The enthalpy of reaction for thr equation
`2H_2(g) +O_2(g) rarr 2H_2O(l)` is `DeltaH^@=-572kj/mol`
what will be the standard enthalpy for the formation of `H_2O`(l)?

Explanation

Given that,
`2H_(2) (g) + O_(2) (g) rarr 2H_(2) O (l), Delta _(r) H^(@) = - 572 kJ mol^(-1)`
Enthalpy of formation is the enthalpy change of the reaction when 1 mole of the compound is formed from its elements then
`H_(2) (g) + (1)/(2) O_(2) (g) rarr H_(2) O (l), Delta_(r) H^(@)=`?
This can be obtained by dividing the given equation by 2.
Therefore, `Delta H^(@) (H_(2)O) = - (572 k J mol^(-1))/(2) = - 286 kJ mol^(-1)`