Question

Given that the temperature coefficient for the saponification of ethyl acetate by $NaOH$ is 1.75. Calculate the activation energy ($T={25}^{o}C$).

• A. 1.0207 kcal/mol
• B. 10.207 kcal/mol
• C. 5.1 kcal/mol
• D. 2.3 kcal/mol

Answer 1

The correct option is B 10.207 kcal/mol

Temperature coefficent $=\frac{kat(t+{10}^{\circ }C)}{kat\left(t^{\circ }C\right)}=\frac{kat\left({35}^{\circ }C\right)}{kat\left({25}^{\circ }C\right)}=\mathrm{1.75.}$
The Arrhenius equation at two different temperatures is $\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\frac{(T_2-T_1)}{T_1{T}_2}.$
Hence, $\log 1.75=\frac{E_a}{2.303\times 1.987}\frac{(308-298)}{298\times 308}.$

$E_a=10.207kcal/mol$.