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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)` |
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