Question

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

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

Answer 1

Answer: (A)

According to the Arrhenius equation.
$\log \frac{K_2}{K_1}=\frac{E_a}{2.303R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$
$\frac{K_2}{K_1}=2$;
$T_1=295K;T_2=305K$
$R=8.314J{K}^{-1}mo{l}^{-1}$
$\therefore \log 2=\frac{E_a}{2.303\times \left(8.314J{K}^{-1}mo{l}^{-1}\right)}\left[\frac{1}{295K}-\frac{1}{305K}\right]$
or $0.3010=\frac{E_a}{8.314J{K}^{-1}mo{l}^{-1}}\times \frac{(10)}{295\times 305}$
$E_a=\frac{0.3010\times 2.303\times 8.314\times 295\times 305}{10}$
$=51860Jmo{l}^{-1}=51.86KJmo{l}^{-1}$