Question:
For a reaction, the rate constant double when temperature is increased from 300 K to 310 K. The activation energy ($E_{a}$) of this reaction is (Take $R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}$)
For a reaction, the rate constant double when temperature is increased from 300 K to 310 K. The activation energy ($E_{a}$) of this reaction is (Take $R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}$)
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A handy approximation for a rate doubling near room temperature ($300\text{ K}$ to $310\text{ K}$) is that $E_{a}$ regularly falls close to $54\text{ kJ/mol}$.
Updated On: Jun 3, 2026
- $53.6\text{ kJ mol}^{-1}$
- $43.6\text{ kJ mol}^{-1}$
- $58.5\text{ kJ mol}^{-1}$
- $62.4\text{ kJ mol}^{-1}$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
The temperature dependence of a reaction rate constant is given by the logarithmic form of the Arrhenius equation: $\log_{10}\left(\frac{k_{2}}{k_{1}}\right) = \frac{E_{a}}{2.303 R} \left(\frac{T_{2} - T_{1}}{T_{1} T_{2}}\right)$.
Step 2: Meaning
Given that the rate constant doubles, we have $\frac{k_{2}}{k_{1}} = 2$. The operational temperatures are $T_{1} = 300\text{ K}$ and $T_{2} = 310\text{ K}$.
Step 3: Analysis
Substituting the values into the Arrhenius equation: $$\log_{10}(2) = \frac{E_{a}}{2.303 \times 8.314} \left(\frac{310 - 300}{300 \times 310}\right)$$ $$0.3010 = \frac{E_{a}}{19.147} \left(\frac{10}{93000}\right)$$ $$E_{a} = \frac{0.3010 \times 19.147 \times 93000}{10} \approx 53598\text{ J mol}^{-1} \approx 53.6\text{ kJ mol}^{-1}$$
Step 4: Conclusion
The calculated activation energy is approximately $53.6\text{ kJ mol}^{-1}$.
Final Answer: (A)
The temperature dependence of a reaction rate constant is given by the logarithmic form of the Arrhenius equation: $\log_{10}\left(\frac{k_{2}}{k_{1}}\right) = \frac{E_{a}}{2.303 R} \left(\frac{T_{2} - T_{1}}{T_{1} T_{2}}\right)$.
Step 2: Meaning
Given that the rate constant doubles, we have $\frac{k_{2}}{k_{1}} = 2$. The operational temperatures are $T_{1} = 300\text{ K}$ and $T_{2} = 310\text{ K}$.
Step 3: Analysis
Substituting the values into the Arrhenius equation: $$\log_{10}(2) = \frac{E_{a}}{2.303 \times 8.314} \left(\frac{310 - 300}{300 \times 310}\right)$$ $$0.3010 = \frac{E_{a}}{19.147} \left(\frac{10}{93000}\right)$$ $$E_{a} = \frac{0.3010 \times 19.147 \times 93000}{10} \approx 53598\text{ J mol}^{-1} \approx 53.6\text{ kJ mol}^{-1}$$
Step 4: Conclusion
The calculated activation energy is approximately $53.6\text{ kJ mol}^{-1}$.
Final Answer: (A)
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