Question

The rate of the chemical reaction doubles for an increase of \(10 \, \mathrm{K}\) in absolute temperature from \(298 \, \mathrm{K}\). What will be the activation energy?

• A. 52.897 kJ mol\(^{-1}\)
• B. 51.897 kJ mol\(^{-1}\)
• C. 42.897 kJ mol\(^{-1}\)
• D. 41.897 kJ mol\(^{-1}\)

Hint:

If the reaction rate doubles for every \(10 \, \mathrm{K}\) rise, use the Arrhenius equation directly with \(\frac{k_2}{k_1}=2\). Keep temperature in Kelvin and convert the final answer into kJ mol\(^{-1}\) if needed.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The temperature dependence of the rate of a chemical reaction is explained by the Arrhenius equation.
Step 2: Formula Application:
$\log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R} \left(\frac{T_2 - T_1}{T_1 T_2}\right)$
Step 3: Explanation:
Given: $k_2/k_1 = 2$, $T_1 = 298$ K, $T_2 = 308$ K, $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$. $\log(2) = \frac{E_a}{2.303 \times 8.314} \left(\frac{10}{298 \times 308}\right)$ $0.3010 = \frac{E_a}{19.147} \times (0.0001089)$ $E_a \approx 52897 \text{ J mol}^{-1} = 52.897 \text{ kJ mol}^{-1}$.
Step 4: Final Answer:
The activation energy is 52.897 kJ mol$^{-1}$.