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 the 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)

$k=Ae\frac{-E_a}{RT}$
For methyl nitrite $k_1=10^{13}{e}^{[-152300/(8.314\times T)]}$
For ethyl nitrite $k_2=10^{14}{e}^{[-157700/(8.314\times T)]}$
If $k_1=k_2$ then
$10^{13}{e}^{[-152300/(8.314\times T)]}=10^{14}{e}^{[-157700/(8.314\times T)]}$
Or $2.303\log 10=\frac{157700-152300}{8.314\times T}$
$T=282K$