Question

A first order reaction is 50% complete in 20 minutes at $27{}^{\circ }C$ and in 5 minutes at $47{}^{\circ }C$. The energy of activation of the reaction is:

• A. 43.85 kJ/mol
• B. 55.14 kJ/mol
• C. 11.97 kJ/mol
• D. ​​​​​​6.65 kJ/mol

Answer 1

The correct option is B

55.14 kJ/mol

We know that half life for first order reaction is
$t_{\frac{1}{2}}=\frac{0.693}{k}$
So rate constant at ${27}^{o}C$
$k_1=\frac{0.693}{20}mi{n}^{-1}$
Rate constant at ${47}^{o}C$
$k_2=\frac{0.693}{5}mi{n}^{-1}$
Arrhenius equation:
$k=A{e}^{(-\frac{E_a}{RT})}$
Ratio of $k_1$ to $k_2$ is equal to

$\frac{k_1}{k_2}=e^{\left[\frac{E_a}{R}\left(\frac{T_1-T_2}{T_1{T}_2}\right)\right]}$
Taking natural log on both sides, we get:

$ln\frac{k_1}{k_2}=\left[\frac{E_a}{R}\left(\frac{T_1-T_2}{T_1{T}_2}\right)\right]$

$ln\frac{1}{4}=\left[\frac{E_a}{8.314}\left(\frac{300-320}{300\times 320}\right)\right]$
On solving, we get:
$E_a=55.14kJ/mol$