Question

The rate of decomposition for methyl nitrite and ethyl nitrite can be given in terms of rate constant k₁ and k₂ respectively. The energy of activation for the two reactions are $152.30kJmo{l}^{-1}and157.7kJmo{l}^{-1}$ as well as frequency factors are $10^{13}$ and $10^{14}$ respectively for the decomposition of methyl and ethyl nitrite. Calculate the temperature at which rate constant will be same for the two reactions.

• A. $256\text{K}$
• B. $354\text{K}$
• C. $282\text{K}$
• D. $674\text{K}$

Answer 1

Answer: (C)

$\mathrm{k}=\mathrm{A}{\mathrm{e}}^{\frac{-{\mathrm{E}}_{\mathrm{a}}}{\mathrm{RT}}}$
For methyl nitrite ${\mathrm{k}}_1=10^{13}{\mathrm{e}}^{[-152300/(8.314\times \mathrm{T})]}$
For ethyl nitrite ${\mathrm{k}}_2=10^{14}{\mathrm{e}}^{[-157700/(8.314\times \mathrm{T})]}$
If ${\mathrm{k}}_1={\mathrm{k}}_2$ then
$10^{13}{\mathrm{e}}^{[-152300/(8.314\times \mathrm{T})]}=10^{14}{\mathrm{e}}^{[-157700/(8.314\times \mathrm{T})]}$
Or $2.303\log 10=\frac{157700-152300}{8.314\times \mathrm{T}}$
$\mathrm{T}=282\mathrm{K}$