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`: 2H_(2)(g) + O_(2)(g) rarr2H_(2)O(l), Delta_(r)H^(@)= - 572kJ mol^(-1)`
Enthalpy of formation is the enthalpy change of the reaction when 1 mole of the compound is formed from its elements, i.e., we aim at
`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.
Hence, `Delta_(2)H^(@) ( H_(2)O) = ( - 572kJ mol^(-1))/( 2) =-286kJ mol^(-1)`