The enthalpy of the reaction
`H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(g)` is `DeltaH_(1)` and that of `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l)` is `DeltaH_(2)`. Then

Options

A. `Delta H_1 lt Delta H_2`
B. `Delta H_1 + Delta H_2 = 0`
C. `Delta H_1 gt Delta H_2`
D. `Delta H_1 = Delta H_2`

Answer

A

Explanation

In formation of `H_2O` , some energy is required to convert `H_2O(l)` into `H_2O(g)`, so less energy is released in the formation of `H_(2)O(g)`.
`:. Delta H_(2) > Delta H_(1)`.