A) 80
B) 120
C) 40
D) 200
E) 280
Correct Answer: E
Solution :
\[{{E}_{a}}\] (activation energy) \[(A\to B)=80\text{ }kJ\text{ }mo{{l}^{-1}}\] Heat of reaction\[(A\to B)=200\text{ }kJ\text{ }mo{{l}^{-1}}\] for\[(B\to A)\]backward reaction \[{{E}_{a}}(B\to A)={{E}_{a}}(A\to B)+\]heat of reaction \[=80+200=280\,kJ\,mo{{l}^{-1}}\]You need to login to perform this action.
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