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_((I))` is `DeltaH_(2)`. Then

Options

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

Answer

A

Explanation

As the physical state is also changed in second reaction, so the value of`DeltaH_(2)`, will be greater than `DeltaH_(1)`