Chemical Kinetics Question 34
Question: The rate constant, the activation energy and the arrhenius parameter of a chemical reaction at $ 25^{o}C $ are $ 3.0\times {10^{-4}}{s^{-1}} $ , $ 104.4,kJ,mo{l^{-1}} $ and $ 6.0\times 10^{14}{s^{-1}} $ respectively. The value of the rate constant as $ T\to \infty $ is [IIT 1996]
Options:
A) $ 2.0\times 10^{18}{s^{-1}} $
B) $ 6.0\times 10^{14}{s^{-1}} $
C) Infinity
D) $ 3.6\times 10^{30}{s^{-1}} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ T_2=T(say),,T=25^{o}C=298K, $
$ E_{a}=104.4kJmo{l^{-1}}=104.4\times 10^{3},J,mo{l^{-1}} $
$ K_1=3\times {10^{-4}},K_2= $ -,
$ \log \frac{K_2}{K_1}=\frac{E_{a}}{2.303R}[ \frac{1}{T_1}-\frac{1}{T_2} ] $
$ \log \frac{K_2}{3\times {10^{-4}}}=\frac{104.4\times 10^{3}Jmo{l^{-1}}}{2.303\times (8.314J{k^{-1}}mo{l^{-1}})} $
$ [ \frac{1}{298K}-\frac{1}{T} ]AsT\to \infty \text{,}\frac{1}{T}\to 0 $
$ \therefore \log \frac{K_2}{3\times {10^{-4}}}=\frac{104.4\times 10^{3}Jmo{l^{-1}}}{2.303\times 8.314\times 298} $
$ \log \frac{K_2}{3\times {10^{-4}}}=18.297,\frac{K_2}{3\times {10^{-4}}} $
$ =1.98\times 10^{18} $ or
$ K_2=(1.98\times 10^{18})\times (3\times {10^{-4}})=6\times 10^{14}{s^{-1}} $
