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Calorific value of H2 is -143KJ g^(-1) ....

Calorific value of `H_2` is `-143KJ g^(-1)` . Hence, `Delta H_(f)^(@)` of `H_2O` is:

A

`-143 kJ mol^(-1)`

B

`-286 kJ mol^(-1)`

C

`+143 kJ mol^(-1)`

D

`+286 kJ mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
`H_(2(g)) + 1/2 O_(2)(g) rarr H_(2)O(l)`
`DeltaH_(f) = ?`
In the above reaction for 1 mole formation of `H_2O` we need to burn `2g H_2` molecule
`:. Delta H_(f) (H_2O) = -2 xx 143 kJ//mol = -286 kJ//mol`.
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