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 $152.30 \, kJ \, mol^{- 1} \, and \, 157.7 \, kJ \, mol^{- 1}$ 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} $