Question

Calculate the temperature at which rate constant will be same for the two reactions, If the rate of decomposition for methyl nitrite and ethyl nitrite can be given in terms of rate constant $k_1$ and $k_2$ respectively. The energy of activation for the two reactions are $152.30kJ/mo{l}^{-1}and157.7kJ/mo{l}^{-1}$ as well as frequency factors are $10^{13}$ and $10^{14}$ respectively for the decomposition of methyl and ethyl nitrite.

• A. 256 K
• B. 354 K
• C. 282 K
• D. 674 K

Answer 1

Answer: (C)

$k=A{e}^{\frac{-E_a}{RT}}$
For methyl nitrite $k_1={\left(10\right)}^{13}{e}^{\left[-152300/\left(8.314\times T\right)\right]}$
For ethyl nitrite $k_2={\left(10\right)}^{14}{e}^{\left[-157700/\left(8.314\times T\right)\right]}$
If $k_1=k_2$ then
${\left(10\right)}^{13}{e}^{\left[-152300/\left(8.314\times T\right)\right]}={\left(10\right)}^{14}{e}^{\left[-157700/\left(8.314\times T\right)\right]}$
Or $2.303log10=\frac{157700-152300}{8.314\times T}$
$T=282K$