Question

Given that the temperature coefficient for the saponification of ethyl acetate by $NaOH$ is $1.75$. Calculate activation energy for the saponification of ethyl acetate.

Answer 1

Given that $K_2/K_1=1.75$

$T_1={25}^{o}C=298$

$T_2={35}^{o}C=308$

$\Rightarrow 2.303\log \frac{K_2}{K_1}=\frac{E_{\alpha }}{R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$

$\Rightarrow 2.303\log 1.75=\frac{E_{\alpha }}{1.987}\left[\frac{308-298}{308\times 298}\right]$

$\Rightarrow E_{\alpha }=10.207Kcal/mole$