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 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 =282 K $