_{t}$ .

At 360 s: $p_{A} = 70.0 - 54.0 = 16.0$ mm: $k = \frac{2.303}{360} lo g \frac{35.0}{16.0} \approx 2.17 \times 1 0^{- 3}$ s⁻¹
At 720 s: $p_{A} = 70.0 - 63.0 = 7.0$ mm: $k = \frac{2.303}{720} lo g \frac{35.0}{7.0} \approx 2.24 \times 1 0^{- 3}$ s⁻¹

Average $k \approx 2.2 \times 1 0^{- 3}$ s⁻¹.

k ≈ $2.2 \times 1 0^{- 3}$ s⁻¹.

3.21

The following data were obtained during the first order thermal decomposition of

SO X_{2} Cl X_{2}

at a constant volume:

SO X_{2} Cl X_{2} (g) SO X_{2} (g) + Cl X_{2} (g)

. Experiment 1: time 0 s, total pressure 0.5 atm; Experiment 2: time 100 s, total pressure 0.6 atm. Calculate the rate of the reaction when total pressure is 0.65 atm.

Solution

$p (SO X_{2} Cl X_{2}) = 2 p_{0} - p_{t}$ .

At 100 s: $p = 1.0 - 0.6 = 0.4$ atm, so

$k = \frac{2.303}{100} lo g \frac{0.5}{0.4} = \frac{2.303 \times 0.0969}{100} \approx 2.23 \times 1 0^{- 3} s^{- 1}$

At $p_{t} = 0.65$ atm: $p (SO X_{2} Cl X_{2}) = 1.0 - 0.65 = 0.35$ atm.

Rate $= k p (SO X_{2} Cl X_{2}) = 2.23 \times 1 0^{- 3} \times 0.35 \approx 7.8 \times 1 0^{- 4}$ atm s⁻¹.

About $7.8 \times 1 0^{- 4}$ atm s⁻¹.

3.22

The rate constant for the decomposition of

N X_{2} O X_{5}

at various temperatures is given below: T/°C: 0, 20, 40, 60, 80;

1 0^{5} \times 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° and 50°C.

Solution

T (K)	1/T (10⁻³ K⁻¹)	k (s⁻¹)	ln k
273	3.663	7.87 × 10⁻⁷	−14.06
293	3.413	1.70 × 10⁻⁵	−10.98
313	3.195	2.57 × 10⁻⁴	−8.27
333	3.003	1.78 × 10⁻³	−6.33
353	2.833	2.14 × 10⁻²	−3.84

The plot of ln k against 1/T is a straight line. Its slope (best fit) is about $- 1.21 \times 1 0^{4}$ K.

$E_{a} = - slope \times R = 1.21 \times 1 0^{4} \times 8.314 \approx 1.0 \times 1 0^{5}$ J mol⁻¹ $\approx 100$ kJ mol⁻¹
Intercept: $ln A \approx 30.3$ , so $A \approx 1.4 \times 1 0^{13}$ s⁻¹

Predictions from the line, $ln k = 30.3 - \frac{12100}{T}$ :

30°C (303 K): $ln k \approx - 9.65$ , $k \approx 6.4 \times 1 0^{- 5}$ s⁻¹
50°C (323 K): $ln k \approx - 7.18$ , $k \approx 7.6 \times 1 0^{- 4}$ s⁻¹

(Values read from a hand-drawn graph will differ slightly.)

$E_{a}$ ≈ 100 kJ mol⁻¹, A ≈ $1.4 \times 1 0^{13}$ s⁻¹; k ≈ $6.4 \times 1 0^{- 5}$ s⁻¹ at 30°C and $7.6 \times 1 0^{- 4}$ s⁻¹ at 50°C.

3.23

The rate constant for the decomposition of hydrocarbons is

2.418 \times 1 0^{- 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?

Solution

$A = k e^{E_{a} / R T}$ , with $\frac{E _{a}}{R T} = \frac{179900}{8.314 \times 546} = 39.63$ .

$A = 2.418 \times 1 0^{- 5} \times e^{39.63} = 2.418 \times 1 0^{- 5} \times 1.63 \times 1 0^{17} \approx 3.9 \times 1 0^{12} s^{- 1}$

A ≈ $3.9 \times 1 0^{12}$ s⁻¹.

3.24

Consider a certain reaction A → Products with

k = 2.0 \times 1 0^{- 2}

s⁻¹. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L⁻¹.

