Question

The rate of the chemical reaction doubles when the temperature is raised from \(298\,K\) to \(308\,K\). Calculate activation energy \((E_a)\) for this reaction assuming that it does not change with temperature. (Given : \(R = 8.314\,J\,mol^{-1}\,K^{-1}\), \(\log 2 = 0.30\))

Hint:

For temperature dependence of reaction rates, remember the Arrhenius equation: \[ \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] If the rate doubles, then \[ \frac{k_2}{k_1}=2 \] and use \[ \log 2 = 0.3010 \approx 0.30 \] to calculate the activation energy.

Answer 1

Step 1: Set up the two-temperature Arrhenius equation.
\[ \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) \] Given: $T_1 = 298\,K$, $T_2 = 308\,K$, rate doubles so $k_2/k_1 = 2$ and $\log 2 = 0.30$.
Step 2: Calculate the temperature factor.
\[ \frac{1}{T_1} - \frac{1}{T_2} = \frac{T_2 - T_1}{T_1 T_2} = \frac{10}{298 \times 308} = \frac{10}{91784} \approx 1.09 \times 10^{-4}\,K^{-1} \]
Step 3: Solve for $E_a$.
\[ 0.30 = \frac{E_a}{2.303 \times 8.314} \times 1.09 \times 10^{-4} \implies E_a = \frac{0.30 \times 19.147}{1.09 \times 10^{-4}} \approx 52720\,J\,mol^{-1} \] \[ \boxed{E_a \approx 52.8\,kJ\,mol^{-1}} \]