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

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

1.	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)`. ThenA. `DeltaH_(1) lt DeltaH_(2)`B. `DeltaH_(1)+DeltaH_(2)=0`C. `DeltaH_(1) gt DeltaH_(2)`D. `DeltaH_(1)=DeltaH_(2)`
Answer» Correct Answer - A 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)`.

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

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