JEE PYQ: Chemical Kinetics Question 17
Question 17 - 2020 (04 Sep Shift 2)
The number of molecules with energy greater than the threshold energy for a reaction increases five fold by a rise of temperature from 27°C to 42°C. Its energy of activation in J/mol is ____. (Take $\ln 5 = 1.6094$; $R = 8.314,\text{J mol}^{-1}\text{K}^{-1}$)
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Answer: 84297.48
Solution
$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$. $\ln 5 = \frac{E_a}{8.314}\left(\frac{1}{300} - \frac{1}{315}\right)$. $E_a = \frac{1.6094 \times 8.314 \times 300 \times 315}{15} = 84297.48,\text{J/mol}$.
BETA
