Class 12 NCERT Solutions Chemistry Chapter 4 – Chemical Kinetics

Q.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) → 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]

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{Given Rate=k}{\left[\text{NO}\right]}^{\text{2}}\\ \text{Therefore, order of the reaction=2}\\ \text{k=}\frac{\text{Rate}}{{\text{[NO]}}^{\text{2}}}\\ \text{Dimension\hspace{0.17em}of k}\\ \text{=}\frac{{\text{mol L}}^{\text{-1}}{\text{s}}^{\text{-1}}}{{{\text{(mol L}}^{\text{-1}}\text{)}}^{\text{2}}}\\ \text{=}\frac{{\text{mol L}}^{\text{-1}}{\text{s}}^{\text{-1}}}{{\text{mol}}^{\text{2}}{\text{L}}^{\text{-2}}}\\ {\text{=L mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\\ \\ \left(\text{ii}\right){\text{Given rate=k[H}}_{\text{2}}{\text{O}}_{\text{2}}{\text{][I}}^{\text{–}}\text{]}\\ \text{Therefore, order of the reaction=2}\\ \text{k=}\frac{\text{Rate}}{\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]{\text{[I}}^{\text{–}}\text{]}}\\ \text{Dimension of k}\\ \text{=}\frac{{\text{mol L}}^{\text{-1}}{\text{s}}^{\text{-1}}}{\left({\text{mol L}}^{\text{-1}}\right){\text{(mol L}}^{\text{-1}}\text{)}}\\ {\text{=L mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\\ \\ \left(\text{iii}\right)\text{Given rate = k}{\left[{\text{CH}}_{\text{3}}\text{CHO}\right]}^{\frac{\text{3}}{\text{2}}}\text{=}\frac{\text{3}}{\text{2}}\\ \text{Therefore,order of reaction=\hspace{0.17em}k=}\frac{\text{Rate}}{{{\text{[CH}}_{\text{3}}\text{CHO]}}^{\frac{\text{3}}{\text{2}}}}\\ \text{Dimension of k}\\ \text{=}\frac{{\text{mol L}}^{\text{-1}}{\text{s}}^{\text{-1}}}{{{\text{(mol L}}^{\text{-1}}\text{)}}^{\frac{\text{3}}{\text{2}}}}\\ \text{=}\frac{{\text{mol L}}^{\text{-1}}{\text{s}}^{\text{-1}}}{{\text{mol}}^{\frac{\text{3}}{\text{2}}}{\text{L}}^{\frac{\text{3}}{\text{2}}}}\\ {\text{=L}}^{\frac{\text{1}}{\text{2}}}{\text{mol}}^{\frac{\text{-1}}{\text{2}}}{\text{s}}^{\text{-1}}\\ \\ \left(\text{iv}\right){\text{Given rate=k[C}}_{\text{2}}{\text{H}}_{\text{5}}\text{Cl]}\\ \text{Therefore, order of the reaction=1}\\ \text{k=}\frac{\text{Rate}}{{\text{[C}}_{\text{2}}{\text{H}}_{\text{5}}\text{Cl]}}\\ \text{Dimension of k}\\ \text{=}\frac{{\text{mol L}}^{\text{-1}}{\text{s}}^{\text{-1}}}{{\text{mol L}}^{\text{-1}}}\\ {\text{=s}}^{\text{-1}}\end{array}$

Q.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.

Ans.

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 L^-1 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

=\frac{1}{2}\times 0.04mo{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

= 3.89×10^-9mol L^-1 s^-1

Q.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?

Ans.

Q.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.,

\text{Rate}=k{({\text{P}}_{{\text{CH}}_3{\text{OCH}}_3})}^{3/2}

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

Ans.

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

Unit of rate = bar min^-1

Q.5 Mention the factors that affect the rate of a chemical reaction.

Ans.

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

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

(ii) Temperature

(iii) Nature of reactants

(iv) Presence of a catalyst

(v) Surface area

Q.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?

Ans.

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

=4R

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

$\begin{array}{l}\begin{array}{l}\left(\text{ii}\right)\text{If the concentration of the reactant is reduced to half, i.e.}\left[\text{A}\right]\text{=}\frac{\text{1}}{\text{2}}\text{a then the rate}\\ \text{of the reaction would be}\end{array}\\ {\text{R}}^{\text{*}}\text{=k}{\left(\frac{\text{1}}{\text{2}}\text{a}\right)}^{\text{2}}\\ \text{=}\frac{\text{1}}{\text{4}}\\ \text{=}\frac{\text{1}}{\text{4}}\text{R}\\ \text{Therefore, the rate of the reaction would be reduced to =}\frac{{\text{1}}^{\text{th}}}{\text{4}}\end{array}$

Q.7 The following data were obtained during the first order thermal decomposition of SO₂Cl₂ at a constant volume.

