JEE PYQ: Chemical Kinetics Question 32
Question 32 - 2019 (10 Apr Shift 2)
For the reaction of $\text{H}_2$ with $\text{I}_2$, the rate constant is $2.5 \times 10^{-4},\text{dm}^3,\text{mol}^{-1},\text{s}^{-1}$ at 327°C and $1.0,\text{dm}^3,\text{mol}^{-1},\text{s}^{-1}$ at 527°C. The activation energy for the reaction, in kJ mol$^{-1}$ is:
($R = 8.314,\text{J K}^{-1}\text{mol}^{-1}$)
(1) 166
(2) 150
(3) 72
(4) 59
Show Answer
Answer: (1)
Solution
$K = e^{-E_a/RT}$ or $\log K = \frac{-E_a}{2.303,RT}$. $\log\frac{K_2}{K_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$. $\log\frac{1}{2.5 \times 10^{-4}} = \frac{E_a}{8.314 \times 2.303}\left(\frac{1}{600} - \frac{1}{800}\right)$. $3.6 = \frac{E_a}{8.314 \times 2.303} \times \frac{200}{600 \times 800}$. $E_a = 165.4,\text{kJ/mol} \approx 166,\text{kJ/mol}$.
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