NCERT Solutions for Class 12 Chemistry Chapter 3 Chemical Kinetics

Chemical Kinetics Chapter 3 is a significant unit of the syllabus from exam point of view and students must make sure to prepare it comprehensively to score well. SimplyAcad offered answers will make sure to give proper insights of various theories and concepts related to the chapter. Our subject experts are always free to support students for several upcoming examinations including the entrance tests of IITs, VITEEE, and many more.

 

These answers will ensure students to strengthen their basics and solve different questions thrown at them with confidence. Scroll below to find the answers according to the latest syllabus of NCERT as prescribed by CBSE for the approaching 12th boards. 

 

NCERT Solutions for Class 12 Chemical Kinetics Exercise 3.1

Question 1. For the reaction \textbf{R} → \textbf{P}, the concentration of the reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction in both minutes and seconds. \textbf

{Solution} The average rate of reaction is given by the formula: [ \text{Average rate of reaction} = -\frac{\Delta [R]}{\Delta t} ] Where: [ \Delta [R] = [R]_2 – [R]_1 = 0.02 \, \text{M} – 0.03 \, \text{M} = -0.01 \, \text{M} ] And: [ \Delta t = t_2 – t_1 = 25 \, \text{minutes} ] So the average rate of reaction is: [ \text{Average rate of reaction} = -\frac{-0.01}{25} = 4 \times 10^{-4} \, \text{M min}^{-1} ] To express the average rate of reaction in seconds, we convert minutes to seconds: [ \text{Average rate of reaction} = 4 \times 10^{-4} \, \text{M min}^{-1} \times \frac{1}{60} \, \text{min sec}^{-1} ] [ = 6.67 \times 10^{-6} \, \text{M sec}^{-1} ] Thus, the average rate of reaction is (4 \times 10^{-4}) M min(^{-1}) or (6.67 \times 10^{-6}) M sec(^{-1}).

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

{Solution} Given: [ \text{Initial concentration, } [A]_1 = 0.5 \, \text{M} ] [ \text{Final concentration, } [A]_2 = 0.4 \, \text{M} ] [ \text{Time interval, } \Delta t = 10 \, \text{min} ] For the reaction \textbf{2A → Products}, the rate of reaction is given by: [ \text{Rate} = -\frac{1}{m} \frac{d[A]}{dt} = \frac{1}{n} \frac{d[B]}{dt} ] For this specific reaction: [ \text{Rate} = -\frac{1}{2} \frac{\Delta [A]}{\Delta t} ] [ = -\frac{1}{2} \frac{[A]_2 – [A]_1}{\Delta t} ] Substituting the given values: [ \text{Rate} = -\frac{1}{2} \frac{0.4 – 0.5}{10 \, \text{min}} ] [ = -\frac{1}{2} \frac{-0.1}{10 \, \text{min}} ] [ = \frac{0.1}{20 \, \text{min}} ] [ = 0.005 \, \text{mol L}^{-1} \, \text{min}^{-1} ] Thus, the rate of the reaction during this interval is (5 \times 10^{-3}) mol L\textsuperscript{−1} min\textsuperscript{−1}.

Question 3. 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? Chemical Kinetics Chapter 3

Solution :
The order of the reaction r  = k [A]½ [B]2

= 1 / 2 + 2

= 2.5

Question 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? : Chemical Kinetics Chapter 3

Solution :
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:

Rate1 = 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.

Question 5. A first-order reaction has a rate constant (1.15 \times 10^{-3} \, \text{s}^{-1}). How long will it take for 5 g of this reactant to reduce to 3 g? \textbf

{Solution} Given: [ \text{Initial amount, } [R]_0 = 5 \, \text{g} ] [ \text{Final amount, } [R] = 3 \, \text{g} ] [ \text{Rate constant, } k = 1.15 \times 10^{-3} \, \text{s}^{-1} ] For a first-order reaction, the time required to reduce the reactant concentration can be calculated using the formula: [ t = \frac{2.303}{k} \log \frac{[R]_0}{[R]} ] Substituting the given values: [ t = \frac{2.303}{1.15 \times 10^{-3}} \log \frac{5}{3} \, \text{s} ] Calculating the logarithm and time: [ t = \frac{2.303}{1.15 \times 10^{-3}} \times 0.2219 \, \text{s} ] [ t \approx 444 \, \text{s} ] Thus, 5 g of the reactant will reduce to 3 g in approximately 444 seconds. 

