Question

The rate of a reaction decreased by $3.555$ times when the temperature was changed from $40^{\circ }C$ to $30^{\circ }C$. The activation energy (in $kJmo{l}^{-1}$ ) of the reaction is--------.
Take; $R=8.314Jmo{l}^{-1}{K}^{-1}$ In $3.555=1.268$

Answer 1

Answer: 99.90

$\ell n\left(\frac{K_{T_2}}{K_{T_i}}\right)=\frac{E_a}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$

$T_1=303K;T_2=313K$

$\frac{K_{T_2}}{K_{T_i}}=3.555$

$\ell (3.555)=\frac{E_a}{8.314}\left[\frac{1}{303}-\frac{1}{313}\right]$

$E_a=99980.715$

$E_a=99.98\frac{kJ}{mole}$