${\mathrm{SO}}_2{\mathrm{Cl}}_2\left(\mathrm{g}\right)\to {\mathrm{SO}}_2\left(\mathrm{g}\right)+\mathrm{}{\mathrm{Cl}}_2\left(\mathrm{g}\right)$

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

Ans.

Q.8 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 E_a for this reaction and at what temperature will its half-period be 256 minutes?

Ans.

Q.9 What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?

Ans.

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

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

k = Ae^-E_a/RT

where, k is 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

Q.10 In a pseudo first order hydrolysis of ester in water, the following results were obtained:

Calculate the average rate of reaction between the time interval 30 to 60 seconds.

Ans.

Average rate of reaction between the time interval, 30 to 60 seconds,

$\begin{array}{l}=\frac{\text{d}\left[\text{Ester}\right]}{\text{dt}}\\ =\frac{0.31-0.17}{60-30}\\ =\frac{0.14}{30}\\ =\mathrm{}\text{4}.\text{67}\times \text{1}{0}^{-\text{3}}{\text{molL}}^{-\text{1}}{\text{s}}^{-\text{1}}\\ \end{array}$

Q.11 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?

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{The differential rate equation will be}\\ \text{–}\frac{\text{d[R]}}{\text{dt}}\text{=k}\left[\text{A}\right]{\text{[ B]}}^{\text{2}}\\ \\ \left(\text{ii}\right)\text{If the concentration of B is increased three times then}\\ \text{–}\frac{\text{d[R]}}{\text{dt}}\text{=k}\left[\text{A}\right]{\text{[3B]}}^{\text{2}}\\ \text{=9.k}\left[\text{A}\right]{\text{[B]}}^{\text{2}}\\ \text{Therefore, the rate of reaction will increase 9 times.}\\ \\ \left(\text{iii}\right)\text{When the concentrations of both A and B are doubled,}\\ \text{–}\frac{\text{d[R]}}{\text{dt}}\text{=k}\left[\text{A}\right]{\text{[B]}}^{\text{2}}\\ \text{=k}[2\text{A}]{\text{[2B]}}^{\text{2}}\\ \text{=8.k}\left[\text{A}\right]{\text{[B]}}^{\text{2}}\\ \text{Therefore, the rate of reaction will increase 8 times.}\end{array}$

Q.12 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:

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

Ans.

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

Therefore,

$\begin{array}{l}{\mathrm{r}}_0=\mathrm{k}{\left[\mathrm{A}\right]}^{\mathrm{x}}{\left[\mathrm{B}\right]}^{\mathrm{y}}\\ 5.07\times {10}^{-5}=\mathrm{k}{[0.20]}^{\mathrm{x}}{[0.30]}^{\mathrm{y}}----\left(\mathrm{i}\right)\\ 5.07\times {10}^{-5}=\mathrm{k}{[0.20]}^{\mathrm{x}}{[0.10]}^{\mathrm{y}}----\left(\mathrm{ii}\right)\\ 1.43\times {10}^{-4}=\mathrm{k}{[0.40]}^{\mathrm{x}}{[0.05]}^{\mathrm{y}}----\left(\mathrm{iii}\right)\\ \text{Dividing equation}\left(\mathrm{i}\right)\text{by}\left(\mathrm{ii}\right)\text{, we obtain}\\ \frac{5.07\mathrm{}\times \mathrm{}{10}^{-5}}{5.07\mathrm{}\times \mathrm{}{10}^{-5}}=\frac{\mathrm{k}{[0.20]}^{\mathrm{x}}{[0.30]}^{\mathrm{y}}}{\mathrm{k}{[0.20]}^{\mathrm{x}}{[0.10]}^{\mathrm{y}}}\\ \Rightarrow 1=\frac{{[0.30]}^{\mathrm{y}}}{{[0.10]}^{\mathrm{y}}}\\ \Rightarrow {\left(\frac{0.30}{0.10}\right)}^0={\left(\frac{0.30}{0.10}\right)}^{\mathrm{y}}\\ \Rightarrow \mathrm{y}=0\\ \text{Dividing equation}\left(\mathrm{iii}\right)\text{by}\left(\mathrm{ii}\right)\text{, we obtain}\\ \frac{1.43\mathrm{}\times \mathrm{}{10}^{-4}}{5.07\mathrm{}\times \mathrm{}{10}^{-5}}=\frac{\mathrm{k}{[0.40]}^{\mathrm{x}}{[0.05]}^{\mathrm{y}}}{\mathrm{k}{[0.20]}^{\mathrm{x}}{[0.30]}^{\mathrm{y}}}\\ \Rightarrow \frac{1.43\mathrm{}\times \mathrm{}{10}^{-4}}{5.07\mathrm{}\times \mathrm{}{10}^{-5}}=\frac{{[0.40]}^{\mathrm{x}}}{{[0.20]}^{\mathrm{x}}}\mathrm{}\left[\begin{array}{c}\mathrm{Since}\mathrm{}\mathrm{y}=0,\\ {[0.05]}^{\mathrm{y}}={[0.30]}^{\mathrm{y}}=1\end{array}\right]\\ \Rightarrow 2.821=\mathrm{}{2}^{\mathrm{x}}\\ \Rightarrow \log 2.821=\mathrm{xlog}\mathrm{}2\left(\mathrm{Takinglogon}\mathrm{}\mathrm{both}\mathrm{}\mathrm{sides}\right)\\ \Rightarrow \mathrm{x}=\frac{\log 2.821}{\log 2}\\ =1.496\\ =1.5\left(\text{approximately}\right)\\ \text{Hence, the order of the reaction with respect to A is 1.5 and with respect to B is zero.}\end{array}$

