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GUJCET · Chemistry · Chemical Kinetics
The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. What will be the activation energy?
- A \(52.897 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- B \(51.897 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- C \(42.897 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- D \(41.897 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(A) \(52.897 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\) \(\ln(2) = \frac{E_a}{8.314 \, \mathrm{J} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1}}\left(\frac{1}{298 \, \mathrm{K}} - \frac{1}{308 \, \mathrm{K}}\right)\)
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