Question 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. : Chemical Kinetics Chapter 3

Solution :
We know that for a 1st order reaction,

t½  = 0.693 / k

It is given that t1/2 = 60 min

k = 0.693 / t½

= 0.693 / 60

= 0.01155 min-1

= 1.155 min-1

Or

k = 1.925 x 10-2 s-1

Question 7. What will be the effect of temperature on rate constant? Chemical Kinetics Chapter 3

Solution :
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 – Ea / RT

Where,

A is the Arrhenius factor or the frequency factor

T is the temperature

R is the gas constant

Ea is the activation energy

Question 8. The rate of a chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate the activation energy ( E_a ). \textbf

{Solution} Given: [ T_1 = 298 \, \text{K} ] [ T_2 = 308 \, \text{K} \quad (\text{as } T_2 = T_1 + 10 \, \text{K}) ] It is stated that the rate of reaction doubles when the temperature is increased by 10 K. Therefore, if ( k_1 = k ) at ( T_1 ), then ( k_2 = 2k ) at ( T_2 ). Using the Arrhenius equation and its logarithmic form: [ \log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303 \cdot R} \left( \frac{1}{T_1} – \frac{1}{T_2} \right) ] Substitute the given values: [ \log 2 = \frac{E_a}{2.303 \cdot 8.314} \left( \frac{1}{298} – \frac{1}{308} \right) ] Calculating the right-hand side: [ \log 2 = \frac{E_a}{2.303 \cdot 8.314} \left( \frac{1}{298} – \frac{1}{308} \right) ] [ E_a = 2.303 \cdot 8.314 \cdot \frac{\log 2}{\left( \frac{1}{298} – \frac{1}{308} \right)} ] [ E_a \approx 2.303 \cdot 8.314 \cdot \frac{0.3010}{\left( \frac{1}{298} – \frac{1}{308} \right)} ] [ E_a \approx 2.303 \cdot 8.314 \cdot \frac{0.3010}{\left( \frac{308 – 298}{298 \cdot 308} \right)} ] [ E_a \approx 2.303 \cdot 8.314 \cdot \frac{0.3010 \cdot 298 \cdot 308}{10} ] [ E_a \approx 2.303 \cdot 8.314 \cdot \frac{0.3010 \cdot 91784}{10} ] [ E_a \approx 2.303 \cdot 8.314 \cdot 2764.48 ] [ E_a \approx 53.7 \, \text{kJ mol}^{-1} ] Therefore, the activation energy ( E_a ) is approximately ( 53.7 \, \text{kJ mol}^{-1} ).

Question 9. The activation energy for the reaction 2HI(g) → H2 + I2(g) is 209.5 kJ mol−1 at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy? : Chemical Kinetics Chapter 3

Solution :
In the given case:

Ea = 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:

x = e−Ea / RT

⇒Inx= −Ea / RT

⇒logx=−Ea / 2.303RT

⇒logx= −209500Jmol−1 / 2.303 × 8.314JK−1mol−1×581

=−18.8323

Now,x= Antilog (−18.8323)

=1.471×10−19

Question 10. From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants. : Chemical Kinetics Chapter 3

(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]

Solution :
(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-1s-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 / mol3/2  L-3/2

= L½ mol-½  s-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

 

Question 11. 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.

Solution :
The initial rate of the reaction is

Rate = k [A][B]2

= (2.0 × 10−6 mol−2 L2 s−1) (0.1 mol L−1) (0.2 mol L−1)2

= 8.0 × 10−9 mol−2 L2 s−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 L2 s−1) (0.06 mol L−1) (0.18 mol L−1)2

= 3.89 mol L−1 s−1

Question 12. The decomposition of (\text{NH}_3) on a platinum surface is a zero-order reaction. Given the rate constant (k = 2.5 \times 10^{-4} \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1}), what are the rates of production of (\text{N}_2) and (\text{H}_2)? \textbf