Q.13 The following results have been obtained during the kinetic studies of the reaction:

2A + B → C + D

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

Ans.

Let the order of the reaction with respect to A be x and with respect to B be y. Therefore, the rate of the reaction is given by,

$\begin{array}{l}\mathrm{Rate}=\mathrm{k}{\left[\mathrm{A}\right]}^{\mathrm{x}}{\left[\mathrm{B}\right]}^{\mathrm{y}}\\ \text{According to the question,}\\ {\text{6.0×10}}^{-3}=\mathrm{k}{[0.1]}^{\mathrm{x}}{[0.1]}^{\mathrm{y}}----\left(\mathrm{i}\right)\\ 7.2\times {10}^{-2}=\mathrm{k}{[0.3]}^{\mathrm{x}}{[0.2]}^{\mathrm{y}}----\left(\mathrm{ii}\right)\\ 2.88\times {10}^{-1}=\mathrm{k}{[0.3]}^{\mathrm{x}}{[0.4]}^{\mathrm{y}}----\left(\mathrm{iii}\right)\\ \text{Dividing equation}\left(\mathrm{iv}\right)\text{by}\left(\mathrm{i}\right)\text{, we obtain}\\ \frac{2.40\mathrm{}\mathrm{x}\mathrm{}{10}^{-2}}{6.0\mathrm{}\mathrm{x}\mathrm{}{10}^{-3}}=\frac{\mathrm{k}{[0.4]}^{\mathrm{x}}{[0.1]}^{\mathrm{y}}}{\mathrm{k}{[0.1]}^{\mathrm{x}}{[0.1]}^{\mathrm{y}}}\\ \Rightarrow 4=\frac{{[0.4]}^{\mathrm{x}}}{{[0.1]}^{\mathrm{x}}}\\ \Rightarrow 4={\left(\frac{0.4}{0.1}\right)}^{\mathrm{x}}\\ \Rightarrow {\left(4\right)}^1={\left(4\right)}^{\mathrm{x}}\\ \Rightarrow \mathrm{x}=1\\ \text{Dividing equation}\left(\mathrm{iii}\right)\mathrm{by}\left(\mathrm{ii}\right),\text{we obtain}\\ \frac{2.88\mathrm{}\mathrm{x}\mathrm{}{10}^{-1}}{7.2\mathrm{}\mathrm{x}\mathrm{}{10}^{-2}}=\frac{\mathrm{k}{[0.3]}^{\mathrm{x}}{[0.4]}^{\mathrm{y}}}{\mathrm{k}{[0.3]}^{\mathrm{x}}{[0.2]}^{\mathrm{y}}}\\ \Rightarrow 4={\left(\frac{0.4}{0.2}\right)}^{\mathrm{y}}\\ \Rightarrow 4=2^{\mathrm{y}}\\ \Rightarrow 2^2=2^{\mathrm{y}}\\ \Rightarrow \mathrm{y}=2\\ \text{Therefore, the rate law is rate = k}\left[\mathrm{A}\right]{\left[\mathrm{B}\right]}^2\\ \Rightarrow \mathrm{k}=\frac{\mathrm{Rate}}{\left[\mathrm{A}\right]{\left[\mathrm{B}\right]}^2}\\ \text{From experiment I, we obtain}\\ \text{k}=\frac{6.0\text{x}{10}^{-3}{\text{mol L}}^{-1}{\text{min}}^{-1}}{(0.1{\text{mol L}}^{-1}){(0.1{\text{mol L}}^{-1})}^2}\\ =6.0\mathrm{}{\mathrm{L}}^2\mathrm{}{\mathrm{mol}}^{-2}\mathrm{}{\min }^{-1}\\ \text{From experiment II, we obtain}\\ \mathrm{k}=\frac{7.2\mathrm{}\mathrm{x}\mathrm{}{10}^{-2}\mathrm{}\mathrm{mol}\mathrm{}{\mathrm{L}}^{-1}\mathrm{}{\min }^{-1}}{(0.3\mathrm{}\mathrm{mol}\mathrm{}{\mathrm{L}}^{-1}){(0.2\mathrm{}\mathrm{mol}\mathrm{}{\mathrm{L}}^{-1})}^2}\\ =6.0\mathrm{}{\mathrm{L}}^2\mathrm{}{\mathrm{mol}}^{-2}\mathrm{}{\min }^{-1}\\ \text{From experiment III, we obtain}\\ \mathrm{k}=\frac{2.88\mathrm{}\mathrm{x}\mathrm{}{10}^{-1}\mathrm{}\mathrm{mol}\mathrm{}{\mathrm{L}}^{-1}\mathrm{}{\min }^{-1}}{(0.3\mathrm{}\mathrm{mol}\mathrm{}{\mathrm{L}}^{-1}){(0.4\mathrm{}\mathrm{mol}\mathrm{}{\mathrm{L}}^{-1})}^2}\\ =6.0\mathrm{}{\mathrm{L}}^2\mathrm{}{\mathrm{mol}}^{-2}\mathrm{}{\min }^{-1}\\ \text{From experiment IV, we obtain}\\ \mathrm{k}=\frac{2.40\mathrm{}\mathrm{x}\mathrm{}{10}^{-2}\mathrm{}\mathrm{mol}\mathrm{}{\mathrm{L}}^{-1}\mathrm{}{\min }^{-1}}{(0.4\mathrm{}\mathrm{mol}\mathrm{}{\mathrm{L}}^{-1}){(0.1\mathrm{}\mathrm{mol}\mathrm{}{\mathrm{L}}^{-1})}^2}\\ =6.0\mathrm{}{\mathrm{L}}^2\mathrm{}{\mathrm{mol}}^{-2}\mathrm{}{\min }^{-1}\\ \mathrm{Therefore},\mathrm{rate}{\text{constant, k=6.0 L}}^2{\mathrm{mol}}^{-2}{\min }^{-1}\end{array}$

