Q.
The rate of a reaction quadruples when the temperature changes from $300 \, $ to $310K$ . The activation energy of this reaction is:
(Assume Activation energy and pre-exponential factor are independent of temperature; $ln\left(\right.2\left.\right)=0.693;R=8.314J\left(mol\right)^{- 1}K^{- 1}$ )
NTA AbhyasNTA Abhyas 2022
Solution:
The activation energy, rate constant and temperature can be related using Arrhenius equation.
$k=Ae^{- \frac{E_{a}}{RT}}$
$ln\left(\frac{k_{2}}{k_{1}}\right)=\frac{E_{a}}{R} \, \left(\frac{1}{T_{1}} - \frac{1}{T_{2}} \, \right)$
$ln\left(4\right)=\frac{E_{a}}{R} \, \left(\frac{1}{300} - \frac{1}{310} \, \right)$
$2\times 0.693=\frac{E_{a}}{R} \, \left(\frac{10}{300 \times 310} \, \right)$
$\text{E}_{\text{a}} = \frac{2 \times 0.693 \times 300 \times 310 \times 8.314}{10} = 107.2 \text{kJ/mol}$
$k=Ae^{- \frac{E_{a}}{RT}}$
$ln\left(\frac{k_{2}}{k_{1}}\right)=\frac{E_{a}}{R} \, \left(\frac{1}{T_{1}} - \frac{1}{T_{2}} \, \right)$
$ln\left(4\right)=\frac{E_{a}}{R} \, \left(\frac{1}{300} - \frac{1}{310} \, \right)$
$2\times 0.693=\frac{E_{a}}{R} \, \left(\frac{10}{300 \times 310} \, \right)$
$\text{E}_{\text{a}} = \frac{2 \times 0.693 \times 300 \times 310 \times 8.314}{10} = 107.2 \text{kJ/mol}$