A) \[6.91\text{ kc}al\text{ }mo{{l}^{-1}}\]
B) \[3.00\text{ kc}al\text{ }mo{{l}^{-1}}\]
C) \[4.15\,kcal\,mo{{l}^{-1}}\]
D) \[5.10kcal\,mo{{l}^{-1}}\]
Correct Answer: A
Solution :
[a] \[{{e}^{-{{E}_{a}}/RT}}={{10}^{-3}}\]% \[={{10}^{-5}};\] \[{{E}_{a}}=2.303\times 2\times 300\times 5\,cal\] \[=6.91 kcal/mo{{\operatorname{l}}^{-1}}\]You need to login to perform this action.
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