Q.14 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:

Ans.

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

Q.15 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

Ans.

$\begin{array}{l}\left(\mathrm{i}\right){\text{Half life, t}}_{1/2}=\frac{0.693}{\mathrm{k}}\\ =\frac{0.693}{200{\text{s}}^{-1}}\\ =3.47\times {10}^{–5}\mathrm{s}\left(\text{approximately}\right)\\ \\ \left(\mathrm{ii}\right){\text{Half life, t}}_{1/2}=\frac{0.693}{\mathrm{k}}\\ =\frac{0.693}{2{\text{min}}^{-1}}\\ =0.35\text{min}\left(\text{approximately}\right)\\ \\ \left(\mathrm{iii}\right){\text{Half life, t}}_{1/2}=\frac{0.693}{\mathrm{k}}\\ =\frac{0.693}{4{\mathrm{years}}^{-1}}\\ =0.173\mathrm{years}\left(\mathrm{approximately}\right)\end{array}$

Q.16 The half-life for radioactive decay of ¹⁴ C 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.

Ans.

$\begin{array}{l}\mathrm{k}=\frac{0.693}{{\mathrm{t}}_{1/2}}\\ \text{Here,}\\ \text{=}\frac{0.693}{5730}{\mathrm{years}}^{-1}\\ \mathrm{It}\text{is known that,}\\ \text{t=}\frac{2.303}{\mathrm{k}}\log \frac{{\left[\mathrm{R}\right]}_0}{\left[\mathrm{R}\right]}\\ =\frac{2.303}{\frac{0.693}{5730}}\log \frac{100}{80}\\ =1845\text{years}\left(\mathrm{approximatelly}\right)\\ \mathrm{Hence},\mathrm{the}\text{age of the sample is 1845 years.}\end{array}$

Q.17 The experimental data for decomposition of N₂O₅

[2N₂O₅ → 4NO₂ + O₂]

In gas phase at 318K are given below:

(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).

Ans.

(i)

(ii) Time corresponding to the concentration, 1.630×10²/2=mol L^-1 =81.5 mol L^-1 is the half life. From the graph, the half life is obtained as 1450 s.

(iii)

Q.18 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?

Ans.