Solution

The unit s⁻¹ shows first order: $[A] = [A]_{0} e^{- k t} = 1.0 \times e^{- 2.0} \approx 0.135$ mol L⁻¹.

About 0.135 mol L⁻¹.

3.25

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?

Solution

$k = \frac{0.693}{3.00} = 0.231$ h⁻¹. Fraction left $= e^{- k t} = e^{- 0.231 \times 8} = e^{- 1.848} \approx 0.158$ .

About 0.158 (15.8%).

3.26

The decomposition of hydrocarbon follows the equation

k = (4.5 \times 1 0^{11} s^{- 1}) e^{- 28000 K / T}

. Calculate

E_{a}

Solution

Comparing with $k = A e^{- E_{a} / R T}$ : $\frac{E _{a}}{R} = 28000$ K.

$E_{a} = 28000 \times 8.314 \approx 2.33 \times 1 0^{5}$ J mol⁻¹ $= 232.8$ kJ mol⁻¹.

About 232.8 kJ mol⁻¹.

3.27

The rate constant for the first order decomposition of

H X_{2} O X_{2}

is given by the following equation:

lo g k = 14.34 - 1.25 \times 1 0^{4} K / T

. Calculate

E_{a}

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

Solution

Comparing with $lo g k = lo g A - \frac{E _{a}}{2.303 R T}$ : $\frac{E _{a}}{2.303 R} = 1.25 \times 1 0^{4}$ K.

$E_{a} = 1.25 \times 1 0^{4} \times 2.303 \times 8.314 \approx 2.39 \times 1 0^{5}$ J mol⁻¹ $= 239.3$ kJ mol⁻¹.

For $t_{1/2} = 256$ min: $k = \frac{0.693}{256 \times 60} = 4.51 \times 1 0^{- 5}$ s⁻¹, $lo g k = - 4.35$ .

$- 4.35 = 14.34 - \frac{1.25 \times 1 0 ^{4}}{T}$ , so $T = \frac{1.25 \times 1 0 ^{4}}{18.69} \approx 669$ K.

$E_{a}$ ≈ 239.3 kJ mol⁻¹; T ≈ 669 K.

3.28

The decomposition of A into product has value of k as

4.5 \times 1 0^{3}

s⁻¹ at 10°C and energy of activation 60 kJ mol⁻¹. At what temperature would k be

1.5 \times 1 0^{4}

s⁻¹?

Solution

$lo g \frac{1.5 \times 1 0 ^{4}}{4.5 \times 1 0 ^{3}} = \frac{60000}{2.303 \times 8.314} (\frac{1}{283} - \frac{1}{T _{2}})$

$0.523 = 3134 (\frac{1}{283} - \frac{1}{T _{2}})$ , so $\frac{1}{T _{2}} = 3.534 \times 1 0^{- 3} - 1.669 \times 1 0^{- 4} = 3.367 \times 1 0^{- 3}$ and $T_{2} \approx 297$ K (about 24°C).

About 297 K (24°C).

3.29

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 \times 1 0^{10}

s⁻¹, calculate k at 318 K and

E_{a}

Solution

For the same time $t$ : $k_{298} t = 2.303 lo g \frac{100}{90}$ and $k_{308} t = 2.303 lo g \frac{100}{75}$ , so

$\frac{k _{308}}{k _{298}} = \frac{lo g ( 4/3 )}{lo g ( 10/9 )} = \frac{0.1249}{0.0458} \approx 2.73$

$E_{a} = \frac{2.303 \times 8.314 \times 298 \times 308}{10} lo g 2.73 \approx 76.6 kJ mol^{- 1}$

At 318 K: $k = A e^{- E_{a} / R T} = 4 \times 1 0^{10} \times exp (- \frac{76640}{8.314 \times 318}) = 4 \times 1 0^{10} \times e^{- 29.0} \approx 1.0 \times 1 0^{- 2}$ s⁻¹.

$E_{a}$ ≈ 76.6 kJ mol⁻¹; k(318 K) ≈ $1.0 \times 1 0^{- 2}$ s⁻¹.

3.30

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.

Solution

$lo g 4 = \frac{E _{a}}{2.303 \times 8.314} (\frac{313 - 293}{293 \times 313})$

$E_{a} = \frac{0.602 \times 2.303 \times 8.314 \times 293 \times 313}{20} \approx 52.9 kJ mol^{- 1}$

About 52.9 kJ mol⁻¹.

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