Question

The rate of decomposition for methyl nitrite and ethylnitrite can be given in terms of rate constant $k_1$ and $k_2$ respectively. The energy of activation for the two reactions are $152.30{\text{kJ mol}}^{–1}$ and $157.7{\text{kJ mol}}^{–1}$ as well as frequency factors are ${10}^{13}$ and ${10}^{14}$ respectively for the decomposition of methyl and ethylnitrite. Calculate the temperature at which rate constant will be same for the two reactions.

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

Answer 1

The correct option is D $\text{282 K}$

According to Arrhenius equation,
$k=A{e}^{-\frac{E_a}{RT}}$
Where,
$E_a=\text{activation energy}$
$A=\text{frequency factor}$
For methyl nitrite $k_1={10}^{13}{e}^{\frac{–152300}{(8.314\times T)}}$
For ethyl nitrite
$k_2={10}^{14}{e}^{\frac{–157700}{(8.314\times T)}}$
If $k_1=k_2$ then,
${10}^{13}{e}^{\frac{–152300}{(8.314\times T)}}={10}^{14}{e}^{\frac{–157700}{(8.314\times T)}}$
$\Rightarrow e^{\frac{–152300}{(8.314\times T)}}=10e^{\frac{–157700}{(8.314\times T)}}$
$\Rightarrow \frac{–152300}{(8.314\times T)}=ln10-\frac{157700}{(8.314\times T)}$
$\Rightarrow 2.303\log 10=\frac{157700-152300}{8.314\times T}$
$T=282.02K$