$\begin{array}{l}\text{It is known that,}\\ \text{t=}\frac{2.303}{\mathrm{k}}\log \frac{{\left[\mathrm{R}\right]}_0}{\left[\mathrm{R}\right]}\\ =\frac{2.303}{60{\mathrm{s}}^{-1}}\log \frac{1}{1/16}\\ =\frac{2.303}{60{\mathrm{s}}^{-1}}\log 16\\ =4.6\times {10}^{-2}\mathrm{s}\left(\text{approximately}\right)\\ {\text{Hence, the required time is 4.6×10}}^{-2}\mathrm{s}\end{array}$

Q.19 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.

Ans.

$\begin{array}{l}\mathrm{k}=\frac{0.693}{{\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{0.693}{28.1}{\mathrm{y}}^{-1}\\ \text{Here, It is known that,}\\ \mathrm{t}=\frac{2.303}{\mathrm{k}}\log \frac{{\left[\mathrm{R}\right]}_0}{\left[\mathrm{R}\right]}\\ \Rightarrow 10=\frac{2.303}{\frac{0.693}{28.1}}\log \frac{1}{\left[\mathrm{R}\right]}\\ \Rightarrow 10=\frac{2.303}{\frac{0.693}{28.1}}(-\log \left[\mathrm{R}\right])\\ \Rightarrow \log \left[\mathrm{R}\right]=-\frac{10\mathrm{}\mathrm{x}\mathrm{}0.693}{2.303\mathrm{}\mathrm{x}\mathrm{}28.1}\\ \Rightarrow \left[\mathrm{R}\right]=\mathrm{antilog}(-0.1071)\\ =\mathrm{antilog}(\stackrel{-}{1}.8929)\\ =0.7814\mathrm{\mu g}\\ \text{Therefore, 0.7814}\mathrm{\mu }{\text{g of}}^{90}\mathrm{Sr}\text{will remain after 10 years.}\\ \text{Again,}\\ \text{t=}\frac{2.303}{\mathrm{k}}\log \frac{{\left[\mathrm{R}\right]}_0}{\left[\mathrm{R}\right]}\\ \mathrm{}\Rightarrow 60=\frac{2.303}{\frac{0.693}{28.1}}\log \frac{1}{\left[\mathrm{R}\right]}\\ \Rightarrow \log \left[\mathrm{R}\right]=-\frac{60\mathrm{}\mathrm{x}\mathrm{}0.693}{2.303\mathrm{}\mathrm{x}\mathrm{}28.1}\\ \Rightarrow \left[\mathrm{R}\right]=\mathrm{antilog}(-0.6425)\\ =\mathrm{antilog}(\stackrel{-}{1}.3575)\\ =0.2278\mathrm{\mu g}\\ \text{Therefore, 0.2278}\mathrm{\mu }{\text{g of}}^{90}\mathrm{Sr}\text{will remain after 60 yeras.}\end{array}$

Q.20 For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

Ans.

Q.21 A first order reaction takes 40 min for 30% decomposition. Calculate t_1/2.

Ans.

$\begin{array}{l}\mathrm{For}\text{a first order reaction,}\\ \text{t=}\frac{2.303}{\mathrm{k}}\log \frac{{\left[\mathrm{R}\right]}_0}{\left[\mathrm{R}\right]}\\ \mathrm{k}=\frac{2.303}{40\min }\log \frac{100}{100-30}\\ =\frac{2.303}{40\min }\log \frac{10}{7}\\ =8.918\times {10}^{-3}{\min }^{-1}\\ {\text{Therefore, t}}_{1/2}\text{of the decomposition reaction is}\\ {\text{t}}_{1/2}=\frac{0.693}{\mathrm{k}}\\ =\frac{0.693}{8.918\times {10}^{-3}}\min \\ =77.7\min \left(\text{approximately}\right)\\ \end{array}$

Q.22 For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

Calculate the rate constant.

Ans.

