Chapter 11, Problem 52QRT

(a)

Interpretation Introduction

Interpretation:

Activation energy and frequency factor for the given reaction has to be calculated.

Concept Introduction:

The temperature dependence of rate constant can be explained through Arrhenius equation.

  $k=A{e}^{\frac{-E_a}{RT}}$

Where $k$ the rate is constant, $A$ is the frequency factor, $E_a$ is the activation energy, $R$ is the universal gas constant and $T$ is the temperature.

The logarithmic form of Arrhenius is given below.

  $\text{ln}\frac{{\text{k}}_{\text{2}}}{{\text{k}}_{\text{1}}}\text{=}\frac{{\text{E}}_{\text{a}}}{\text{R}}\left(\frac{\text{1}}{{\text{T}}_{\text{1}}}-\frac{\text{1}}{{\text{T}}_{\text{2}}}\right)$

Where, ${\text{k}}_{\text{1}}$ and ${\text{k}}_{\text{2}}$ are the rate constants at temperature ${\text{T}}_{\text{1}}$ and ${\text{T}}_{\text{2}}$ respectively.

Answer to Problem 52QRT

The activation energy is calculated as $29.3kJ/mol$ and the value of frequency factor is $7.55\times 10^{3}Lmo{l}^{-1}{s}^{-1}.$

Explanation of Solution

The activation energy can be calculated using following formula,

  $\text{ln}\frac{{\text{k}}_{\text{2}}}{{\text{k}}_{\text{1}}}\text{=}\frac{{\text{E}}_{\text{a}}}{\text{R}}\left(\frac{\text{1}}{{\text{T}}_{\text{1}}}-\frac{\text{1}}{{\text{T}}_{\text{2}}}\right)$

Given,

    $\begin{array}{c}{\text{k}}_{\text{1}}=2.34\times {10}^{-2}Lmo{l}^{-1}{s}^{-1}\\ \\ {\text{k}}_{\text{2}}=1.52\times {10}^{-1}Lmo{l}^{-1}{s}^{-1}\\ \\ {\text{T}}_{\text{1}}={10}^{o}C=283.15K\\ \\ {\text{T}}_{\text{2}}={60}^{o}C=333.15K\end{array}$

On substituting these values in the above equation gives the activation energy as shown below.

    $\begin{array}{c}\text{ln}\frac{{\text{k}}_{\text{2}}}{{\text{k}}_{\text{1}}}\text{=}\frac{{\text{E}}_{\text{a}}}{\text{R}}\left(\frac{\text{1}}{{\text{T}}_{\text{1}}}-\frac{\text{1}}{{\text{T}}_{\text{2}}}\right)\\ \\ \text{ln}\frac{\left(1.52\times {10}^{-1}Lmo{l}^{-1}{s}^{-1}\right)}{\left(2.34\times {10}^{-2}Lmo{l}^{-1}{s}^{-1}\right)}\text{=}\frac{{\text{E}}_{\text{a}}}{0.008314kJ/K/mol}\left(\frac{\text{1}}{283.15K}-\frac{\text{1}}{333.15K}\right)\\ \\ \ln \left(6.49\right)={\text{E}}_{\text{a}}\times \left(0.0637k{J}^{-1}mol\right)\\ \\ {\text{E}}_{\text{a}}=29.3kJ/mol\end{array}$

The activation energy is calculated as $29.3kJ/mol.$

Frequency factor can be calculated using the two given temperature.

Frequency factor calculated using $T_1$ temperature is given below.

  $\begin{array}{c}{k}_1=A{e}^{\frac{-E_a}{R{T}_1}}\\ \\ 2.34\times {10}^{-2}Lmo{l}^{-1}{s}^{-1}=A{e}^{\frac{-29.3kJ/mol}{\left(0.008134kJ/K/mol\right)\times 283.15K}}\\ \\ A=\left(2.34\times {10}^{-2}Lmo{l}^{-1}{s}^{-1}\right)e^{\frac{29.3kJ/mol}{\left(0.008134kJ/K/mol\right)\times 283.15K}}\\ \\ A=\left(2.34\times {10}^{-2}Lmo{l}^{-1}{s}^{-1}\right)\times e^{12.7}\\ \\ =7.66\times 10^{3}Lmo{l}^{-1}{s}^{-1}\end{array}$

Frequency factor calculated using $T_2$ temperature is given below.

  $\begin{array}{c}{k}_2=A{e}^{\frac{-E_a}{R{T}_2}}\\ \\ 1.52\times {10}^{-1}Lmo{l}^{-1}{s}^{-1}=A{e}^{\frac{-29.3kJ/mol}{\left(0.008134kJ/K/mol\right)\times 333.15K}}\\ \\ A=\left(1.52\times {10}^{-1}Lmo{l}^{-1}{s}^{-1}\right)e^{\frac{29.3kJ/mol}{\left(0.008134kJ/K/mol\right)\times 333.15K}}\\ \\ A=\left(1.52\times {10}^{-1}Lmo{l}^{-1}{s}^{-1}\right)\times e^{10.8}\\ \\ =7.45\times 10^{3}Lmo{l}^{-1}{s}^{-1}\end{array}$

Therefore the average value of frequency factor is $7.55\times 10^{3}Lmo{l}^{-1}{s}^{-1}.$

(b)

Interpretation Introduction

Interpretation:

Rate constant of the reaction at a temperature of ${100}^{o}C$ has to be estimated.

Concept Introduction:

Refer to part (a).

Answer to Problem 52QRT

The value of the rate constant is calculated as $0.597Lmo{l}^{-1}{s}^{-1}.$

Explanation of Solution

Rate constant can be calculated using Arrhenius equation.

Given,

    $\begin{array}{l}A=\text{7}{\text{.55×10}}^{\text{3}}\text{L}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\\ \\ E_a=\text{29}\text{.3}\text{kJ/mol}\\ \\ R=0.008314kJ/K/mol\\ \\ T={100}^{o}C=373.15K\end{array}$

Substituting these values gives the value of rate constant as follows,

    $\begin{array}{c}k=A{e}^{\frac{-E_a}{RT}}\\ \\ =\left(\text{7}{\text{.55×10}}^{\text{3}}\text{L}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\right)e^{\frac{-\left(\text{29}\text{.3}\text{kJ/mol}\right)}{\left(0.008314kJ/K/mol\right)\times \left(373.15K\right)}}\\ \\ =0.597Lmo{l}^{-1}{s}^{-1}.\end{array}$

The value of the rate constant is calculated as $0.597Lmo{l}^{-1}{s}^{-1}.$