1.

Consider the following reaction : (A) H^(+)(aq) + OH^(-)(aq) = H_(2)O(l): Delta H = -X_(1) kJ mol^(-1) (B) H_(2)(g) + (1)/(2) O_(2)(g) = H_(2)O(l): Delta H = -X_(2) kJ mol^(-1) ( C ) CO_(2)(g) + H_(2)(g) = CO(g) + H_(2)O(l): Delta H = -X_(3) kJ mol^(-1) (D) C_(2)H_(2)(g) + (5)/(2) O_(2)(g) = 2CO_(2) + H_(2)O(l): Delta H = +X_(4) kJ mol^(-1) Enthalpy of formation of H_(2)O(l) is :

Answer»

`+ X_(1) KJ mol^(-1)`
`- X_(2) kJ mol^(-1)`
`+ X_(3) kJ mol^(-1)`
`- X_(4) kJ mol^(-1)`

Solution :ENTHALPY of FORMATION of `H_(2)O(l)` is
`H_(2)(g) + (1)/(2) O_(2)(g) = H_(2)O(l)`
:. `Delta H = - X_(2) kJ mol^(-1)`


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