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. `DeltaH_(1) lt DeltaH_(2)`
B. `DeltaH_(1) + DeltaH_(2)=0`
C. `DeltaH_(1) gt DeltaH_(2)`
D. `DeltaH_(1) = DeltaH_(2)`

Answer

A

Explanation

`H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O`, `DeltaH_(1)`
`H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)(O(l)`, `H_(2)`
`|DeltaH_(1)| gt |DeltaH_(2)|`
i.e., amount of heat evolved in the first reaction will be less than that in the second reaction. This is because additional amount of heat is evolved in the conversation of `H_(2)O(g)` into `H_(2)O(l)` in the second case.