JEE PYQ: Chemical Kinetics Question 40
Question 40 - 2019 (12 Jan Shift 1)
For a reaction, consider the plot of $\ln k$ versus $1/T$ given in the figure. If the rate constant of this reaction at 400 K is $10^{-5},\text{s}^{-1}$, then the rate constant at 500 K is:
[Graph shows slope $= -4606$ K]
(1) $10^{-6},\text{s}^{-1}$
(2) $2 \times 10^{-4},\text{s}^{-1}$
(3) $10^{-4},\text{s}^{-1}$
(4) $4 \times 10^{-4},\text{s}^{-2}$
Show Answer
Answer: (3)
Solution
$\ln K = \ln A - \frac{E_a}{RT}$. Slope $= \frac{-E_a}{R} = -4606$ K. $\log\left(\frac{K_2}{K_1}\right) = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$. $\log\left(\frac{K_2}{10^{-5}}\right) = \frac{4606}{2.303} \times 4606 \times \left(\frac{1}{400} - \frac{1}{500}\right)$. $\log\left(\frac{K_2}{10^{-5}}\right) = 1$. $\frac{K_2}{10^{-5}} = 10$. $K_2 = 10^{-5} \times 10 = 10^{-4},\text{s}^{-1}$.
BETA