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

$\begin{array}{l}\underset{\begin{array}{l}\mathrm{At}\text{t=0}\\ \text{At t=t}\end{array}}{{\left({\mathrm{CH}}_3\right)}_2\mathrm{CHN}}=\underset{\begin{array}{l}{\mathrm{P}}_0\\ {\mathrm{P}}_{0-\mathrm{p}}\end{array}}{\mathrm{NCH}{\left({\mathrm{CH}}_3\right)}_{2\left(\mathrm{g}\right)}}\to \underset{\begin{array}{l}0\\ \mathrm{p}\end{array}}{{\mathrm{N}}_{2\left(\mathrm{g}\right)}}+\underset{\begin{array}{l}0\\ \mathrm{p}\end{array}}{{\mathrm{C}}_6{\mathrm{H}}_{14\left(\mathrm{g}\right)}}\\ \mathrm{After}\mathrm{time},\mathrm{t},\mathrm{total}\mathrm{pressure},{\text{P}}_{\mathrm{t}}=({\mathrm{P}}_0-\mathrm{p})+\mathrm{p}+\mathrm{p}\\ \Rightarrow {\mathrm{P}}_{\mathrm{t}}={\mathrm{P}}_0+\mathrm{p}\\ \Rightarrow {\mathrm{P}}_{\mathrm{}}={\mathrm{P}}_{\mathrm{t}}-{\mathrm{P}}_0\\ {\text{Therefore, P}}_0-\mathrm{p}={\mathrm{P}}_0-({\mathrm{P}}_{\mathrm{t}}-{\mathrm{P}}_0)\\ =2{\mathrm{P}}_0-{\mathrm{P}}_{\mathrm{t}}\\ \text{for a first order reaction,}\\ \text{k=}\frac{2.303}{\mathrm{t}}\log \frac{{\mathrm{P}}_0}{{\mathrm{P}}_0-\mathrm{p}}\\ =\frac{2.303}{\mathrm{t}}\log \frac{{\mathrm{P}}_0}{2{\mathrm{P}}_0-{\mathrm{p}}_{\mathrm{t}}}\\ \text{When t= 360 s, k=}\frac{2.303}{360\mathrm{s}}\log \frac{35.0}{2\times 35.0-54.0}\\ =2.175\times {10}^{-3}{\mathrm{s}}^{-1}\\ \mathrm{k}=\frac{2.303}{720\mathrm{s}}\log \frac{35.0}{2\times 35.0-63.0}\\ \text{When t = 720s,}\\ \text{=2}..{\text{235×10}}^{-3}{\mathrm{s}}^{-1}\\ \text{Hence, the average value of rate constant is}\\ \text{k=}\frac{(2.175\times {10}^{-3})+(2.235\times {10}^{-3})}{2}{\mathrm{s}}^{-1}\\ =2.21\times {10}^{-3}{\mathrm{s}}^{-1}\end{array}$

Note: There is a slight variation in this Solution and the one given in the NCERT textbook.

Q.23 The rate constant for the decomposition of N₂O₅ at various temperatures is given below:

Draw a graph between ln k and 1/T and calculate the values of A and E_a.
Predict the rate constant at 30⁰ and 50⁰C.

Ans.

From the given data, we obtain

$\begin{array}{l}\text{Slope of the line,}\\ \frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}=–12.301\mathrm{}\mathrm{K}\\ \text{According to Arrhenius equation,}\\ \mathrm{Slope}=-\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\\ \Rightarrow {\mathrm{E}}_{\mathrm{a}}=-\mathrm{Slope}\mathrm{}\times \mathrm{}\mathrm{R}\\ =-(-12.301\mathrm{}\mathrm{K})\times \mathrm{}(8.314\mathrm{}{\mathrm{JK}}^{-1}\mathrm{}{\mathrm{mol}}^{-1})\\ \text{Again,}\\ \ln \mathrm{}\mathrm{k}=\ln \mathrm{}\mathrm{A}-\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{RT}}\\ \ln \mathrm{}\mathrm{A}=\ln \mathrm{}\mathrm{K}+\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{RT}}\\ \text{When T=273k,}\\ \text{ln k =-7.147}\\ \text{Then, In A=-7.147+}\frac{102.27\times {10}^3}{8.314\times 273}\\ =37.911\\ {\text{Therefore, A=2.91×10}}^6\\ \text{When T=30+273k=303k,}\\ \frac{1}{\mathrm{T}}=0.00333\mathrm{k}=3.3\times {10}^{-3}\mathrm{K}\\ \text{Then, at}\frac{1}{\mathrm{T}}=3.3\times {10}^{-3}\mathrm{k}\\ \mathrm{lnk}=\mathrm{}-2.8\\ {\text{Therefore\hspace{0.17em}k=6.08×10}}^{-2}{\mathrm{s}}^{-1}\\ \text{Again when t=50+273k=323k,}\\ \frac{1}{\mathrm{T}}=0.0031\mathrm{}\mathrm{K}=3.1\mathrm{}\times \mathrm{}{10}^{-3}\mathrm{}\mathrm{K}\\ \mathrm{Then},\mathrm{at}\frac{1}{\mathrm{T}}=3.1\mathrm{}\times \mathrm{}{10}^{-3}\mathrm{}\mathrm{K}\\ \mathrm{lnk}=\mathrm{}-0.5\\ \mathrm{Therefore},\mathrm{K}=0.607{\mathrm{s}}^{-1}\end{array}$

Q.24 The rate constant for the decomposition of hydrocarbons is 2.418 × 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.

