A) \[{{E}_{1}}>{{E}_{2}}\]
B) \[{{E}_{1}}<{{E}_{2}}\]
C) \[{{E}_{1}}={{E}_{2}}\]
D) \[\Delta H\] is not related to \[{{E}_{1}}\] and \[{{E}_{2}}\]
Correct Answer: B
Solution :
\[{{H}_{2}}+C{{l}_{2}}2HCl,\] \[\Delta H=-44.2kcal\] Above reaction is exothermic, so the activation energy of forward reaction \[({{E}_{1}})\] is less than activation energy of reverse reaction \[({{E}_{2}})\], ie, \[{{E}_{1}}<{{E}_{2}}\]You need to login to perform this action.
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