The enthalpy change at `298 K` of the reaction
`H_(2)O_(2)(l)toH_(2)O(l)+1//2O_(2)(g)` is `-23.5 kcal mol^(-1)` and enthalpy of formation of `H_(2)O_(2)(l)` is `-44.8 kcal mol^(-1)`. The enthalpy of formation of `H_(2)O(l)` is

Options

A. `-68. " kcal " // mol`
B. 68.3 " kcal " / mol
C. `-91.8 " kcal " // mol`
D. `91.8 " kcal " // mol`

Answer

A

Explanation

`Delta_fH^@=Delta_fH^@(H_2O)-Delta_fH^@(H_2O_2)`
`-23.5 = Delta_fH^@(H_2O)-(-44.8)`
or `Delta_fH^(@)(H_2O)=-68.3 kcal `