Question

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

• A. $E_a=42.7kJmo{l}^{-1}$
• B. $E_a=14.0kJmo{l}^{-1}$
• C. $E_a=64.4kJmo{l}^{-1}$
• D. $E_a=2.6kJmo{l}^{-1}$

Answer 1

The correct option is A $E_a=42.7kJmo{l}^{-1}$

Temperature coefficient is the ratio of the rate of reaction/rate constant at two different temperature differing by ​${10}^{o}C$.
$\text{Temperature coefficient (T.C.)}=\frac{K_{t+10}}{K_t}$

Expression for rate constant at two different temperature is given by
$ln\frac{k_2}{k_1}=\frac{E_a}{R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$​

$log\frac{k_{{35}^{\circ }}}{k_{{25}^{\circ }}}=\frac{E_a}{2.303R}\left[\frac{308-298}{308\times 298}\right]$

$\frac{k_{{35}^{\circ }}}{k_{{25}^{\circ }}}=1.75$

$log1.75=\frac{E_a}{2.303R}\left[\frac{308-298}{308\times 298}\right]$

$0.243=\frac{E_a}{2.303R}\left[\frac{308-298}{308\times 298}\right]$

$E_a=\frac{0.243\times 2.303\times 8.314\times 308\times 298}{10}$
$\Rightarrow E_a=42704.8Jmo{l}^{-1}$
$\Rightarrow E_a=42.7048kJmo{l}^{-1}$