JEE PYQ: Chemical Kinetics Question 27
Question 27 - 2020 (09 Jan Shift 2)
A sample of milk splits after 60 min. at 300 K and after 40 min. at 400 K when the population of lactobacillus acidophilus in it doubles. The activation energy (in kJ/mol) for this process is closest to ____.
(Given, $R = 8.3,\text{J mol}^{-1}\text{K}^{-1}$, $\ln\left(\frac{2}{3}\right) = 0.4$, $e^{-3} = 4.0$)
Show Answer
Answer: 3.98
Solution
For a first order reaction: at 300 K, $k_1 \times 60 = \ln\frac{[A]}{[A_0]}$. At 400 K, $k_2 \times 40 = \ln\frac{[A]}{[A_0]}$. $\frac{k_2}{k_1} = \frac{60}{40}$. $\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left[\frac{1}{T_1} - \frac{1}{T_2}\right]$. $\ln\left(\frac{3}{2}\right) = \frac{E_a}{8.3} \times \frac{100}{400 \times 300}$. $0.4 \times 8.3 \times 1200 = E_a$. $E_a = 3984,\text{J/mol} = 3.984,\text{kJ/mol} \approx 3.98$.
BETA
