JEE PYQ: Chemical Kinetics Question 9
Question 9 - 2021 (25 Feb Shift 2)
The rate constant of a reaction increases by five times on increase in temperature from 27°C to 52°C. The value of activation energy in $\text{kJ mol}^{-1}$ is ____. (Rounded off to the nearest integer)
[$R = 8.314,\text{J K}^{-1}\text{mol}^{-1}$]
Show Answer
Answer: 52
Solution
$\ln\left{\frac{k_2}{k_1}\right} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$. $\ln(5) = \frac{E_a}{R}\left(\frac{1}{300} - \frac{1}{325}\right)$. $\frac{2.303 \times 0.7 \times 8.314 \times 300 \times 325}{25} = E_a$. $E_a = 51524.96,\text{J/mol} \approx 51.524,\text{kJ/mol} \approx 52$.
BETA
