question_answer

If the activation energy for the forward reaction is \[250\,kJ\,mo{{l}^{-1}}\] and that of the reverse  reaction is \[360\,kJ\,mo{{l}^{-1}},\] what is the enthalpy change for the reaction?

A)  \[+\,110\,k\,mo{{l}^{-1}}\]

B)  \[-\,110\,kJ\,mo{{l}^{-1}}\]

C)  \[+\,610\,kJ\,mo{{l}^{-1}}\]

D)  \[-\,610\,kJ\,mo{{l}^{-1}}\]

E) None option is correct

Correct Answer: E

Solution :

\[\Delta H={{E}_{f}}-{{E}_{b}}=250-360\,kJ=110\,kJ.\]