Question

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

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

Answer 1

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

The Arrhenius equation is:
$k=A{e}^{-\frac{E_a}{RT}}$
Then, $log\left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$

When reaction rate becomes double then $\frac{k_2}{k_1}$ will be equal to 2

Then, $lo{g}_{10}2=\frac{E_a}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$
Hence, (d) is correct.