A) \[410\text{ }kJ\text{ }mo{{l}^{-1}}\]
B) \[110\,kJ\,mo{{l}^{-1}}\]
C) \[-110\,kJ\,mo{{l}^{-1}}\]
D) \[-\,410\,kJ\,mo{{l}^{-1}}\]
E) \[90\,kJ\,mo{{l}^{-1}}\]
Correct Answer: C
Solution :
Enthalpy change for a reversible reaction, \[\Delta H={{E}_{a}}(forward)-{{E}_{a}}(backward)\] \[\Delta H=150-260=-110\text{ }kJ\text{ }mo{{l}^{-1}}\]
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