{Solution:} The reaction is zero-order. Therefore, the rate of the reaction is equal to the rate constant. The reaction is given by: [ 2\text{NH}_3 \rightarrow \text{N}_2 + 3\text{H}_2 ] The rate of production of (\text{N}_2) is given by: [ \frac{d[\text{N}_2]}{dt} = \frac{1}{2} \times 2.5 \times 10^{-4} \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} = 1.25 \times 10^{-4} \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} ] The rate of production of (\text{H}_2) is given by: [ \frac{d[\text{H}_2]}{dt} = \frac{3}{2} \times 2.5 \times 10^{-4} \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} = 3.75 \times 10^{-4} \, \text{mol} \, \text{L}^{1} \, \text{s}^{-1} ] Therefore, the rates of production of (\text{N}_2) and (\text{H}_2) are (1.25 \times 10^{-4}) and (3.75 \times 10^{-4}) (\text{mol} \, \text{L}^{-1} \, \text{s}^{-1}), respectively. 

Question 13. 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 andtime in minutes, then what are the units of rate and rate constants?

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

Unit of rate = bar min−1

Rate = k [CH3OCH3]3/2

⇒ k =Rate / [CH3OCH3]3/2

Therefore, unit of rate constants(k) = bar min−1 / bar3/2

= bar-½ min -1

Question 14. Mention the factors that affect the rate of a chemical reaction.

Solution :
The factors that affect the rate of a reaction are as follows.

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

(ii) Temperature

(iii) Presence of a catalyst

Question 15. 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 :
Letthe 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

Therefore, the rate of the reaction would be reduced to

Question 16. What change would happen in the rate constant of a reaction when there is a change in its temperature? How can this temperature effect on rate constant be represented quantitatively?

Solution :
When a temperature of 10∘ rises for a chemical reaction then the rate constant increases and becomes near to double of its original value.

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

k=Ae−Ea/RT

Where,

k = rate constant,

A = Frequency factor / Arrhenius factor,

R = gas constant

T = temperature

Ea = activation energy for the reaction.

Question 17. 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.

Solution :
(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 × 10−3 mol L−1 s−1

(ii) For a pseudo first order reaction,

k = 2.303/ t log [R]º / [R]

For t = 30 s, k1

= 1.911 × 10−2 s−1

For t = 60 s, k1 = 2.303/ 30 log 0.55 / 0.31

= 1.957 × 10−2 s−1

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

= 2.075 × 10 – 2s – 1

= 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 10-2 s – 1

Question 18. {Question:} A reaction is first order in ( A ) and second order in ( B ): (i) Write the differential rate equation. (ii) How is the rate affected when the concentration of ( B ) is tripled? (iii) How is the rate affected when the concentration of both ( A ) and ( B ) is doubled? \textbf

{Solution:} Given that the reaction is first order in ( A ) and second order in ( B ): (i) The differential rate equation is: [ \text{Rate} = k[A][B]^2 ] where: \begin{itemize} \item ( k ) is the rate constant, \item ([A]) is the concentration of reactant ( A ), \item ([B]) is the concentration of reactant ( B ). \end{itemize} (ii) When the concentration of ( B ) is tripled: The rate is proportional to the square of the concentration of ( B ). If ([B]) is tripled, the rate becomes: [ \text{Rate} \propto [B]^2 ] [ \text{Rate} \propto (3[B])^2 ] [ \text{Rate} \propto 9[B]^2 ] Thus, the rate increases by a factor of 9. (iii) When the concentrations of both ( A ) and ( B ) are doubled: The rate equation becomes: [ \text{Rate} = k(2[A])(2[B])^2 ] [ \text{Rate} = k(2[A])(4[B]^2) ] [ \text{Rate} = 8k[A][B]^2 ] Therefore, the rate becomes 8 times the original rate. 

Question 19. 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?

Solution :
Let the order of the reaction with respect to A be x and with respect to B be y.

Therefore,

Dividing equation (iii) by (ii), we obtain

= 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.

Question 20. 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.

Solution :
Let the order of the reaction with respect to A be x and with respect to B be y.