Ans.

$\begin{array}{l}\mathrm{K}=\text{2}.\text{418}\times \text{1}{0}^{-\text{5}}{\text{s}}^{-\text{1}}\\ \mathrm{T}\mathrm{}=\mathrm{}\text{546K}\\ {\mathrm{E}}_{\text{a}}=\text{179}.{\text{9 kJmol}}^{-\text{1}}\mathrm{}=\text{179}.\text{9}\times \mathrm{}\text{1}{0}^{\text{3}}{\text{Jmol}}^{-\text{1}}\\ \mathrm{According}\text{to the Arrhenius equation,}\\ \mathrm{k}={\mathrm{Ae}}^{-{\mathrm{E}}_{\mathrm{a}}/\mathrm{RT}}\\ \Rightarrow \mathrm{lnk}=\mathrm{lnA}-\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{RT}}\\ \Rightarrow \mathrm{logk}=\mathrm{logA}-\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{}\mathrm{RT}}\\ \Rightarrow \mathrm{logA}=\mathrm{logK}+\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{}\mathrm{RT}}\\ =\log (2.418\mathrm{}\times \mathrm{}{10}^{-5}{\mathrm{s}}^{-1})+\frac{179.9\mathrm{}\times \mathrm{}{10}^3\mathrm{}\mathrm{J}\mathrm{}{\mathrm{mol}}^{-1}}{2.303\mathrm{}\times \mathrm{}8.314\mathrm{}{\mathrm{JK}}^{-1}\mathrm{}{\mathrm{mol}}^{-1}\mathrm{}\times \mathrm{}546\mathrm{}\mathrm{K}}\\ =\mathrm{}(0.\text{3835}-\mathrm{}\text{5})+\mathrm{}\text{17}.\text{2}0\text{82}\\ =\mathrm{}\text{12}.\text{5917}\\ \text{Therefore},\text{A}=\text{antilog}(\text{12}.\text{5917})\\ =\text{3}.\text{9}\times \text{1}{0}^{\text{12}}{\text{s}}^{-\text{1}}\mathrm{}\left(\text{approximately}\right)\end{array}$

Q.25 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.

Ans.

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.

$\begin{array}{l}\mathrm{K}=\frac{2.303}{\mathrm{t}}\log \frac{{\left[\mathrm{A}\right]}_0}{\left[\mathrm{A}\right]}\\ \text{Therefore,}\\ \Rightarrow 2.0\mathrm{}\times \mathrm{}{10}^{-2}{\mathrm{s}}^{-1}=\frac{2.303}{100\mathrm{}\mathrm{s}}\log \frac{1.0}{\left[\mathrm{A}\right]}\\ \Rightarrow 2.0\mathrm{}\times \mathrm{}{10}^{-2}{\mathrm{s}}^{-1}=\frac{2.303}{100\mathrm{}\mathrm{s}}(-\log \left[\mathrm{A}\right])\\ \Rightarrow -\log \left[\mathrm{A}\right]=\frac{2.0\mathrm{}\times \mathrm{}{10}^{-2}\mathrm{}\times \mathrm{}100}{2.303}\\ \Rightarrow \left[\mathrm{A}\right]=\mathrm{antilog}(-\frac{2.0\mathrm{}\mathrm{x}\mathrm{}{10}^{-2}\mathrm{}\times \mathrm{}100}{2.303})\end{array}$

= 0.135 mol L^-1 (approximately)

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

Q.26 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?

Ans.

Q.27 The decomposition of hydrocarbon follows the equation k = (4.5 × 10¹¹ s^-1) e^{-28000 K/T} Calculate E_a.

Ans.

The given equation is

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

Arrhenius equation is given by,

$\begin{array}{l}\mathrm{k}={\mathrm{Ae}}^{-{\mathrm{E}}_{\mathrm{a}}/\mathrm{RT}}----\left(\mathrm{ii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and}\left(\mathrm{ii}\right),\text{we obtain}\\ \frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{RT}}=\frac{28000\mathrm{}\mathrm{K}}{\mathrm{T}}\\ \Rightarrow {\mathrm{E}}_{\mathrm{a}}=\mathrm{R}\mathrm{}\times \mathrm{}28000\mathrm{}\mathrm{K}\\ =\mathrm{}\text{8}.{\text{314JK}}^{-\text{1}}{\text{mol}}^{-\text{1}}\times \text{28}000\text{K}\\ =\mathrm{}{\text{232792Jmol}}^{-\text{1}}\\ =\mathrm{}\text{232}.{\text{792kJmol}}^{-\text{1}}\end{array}$

Q.28 The decomposition of A into product has value of k as 4.5 × 10³ s^-1 at 10⁰C and energy of activation 60 kJ mol^-1. At what temperature would k be 1.5 × 10⁴ s^-1?

