Question

If the activation energy for the forward reaction is $ 150\text{ }kJ\text{ }mo{{l}^{-1}} $ and that of the reverse reaction is $ 260\text{ }kJ\text{ }mo{{l}^{-1}}, $ what is the enthalpy change for the reaction?

• 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}} $

Answer 1

Answer: (C)

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}} $