Question

The rate constant of a reaction at $300$ K and $310$ K is $1.6\times {10}^{-3}$ $s^{-1}$ and $3.2\times {10}^{-3}$ $s^{-1}$ respectively. The activation energy of the reaction (in kcal) is in the range of:

• A. $12-13$
• B. $20-25$
• C. $30-40$
• D. $40-50$

Answer 1

The correct option is A $12-13$

Given that at $T_1=300K$, $K_1=1.6\times {10}^{-3}{s}^{-1}$ and at $T_2=310K$, $K_2=3.2\times {10}^{-3}{s}^{-1}$.

$ln\left(\frac{K_2}{K_1}\right)=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

Substituting the values in the above expression, we get

$ln\left[\frac{3.2}{1.6}\right]=\frac{E_a}{2}[\frac{1}{300}-\frac{1}{310}]$
$E_a=\frac{2\left(ln2\right)300\times 310}{10}$ $=12892.5cal$ $=12.892Kcal$.