Question

For the reaction, $ \mathrm{aA} + \mathrm{bB} \underset{ }{\overset{ }{\to }} \mathrm{cC} + \mathrm{dD}$, the plot of $ \mathrm{log} k$ vs $ 1/\mathrm{T}$ is given below:

The temperature at which the rate constant of the reaction is $ {10}^{-4}{\mathrm{s}}^{-1}$is ________ $ \mathrm{K}$. [Rounded off to the nearest integer)

[Given: The rate constant of the reaction is $ {10}^{-5 }{\mathrm{s}}^{-1}$ at $ 500 \mathrm{K}$]

slope = -10000 K 1T —» 

Answer 1

The temperature of the reaction can be calculated as:

Step 1:

${\log }_{10}K={\log }_{10}A-\frac{E_a}{2.303RT}$

Step 2:

Slope $=\frac{E_a}{2.303R}=10000$

Step 3:

$$\begin{gathered} {\log }_{10}(k_1/k_2)=\frac{E_a}{2.303R}\times \left[\frac{1}{T_1}-\frac{1}{T_2}\right] \\ {\log }_{10}(k_1/k_2)=10000\times \left[\frac{1}{500}-\frac{1}{T}\right] \\ 1=10000\times \left[\frac{1}{500}-\frac{1}{T}\right] \\ \frac{1}{10000}=\frac{1}{500}-\frac{1}{T} \\ \frac{1}{T}=\frac{20-1}{10000} \\ =\frac{19}{10000} \\ T=\raisebox{1ex}{$10000$}\!\left/ \!\raisebox{-1ex}{$19$}\right. \\ T=526K \end{gathered}$$

The temperature at which the rate constant of the reaction is ${10}^{-4}{\mathrm{s}}^{-1}$is $526\mathrm{K}$