Chapter 13, Problem 13.39QP

Some reactions are described as parallel in that the reactant simultaneously forms different products with different rate constants. An example is

$\text{A}\underset{k_2}{\overset{k_1}{\to }}\text{B}$

and

$\text{A}\stackrel{-}{\to }\text{C}$

The activation energies are 45.3 kJ/mol for k₁ and 69.8 kJ/mol for k₂. If the rate constants are equal at 320 K, at what temperature will k₁/k₂ = 2.00?

Interpretation Introduction

Interpretation:

If the rate constants of given reactions are equal at $320K$, then the temperature will $\frac{{\text{k}}_{\text{1}}}{{\text{k}}_{\text{2}}}=\text{2}\text{.00}$ has to be identified.

Concept introduction:

Arrhenius equation:

Arrhenius equation is a formula that represents the temperature dependence of reaction rates

The Arrhenius equation can be represented as follows

$k=A{e}^{-E_a/RT}$

$E_a$ is the activation energy ( it is the amount of energy needed to start reaction) and it’s unit is kJ/mol

R represents the universal gas constant and it has the value of 8.314 J/K.mol

T represents the absolute temperature

If  two rate constants at two temperture are known then the activation energy $\left(E_a\right)$ can be  calculated using the following equation,

$\ln \left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$  $\text{R}\text{is}\text{the}\text{gas}\text{constant}\text{=}\text{8}\text{.3145}\text{J/mol}\text{K}$

Explanation of Solution

Some reactions are described as parallel in that the reactant simultaneously forms different products with different rate constants.  Given example for such type of reactions are,

$\begin{array}{l}A\stackrel{k_1}{\to }B\\ \\ A\to C\end{array}$

Activation energy for $k_{1}is45.3kJ/mol$ and $69.8kJ/molfor{k}_2$

If the rate constants of given reactions are equal at $320K$, then the temperature will $\frac{{\text{k}}_{\text{1}}}{{\text{k}}_{\text{2}}}=\text{2}\text{.00}$ can be determined as follows,

Arrhenius equation is a formula that represents the temperature dependence of reaction rates

The Arrhenius equation can be represented as follows

$k=A{e}^{-E_a/RT}$

The frequency factor A for each of the reaction can be determined as follows using the above equation,

$\begin{array}{c}{k}_1=A_1{e}^{-E_a/RT}=A_1{e}^{-\left(\frac{45300J/mol}{\left(8.314J/molK\right)\left(320K\right)}\right)}\\ =\left(4.03\times {10}^{-8}\right)A_1\\ A_1=\frac{k_1}{4.03\times {10}^{-8}}\end{array}$

Similarly frequency factor for reaction $k_2$ can be determined.

$\begin{array}{c}{k}_2=A_2{e}^{-E_a/RT}=A_2{e}^{-\left(\frac{69800J/mol}{\left(8.314J/molK\right)\left(320K\right)}\right)}\\ =\left(4.04\times {10}^{-12}\right)A_1\\ A_2=\frac{k_2}{4.04\times {10}^{-12}}\end{array}$

Since $k_1=k_2$ at $320K$, we can write,

$A_2=\frac{k_1}{4.04\times {10}^{-12}}$

Assuming, the frequency factor is temperature independent and dividing $k_1$ by $k_2$ to solve for the temperature at which ${\text{k}}_{\text{1}}{\text{/k}}_{\text{2}}\text{=}\text{2}\text{.00}$

$\frac{k_1}{k_2}=\frac{A_1{e}^{-E_{a1}/RT}}{A_2{e}^{-E_{a2}/RT}}=\left(\frac{\frac{k_1}{4.03\times {10}^{-8}}}{\frac{k_1}{4.04\times {10}^{-12}}}\right)\left(\frac{e^{-E_{a1}/RT}}{e^{-E_{a2}/RT}}\right)$

$\begin{array}{c}2.00=\left(1.002\times {10}^{-4}\right)\left(\frac{e^{-E_{a1}/RT}}{e^{-E_{a2}/RT}}\right)\\ 1.996\times {10}^4=\left(\frac{e^{-E_{a1}/RT}}{e^{-E_{a2}/RT}}\right)\end{array}$

Take natural logarithm of both sides of the equation.

$\begin{array}{c}\text{ln}\left(\text{1}\text{.996}\text{×}{\text{10}}^{\text{4}}\right)=\frac{{\text{-E}}_{\text{a1}}}{\text{RT}}\text{-}\left(\frac{{\text{-E}}_{\text{a2}}}{\text{RT}}\right)\text{=}\frac{\text{1}}{\text{RT}}\left({\text{E}}_{\text{a2}}\text{-}{\text{E}}_{\text{a1}}\right)\\ \text{9}\text{.901}=\frac{\text{1}}{\left(\text{8}\text{.314}\text{J/mol}\text{K}\right)\left(\text{T}\right)}\left(\text{69800}\text{-}\text{45300}\right)\text{J/mol}\\ \text{82}\text{.32}\text{T}=\text{24500}\\ \\ T=298K\end{array}$

If the rate constants of given reactions are equal at $320K$, then at $298\text{}K$ $\frac{{\text{k}}_{\text{1}}}{{\text{k}}_{\text{2}}}=\text{2}\text{.00}$.