Therefore, rate of the reaction is given by,

Dividing equation (iv) by (i), we obtain

Dividing equation (iii) by (ii), we obtain

Question 21. 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

Solution :
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 × 10−2 mol L−1 min−1 = k (0.1 mol L−1)

⇒ k = 0.2 min−1

From experiment II, we obtain

4.0 × 10−2 mol L−1 min−1 = 0.2 min−1 [A]

⇒ [A] = 0.2 mol L−1

From experiment III, we obtain

Rate = 0.2 min−1 × 0.4 mol L−1

= 0.08 mol L−1 min−1

From experiment IV, we obtain

2.0 × 10−2 mol L−1 min−1 = 0.2 min−1 [A]

⇒ [A] = 0.1 mol L−1

Question 22. 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 :
(i) Half life, t 1/2 = 0.693 / k

= 0.693 / 200 s-1

= 3.47×10 -3 s (approximately)

(ii) Half life, t 1/2 = 0.693 / k

= 0.693 / 2 min-1

= 0.35 min (approximately)

(iii) Half life, t 1/2 = 0.693 / k

= 0.693 / 4 years-1

= 0.173 years (approximately)

Question 23. The half-life for radioactive decay of $^{14}\text{C}$ is 5730 years. An archaeological artifact containing wood had only 80\% of the $^{14}\text{C}$ found in a living tree. Estimate the age of the sample. \textbf

{Solution} The decay constant, $k$, for a radioactive substance is related to its half-life, $t_{1/2}$, by the equation: [ k = \frac{0.693}{t_{1/2}} ] Given: [ t_{1/2} = 5730 \, \text{years} ] Therefore: [ k = \frac{0.693}{5730 \, \text{years}} = 1.209 \times 10^{-4} \, \text{years}^{-1} ] The percentage of $^{14}\text{C}$ remaining in the sample compared to the original amount in a living tree is 80\%. Let $N_0$ be the original number of $^{14}\text{C}$ atoms and $N$ be the remaining number. Since $N = 0.80 \times N_0$, we can use the decay formula to estimate the age of the sample, $t$: [ t = \frac{2.303}{k} \log \left(\frac{N_0}{N}\right) ] Substituting the values: [ t = \frac{2.303}{1.209 \times 10^{-4} \, \text{years}^{-1}} \log \left(\frac{100}{80}\right) ] [ t = \frac{2.303}{1.209 \times 10^{-4}} \log (1.25) ] [ t = \frac{2.303}{1.209 \times 10^{-4}} \times 0.09691 ] [ t \approx \frac{2.303 \times 0.09691}{1.209 \times 10^{-4}} ] [ t \approx \frac{0.22338}{1.209 \times 10^{-4}} ] [ t \approx 1848 \, \text{years} ] Therefore, the estimated age of the sample is approximately 1848 years.

Question 24. 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).

Solution :

(ii) Time corresponding to the concentration, 1630×102 / 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

– k /2.303

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

– k / 2.303. = -0.67 / 3200

Therefore, we obtain,

– k / 2.303  = – 0.67 / 3200

⇒ k = 4.82 x 10-4 s-1

(vi) Half-life is given by,

t½ = 0.693 / k

= 0.639 / 4.82×10-4 s

=1.438 x 103

This value, 1438 s, is very close to the value that was obtained from the graph.

Question 25. The rate constant for the first order reaction is $60 \, \text{s}^{-1}$. How much time will it take to reduce the concentration of the reactant to $\frac{1}{16}$th of its initial value? \section*

{Solution} To find the time required for the concentration to decrease to $\frac{1}{16}$th of its initial value, we use the formula for the first-order reaction rate law: [ t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]} \right) ] Given: \begin{align*} k &= 60 \, \text{s}^{-1} \ [A]_0 &= \text{initial concentration} \ [A] &= \frac{[A]_0}{16} \end{align*} Substituting the values into the equation, we get: [ t = \frac{2.303}{60} \log \left( \frac{[A]_0}{\frac{[A]_0}{16}} \right) ] [ t = \frac{2.303}{60} \log (16) ] Since $\log(16) = \log(2^4) = 4\log(2)$ and $\log(2) \approx 0.3010$, we find: [ t = \frac{2.303}{60} \times 4 \times 0.3010 ] [ t = \frac{2.303 \times 1.204}{60} ] [ t = \frac{2.773812}{60} ] [ t \approx 0.0462 \, \text{s} ] Therefore, the time required for the concentration of the reactant to reduce to $\frac{1}{16}$th of its initial value is approximately $0.0462 \, \text{s}$. 

