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)`.
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
A. `DeltaH_(1) gt DeltaH_(2)`
B. `DeltaH_(1) = DeltaH_(2)`
C. `DeltaH_(1) lt DeltaH_(2)`
D. `DeltaH_(1)+DeltaH_(2)=0`
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Correct Answer - C
Gaseous state is more energetic than liquid state, therefore `DeltaH_(1) lt DeltaH_(2)`
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