Question

The temperature coefficient for the saponification of ethyl acetate by $NaOH$ is $1.75$, when the temperatures rises from $25{}^{o}C$ to $35{}^{o}C$. What is the activation energy for the saponification of ethyl acetate?
(Take ${\log }_{10}\left(1.75\right)=0.243$)

• A. $E_a=42.7{\text{kJ mol}}^{-1}$
• B. $E_a=14.0{\text{kJ mol}}^{-1}$
• C. $E_a=64.4{\text{kJ mol}}^{-1}$
• D. $E_a=2.6{\text{kJ mol}}^{-1}$

Answer 1

The correct option is A $E_a=42.7{\text{kJ mol}}^{-1}$

$\text{Temperature coefficient (T.C.)}=\frac{k_{t+10}}{k_t}$
Where, $k=\text{rate constant}$
According to Arrhenius equation,
$\ln \frac{k_2}{k_1}=\frac{E_a}{R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$​
$\Rightarrow \log \frac{k_{{35}^{\circ }}}{k_{{25}^{\circ }}}=\frac{E_a}{2.303R}\left[\frac{308-298}{308\times 298}\right]$
Given,
$\frac{k_{{35}^{\circ }}}{k_{{25}^{\circ }}}=1.75$
$\therefore \log 1.75=\frac{E_a}{2.303R}\left[\frac{308-298}{308\times 298}\right]$
$\Rightarrow 0.243=\frac{E_a}{2.303R}\left[\frac{308-298}{308\times 298}\right]$
$\Rightarrow E_a=\frac{0.243\times 2.303\times 8.314\times 308\times 298}{10}$
$\Rightarrow E_a=42704.8{\text{J mol}}^{-1}$
$\Rightarrow E_a=42.7048{\text{kJ mol}}^{-1}$