The enthalpy of reaction,
`H_(2)(g)+(1)/(2)O_(2)(g)to H_(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

The enthalpy of `H_(2)O(l)` is less than that of `H_(2)O(g)`, hence, more energy will be released when `H_(2)O(l)` is formed, therefore `DeltaH_(1) lt DeltaH_(2)`.