Question

Which equation is the most appropriate to calculate the energy of activation, if the rate of reaction is doubled by increasing temperature from $T_{1}K$ to $T_{2}K$?

• A. ${\log }_{10}\left(\frac{k_1}{k_2}\right)=\frac{E_a}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$
• B. ${\log }_{10}\left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303R}[\frac{1}{T_2}-\frac{1}{T_1}]$
• C. ${\log }_{10}\frac{1}{2}=\frac{E_a}{2.303}[\frac{1}{T_2}-\frac{1}{T_1}]$
• D. ${\log }_{10}2=\frac{E_a}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$

Answer 1

The correct option is D ${\log }_{10}2=\frac{E_a}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$

We know, $\log \left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$
When reaction rate becomes double then $\frac{k_2}{k_1}$ will be equal to $2$.
Then, ${\log }_{10}2=\frac{E_a}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$