JEE PYQ: Chemical Kinetics Question 21
Question 21 - 2020 (06 Sep Shift 2)
The rate of a reaction decreased by 3.555 times when the temperature was changed from 40°C to 30°C. The activation energy (in kJ mol$^{-1}$) of the reaction is ____. Take; $R = 8.314,\text{J mol}^{-1}\text{K}^{-1}$, $\ln 3.555 = 1.268$
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Answer: 100
Solution
Assuming $A$ and $E_a$ to be independent of temperature: $\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$. $\ln 3.555 = \frac{E_a}{8.314}\left(\frac{1}{303} - \frac{1}{313}\right)$. $E_a = \frac{1.268 \times 8.314 \times 303 \times 313}{10} = 99980.7 \approx 99.98,\text{kJ/mol} \approx 100$.
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