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Statement-1: The enthalpy of formation o...

Statement-1: The enthalpy of formation of H_(2)O(l)is greater than of H_(2)O(g) in magnitude.
Statement -2: Enthalpy chnge Is negative for the condensation reaction
`H_(2)O(g)toH_(2)O(l)`

A

Statement-1 is True, Statement -2 is True, Statement-2 is a correct explanation for Statement-1.

B

Statement -1 is True ,Starement -2 is True ,Statement-2 is not a correct explanation for Statement-1

C

Statement-1 is True ,Statement-2 is False.

D

Statement-1 is False ,Statement-2 is True.

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Assertion (A): The enthalpy of formation of H_(2)O(l) is greater than that of H_(2)O(g) . Reason (R ) : Enthalpy change is negative for the condensation reaction H_(2)O(g) rarr H_(2)O(l) .

What will be standard enthalpy of formation of H_(2)O(l) ?

The heat of formation of H_(2)O_((l)) is -286.2 kJ. The heat of formation of H_(2)O_((g)) is likely to be

For the process, H_(2)O(l) to H_(2)O(g)

The standard molar enthalpies of formation of H_(2)O(l) " and " H_(2)O_(2)(l) are -286 and -188 "kJ"//"mol", respectively. Molar enthalpies of vaporisation of H_(2)O(l) " and "H_(2)O_(2)(l) are 44 and 53 kJ respectively. The bond dissociation enthalpy of O_(2)(g) is 498 "kJ"//"mol" . calculate the bond dissociation enthalphy ("in" "kJ"//"mol" ) of O-O bond in H_(2)O_(2) , assuming that the bond dissociation ethalpy of O-H bond is same in both H_(2) " and " H_(2)O_(2) .

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