Question

The rate constant of the chemical reaction doubled for an increase of 10 K in absolute temperature from 295 K. Calculate E.

• A. 51.8 kJ mol${}^{-1}$
• B. 82.1 kJ mol${}^{-1}$
• C. 23.8 kJ mol${}^{-1}$
• D. 62.1 kJ mol${}^{-1}$

Answer 1

The correct option is A 51.8 kJ mol${}^{-1}$

According to the Arrhenius equation.

log $\frac{K_2}{K_1}=\frac{E_a}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$

$\frac{K_2}{K_1}=2;$

$T_1=295K;T_2=305K$

$R=8.314J{K}^{-1}mo{l}^{-1}$

$\therefore log2=\frac{E_a}{2.303\times \left(8.314J{K}^{-1}mo{l}^{-1}\right)}$ $[\frac{1}{295K}-\frac{1}{305K}]$

or $0.3010=\frac{E_a}{8.314J{K}^{-1}mo{l}^{-1}}\times \frac{\left(10\right)}{295\times 305}$

$E_a=\frac{0.3010\times 2.303\times 8.314\times 295\times 305}{10}$

$=51860Jmo{l}^{-1}=51.86KJmo{l}^{-1}$

Hence, the correct option is $\text{A}$