Chemical Kinetics Question 283
Question: In the Arrhenius plot of ln k Vs $ \frac{1}{T} $ , a linear plot is obtained with a slope of $ -2\times 10^{4}K. $ The energy of activation of the reaction (in $ kJmol{e^{-1}} $ ) is (R value is $ 8.3J{K^{-1}}mo{l^{-1}} $ )
Options:
A) 83
B) 166
C) 249
D) 332
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ k=A{e^{-E_{a}}}^{/RT} $
$ lnk=lnA-E_{a}/RT $ For ln k Vs 1/T In A= intercept $ -E_{a}/R=slope=-2\times 10^{4}K $
$ \therefore E_{a}=8.3\times 2\times 10^{4}J,mo{l^{-1}} $
$ =16.6\times 10^{4}J,mo{l^{-1}}or166,kJ,mo{l^{-1}} $
