JEE PYQ: Chemical Kinetics Question 8
Question 8 - 2021 (25 Feb Shift 1)
For the reaction, $a\text{A} + b\text{B} \to c\text{C} + d\text{D}$, the plot of $\log k$ vs $\frac{1}{T}$ is given below:
[Graph shows a straight line with slope $= -10000$ K]
The temperature at which the rate constant of the reaction is $10^{-4},\text{s}^{-1}$ is ____ K. (Rounded off to the nearest integer)
[Given: The rate constant of the reaction is $10^{-8},\text{s}^{-1}$ at 500 K]
Show Answer
Answer: 526
Solution
$\log_{10} K = \log_{10} A - \frac{E_a}{2.303RT}$. Slope $= \frac{-E_a}{2.303R} = -10000$. $\log_{10}\frac{K_2}{K_1} = 10000 \times \left[\frac{1}{T_1} - \frac{1}{T_2}\right]$. $\log_{10}\frac{10^{-4}}{10^{-8}} = 10000 \times \left[\frac{1}{500} - \frac{1}{T}\right]$. $4 = 10000 \times \left[\frac{1}{500} - \frac{1}{T}\right]$. $\frac{1}{T} = \frac{1}{500} - \frac{4}{10000} = \frac{20 - 4}{10000} = \frac{16}{10000}$. $T = \frac{10000}{16} \approx 526$ K.
BETA