Question 26. 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.

Solution :

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

Again,

Therefore, 0.2278 μg of 90Sr will remain after 60 years.

Question 27. 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 :
For a first order reaction, the time required for 99% completion is

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.

Question 28. A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.

Solution :
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, t1/2 of the decomposition reaction is

t1/2 = 0.693/k

=  0.693 / 8.918 x 10-3  min

= 77.7 min (approximately)

= 77.7 min (approximately)

Question 29. 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.

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

After time, t, total pressure, Pt = (Pº – p) + p + p

⇒ Pt = (Pº + p)

⇒ p = Pt  – P0

therefore, Pº – p = P0  – Pt  – P0

= 2P0 − Pt

For a first order reaction,

k = 2.303/t  Log P0 /P0  – p

=   2.303/t  Log P0 / 2 P0  –  Pt

When t = 360 s, k = 2.303 / 360s log 35.0 / 2×35.0 – 54.0

= 2.175 × 10−3 s−1

When t = 720 s, k = 2.303 / 720s log 35.0 / 2×35.0 – 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

Question 30. The following data were obtained during the first order thermal decomposition of SO2Cl2 at 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.

Solution :
The thermal decomposition of SO2Cl2 at a constant volume is represented by the following equation.

After time, t, total pressure, Pt = (Pº – p) + p + p

⇒ Pt = (Pº + p)

⇒ p = Pt  – Pº

therefore, Pº – p = Pº  – Pt  – Pº

= 2 Pº –  Pt

For a first order reaction,

k = 2.303/t  Log  Pº / Pº  – p

=   2.303/t  Log  Pº / 2 Pº  –  Pt

When t= 100 s,

k = 2.303 / 100s log 0.5 / 2×0.5 – 0.6

= 2.231 × 10 – 3s – 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

PSOCL2 = 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(pSOCL2)

= (2.23 × 10 – 3s – 1) (0.35 atm)

= 7.8 × 10 – 4 atm s – 1

Question 31. 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 Ea.

Predict the rate constant at 30º and 50ºC.

Solution :
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
ln k	−7.147	− 4.075	−1.359	−0.577	3.063

Slope of the line,

In k= – 2.8

Therefore, k = 6.08×10-2s-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

Question 32. The rate constant for the decomposition of hydrocarbons is ( 2.418 \times 10^{-5} \, \text{s}^{-1} ) at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of the pre-exponential factor? \textbf

{Solution:} According to the Arrhenius equation: [ \log k = \log A – \frac{E_a}{2.303 \, RT} ] where: \begin{itemize} \item ( k ) is the rate constant, \item ( A ) is the pre-exponential factor, \item ( E_a ) is the activation energy, \item ( R ) is the gas constant, \item ( T ) is the temperature in Kelvin. \end{itemize} Given: \begin{align} k &= 2.418 \times 10^{-5} \, \text{s}^{-1},\ E_a &= 179.9 \, \text{kJ/mol} = 179900 \, \text{J/mol},\ R &= 8.314 \, \text{J/K} \cdot \text{mol},\ T &= 546 \, \text{K}. \end{align} We can rearrange the Arrhenius equation to solve for ( \log A ): [ \log A = \log k + \frac{E_a}{2.303 \, RT} ] Substituting the known values: [ \log A = \log(2.418 \times 10^{-5} \, \text{s}^{-1}) + \frac{179900 \, \text{J/mol}}{2.303 \times (8.314 \, \text{J/K} \cdot \text{mol}) \times 546 \, \text{K}} ] [ \log A = -4.6184 + 17.21 ] [ \log A = 12.5916 ] To find ( A ), we take the antilog: [ A = \text{antilog}(12.5916) = 3.9 \times 10^{12} \, \text{s}^{-1} ] Thus, the pre-exponential factor ( A ) is ( 3.9 \times 10^{12} \, \text{s}^{-1} ). 

