JEE PYQ: Chemical Kinetics Question 26
Question 26 - 2020 (09 Jan Shift 1)
For following reactions:
$\text{A} \xrightarrow{700\text{K}} \text{Product}$
$\text{A} \xrightarrow{500\text{K}}_{\text{catalyst}} \text{Product}$
it was found that the $E_a$ is decreased by 30 kJ/mol in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre-exponential factor is same):
(1) 75 kJ/mol
(2) 105 kJ/mol
(3) 135 kJ/mol
(4) 198 kJ/mol
Show Answer
Answer: (1)
Solution
Given $k_1 = k_2$. $Ae^{-E_{a_1}/RT_1} = Ae^{-E_{a_2}/RT_2}$. $\frac{E_{a_1}}{T_1} = \frac{E_{a_2}}{T_2}$. $E_{a_2} = E_{a_1} - 30$. $\frac{E_{a_2} + 30}{700} = \frac{E_{a_2}}{500}$. $500(E_{a_2} + 30) = 700 E_{a_2}$. $150 = 2E_{a_2}$. $E_{a_2} = 75,\text{kJ/mol}$.
BETA