Ans.

Q.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 × 10¹⁰ s^-1. Calculate k at 318 K and E_a.

Ans.

$\begin{array}{l}\mathrm{For}\text{a first order reaction,}\\ \mathrm{t}=\frac{2.303}{\mathrm{k}}\log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\\ \text{At 298k,}\mathrm{t}=\frac{2.303}{\mathrm{k}}\log \frac{100}{90}\\ =\frac{0.1054}{\mathrm{k}}\\ \text{At 308k,}{\mathrm{t}}^1=\frac{2.303}{{\mathrm{k}}^1}\log \frac{100}{75}\\ =\frac{2.2877}{\mathrm{k}}\\ \text{According to the question,}\\ \mathrm{t}=\mathrm{}{\mathrm{t}}^1\\ \Rightarrow \frac{0.1054}{\mathrm{k}}=\frac{0.2877}{{\mathrm{k}}^1}\\ \Rightarrow \frac{{\mathrm{k}}^1}{\mathrm{k}}=2.7296\\ \text{From Arrhenius equation, we obtain}\\ \log \frac{{\mathrm{k}}^1}{\mathrm{k}}=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{}\mathrm{R}}\left(\frac{{\mathrm{T}}^1-\mathrm{T}}{{\mathrm{TT}}^1}\right)\\ \text{log}(2.7296)=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{}\times \mathrm{}8.314}\left(\frac{308-298}{298\mathrm{}\times \mathrm{}308}\right)\\ {\mathrm{E}}_{\mathrm{a}}=\frac{2.303\mathrm{}\times \mathrm{}8.314\mathrm{}\times \mathrm{}298\mathrm{}\times \mathrm{}308\mathrm{}\times \log (2.7296)}{308-298}\\ =76640.096\mathrm{}\mathrm{J}\mathrm{}{\mathrm{mol}}^{-1}\\ =76.64\mathrm{}\mathrm{kJ}\mathrm{}{\mathrm{mol}}^{-1}\\ \mathrm{To}\text{calculate k at 318 k,}\\ {\text{It is given that, A=4×10}}^{10}{\mathrm{s}}^{-1},\mathrm{T}=318\mathrm{K}\\ \text{Again, from Arrhenius equation, we obtain}\\ \mathrm{logk}=\mathrm{logA}-\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{}\mathrm{RT}}\\ =\log (4\mathrm{}\times \mathrm{}{10}^{10})-\frac{76.64\mathrm{}\times \mathrm{}{10}^3}{2.303\mathrm{}\times \mathrm{}8.314\mathrm{}\times \mathrm{}318}\\ =(0.6021+10)-12.5876\\ =\; –1.9855\\ \mathrm{Therefore}\text{, k = Antilog}(-1.9855)\\ =1.034\times {10}^{-2}{\mathrm{s}}^{-1}\end{array}$

Q.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.

Ans.

$\begin{array}{l}\text{From Arrhenius equation, we obtain}\\ \log \frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{}\mathrm{R}}\left(\frac{{\mathrm{T}}_2-{\mathrm{T}}_1}{{\mathrm{T}}_1{\mathrm{T}}_2}\right)\\ {\text{It is given that, k}}_2=4{\mathrm{k}}_1\\ {\text{T}}_{\text{1}}\mathrm{}=\text{293K}\\ {\text{T}}_{\text{2}}\mathrm{}=\mathrm{}\text{313 K}\\ \text{Therefore,}\log \frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{}\times \mathrm{}8.314}\left(\frac{313-293}{293\mathrm{}\times \mathrm{}313}\right)\\ \Rightarrow 0.6021=\frac{20\mathrm{}\times \mathrm{}{\mathrm{E}}_{\mathrm{a}}\mathrm{}}{2.303\times \mathrm{}8.314\mathrm{}\times \mathrm{}293\mathrm{}\times 313\mathrm{}}\\ \Rightarrow {\mathrm{E}}_{\mathrm{a}}=\frac{0.6021\mathrm{}\times \mathrm{}2.303\mathrm{}\times \mathrm{}8.314\mathrm{}\times \mathrm{}293\mathrm{}\times \mathrm{}313\mathrm{}}{20}\\ =52863.33\mathrm{}\mathrm{J}\mathrm{}{\mathrm{mol}}^{-1}\\ =52.86\mathrm{}{\mathrm{kJmol}}^{-1}\\ {\text{Hence, the required energy of activation is 52.86 kJmol}}^{-1}.\end{array}$