Question 33. Consider a certain reaction A → Products with k = 2.0 × 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.  : Chemical Kinetics Chapter 3

Solution :
k = 2.0 × 10−2 s−1

T = 100 s

[A]o = 1.0 moL−1

Since the unit of k is 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.

Question 34. Sucrose decomposes in an acid solution into glucose and fructose according to the first-order rate law, with ( t_{1/2} = 3.00 \, \text{h} ). What fraction of a sample of sucrose remains after 8 hours? \textbf

{Solution:} For first-order reactions, the rate constant ( k ) is related to the half-life ( t_{1/2} ) by the equation: [ k = \frac{0.693}{t_{1/2}} ] Given: [ t_{1/2} = 3.00 \, \text{h} ] The rate constant ( k ) is: [ k = \frac{0.693}{3.0 \, \text{h}} = 0.231 \, \text{h}^{-1} ] The integrated rate law for a first-order reaction is: [ \log \frac{[A]_0}{[A]} = \frac{kt}{2.303} ] Where: \begin{itemize} \item ([A]_0) is the initial concentration of the reactant, \item ([A]) is the concentration of the reactant at time ( t ), \item ( k ) is the rate constant, \item ( t ) is the time. \end{itemize} Substituting the values, we get: [ \log \frac{[A]_0}{[A]} = \frac{0.231 \, \text{h}^{-1} \times 8 \, \text{h}}{2.303} ] [ \log \frac{[A]_0}{[A]} = 0.8024 ] Now, calculate the fraction of sucrose remaining after 8 hours: [ \frac{[A]}{[A]_0} = 10^{-\log \frac{[A]_0}{[A]}} = 10^{-0.8024} ] [ \frac{[A]}{[A]_0} \approx 0.1576 ] Thus, approximately 0.1576 (or 15.76\%) of the initial sucrose remains after 8 hours.

Question 35. The decomposition of hydrocarbon follows the equation k = (4.5 × 1011 s−1) e−28000 K/T Calculate Ea.  : Chemical Kinetics Chapter 3

Solution :
The given equation is

k = (4.5 × 1011 s−1) e−28000 K/T (i)

Arrhenius equation is given by,

k= Ae -Ea/RT  (ii)

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

Ea  / RT  =  28000K / T

⇒ Ea  = R x 28000K

= 8.314 J K−1 mol−1 × 28000 K

= 232792 J mol−1

= 232.792 kJ mol−1

Question 36. The rate constant for the first order decomposition of H2O2 is given by the following equation:

log k = 14.34 − 1.25 × 104 K/T

Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?  : Chemical Kinetics Chapter 3

Solution :
Arrhenius equation is given by,

k= Ae -Ea/RT

⇒In k = In A – Ea/RT

⇒In k = Log A – Ea/RT

⇒ Log k = Log A – Ea/2.303RT     (i)

The given equation is

Log k = 14.34 – 1.25 104 K/T         (ii)

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

Ea/2.303RT  = 1.25 104 K/T

= 1.25 × 104 K × 2.303 × 8.314 J K−1 mol−1

= 239339.3 J mol−1 (approximately)

= 239.34 kJ mol−1

Also, when t1/2 = 256 minutes,

k = 0.693 / t1/2

= 0.693 / 256

= 2.707 × 10 – 3 min – 1

= 4.51 × 10 – 5s – 1

= 2.707 × 10−3 min−1

= 4.51 × 10−5 s−1

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

= 668.95 K

= 669 K (approximately)

Question 37. 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 × 104 s−1?  : Chemical Kinetics Chapter 3

Solution :
From Arrhenius equation, we obtain

log k2/k1 = Ea / 2.303 R (T2 – T1) / T1T2

Also, k1 = 4.5 × 103 s−1

T1 = 273 + 10 = 283 K

k2 = 1.5 × 104 s−1

Ea = 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.

Question 38. 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 × 1010 s−1. Calculate k at 318 K and Ea. : Chemical Kinetics Chapter 3

Solution :
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

Question 39. 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. : Chemical Kinetics Chapter 3

Solution :
From Arrhenius equation, we obtain

Hence, the required energy of activation is 52.86 kJmol−1.