1.

Consider the following reaction, `N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g)+22kcal" mol"^(-1)` The activation energy for the forward reaction is 50 kcal. What is the activation energy for the backward reaction?A. `28kcal" mol"^(-1)`B. `+28kcal"mol"^(-1)`C. `72kcal"mol"^(-1)`D. `+72kcal" mol"^(-1)`

Answer» Correct Answer - D
The activation energy for forward reaction
`=50kcal" mol"^(-1)`
Energy released during the reaction =`22"kcal mol"^(-1)`
Therefore, the acativation energy for the backward reaction `=50+22"kcal"=72"kcal mol"^(-1)`


Discussion

No Comment Found

Related InterviewSolutions