Chemical Kinetics Question 10
Question 10 - 2024 (31 Jan Shift 2)
$ A \to B $
The rate constants of the above reaction at $200 K$ and $300 K$ are $0.03 min^{-1}$ and $0.05 min^{-1}$ respectively. The activation energy for the reaction is $J$ (Nearest integer)
(Given : In $10=2.3$
$R=8.3 J K^{-1} mol^{-1}$
$\log 5=0.70$
$\log 3=0.48$
$\log 2=0.30$
Show Answer
Answer: (2520)
Solution:
$\log \frac{K _{300}}{K _{200}}=\frac{E_a}{2.3 \times 8.314}(\frac{1}{T_1}-\frac{1}{T_2})$
$\log \frac{0.05}{0.03}=\frac{Ea}{2.305 \times 8.314} \times[\frac{1}{200}-\frac{1}{300}]$
$E_a=2519.88 J \Rightarrow E_a=2520 J$
