The enthalpy of reaction for the reaction: `2H _(2 (g)) + O _(2 (g)) to 2 H _(2) O _((l)) is Delta , H ^(-) = -572kJ mol ^(-1)` What will be the standard enthalpy for the formation of `H _(2) O _((l))`?

Options

A. `-572kJ mol ^(-1)`
B. `+ 57 kJ mol ^(-1)`
C. `+ 286 kJ mol ^(-1)`
D. `-286 kJ mol ^(-1)`

Explanation

According to the definition of standard enthalpy of formation, the enthalpy change for the following reaction will be standard enthalpy of formation of `H _(2) O (l) . H _(2 (g)) + (1)/(2) O _(2 (g)) to H _(2) O_((l))`
of the standard enthalpy of formation of `H _(2)O _((l))` will be half of the enthalpy of the given equation `i.e. Delta , H` is also halved. `Delta _(f) H _(H_(2)O)^(-) (l) = (1)/(2) xx Delta _(r) H ^(-) = ( - 5 kJ mol ^(-1))/(2) = 286 kJ // mol`