Question

A reaction proceeds five times faster at $57{}^{\circ }C$ as it does at $27{}^{\circ }C$. The energy of activation is:
(take $ln3=1.1,ln5=1.6$)

• A. $10.56kcalmo{l}^{-1}$
• B. $10.80kcalmo{l}^{-1}$
• C. $10.64kcalmo{l}^{-1}$
• D. $10.90kcalmo{l}^{-1}$

Answer 1

The correct option is A $10.56kcalmo{l}^{-1}$

We know,
$2.303log\frac{k_2}{k_1}=\frac{E_a}{R}\lfloor \frac{T_2-T_1}{T_1{T}_2}\rfloor$
Here, $T_2=57+273=330K$
$T_1=27+273=300K$
$R=2\times {10}^{-3}kcal.E_a=?$
As $r=k[A{]}^n$ (at a temperature T)
Hence $\frac{r_2}{r_1}=\frac{k_2}{k_1}$ (at a temperature $T_2$ and$T_1)$
But $\frac{r_2}{r_1}=5$ (Given)
$\therefore \frac{k_2}{k_1}=5$
$\therefore ln5=\frac{E_a}{2\times {10}^{-3}}\lfloor \frac{330-300}{330\times 300}\rfloor$
$\therefore E_a=10.56Kcal/mol$
(Using ln 5 = 1.6)