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The enthalpy of the reaction : `H_(2)O_(2)(l)toH_(2)O(l)+(1)/(2)O_(2)(g)` is `-23.5 kcal mol^(-1)` and the enthalpy of formation of `H_(2)O(l)` is `-68.3 kcal mol^(-1)`. The enthalpy of formation of `H_(2)O_(2)(l)` isA. `-44.8 kcal mol^(-1)`B. `44.8 kcal mol^(-1)`C. `-91.8 kcal mol^(-1)`D. `91.8 kcal mol^(-1)` |
Answer» Correct Answer - A `H_(2)+O_(2)toH_(2)O_(2)` is `DeltaH_(f)` of `H_(2)O_(2)`. To get `H_(f)` for this reaction subtract equation `(i)` from `DeltaH_(f)` of `H_(2)O`. |
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