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For $A + B \longrightarrow C + D$; $\Delta H = -20 \ kJ \ mol^{-1}$,the activation energy of the forward reaction is $85 \ kJ \ mol^{-1}$. The activation energy for the backward reaction is.....$kJ \ mol^{-1}$
- A$65$
- B$105$
- C$85$
- D$40$
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