1.

H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(l), DeltaH " at "298 K=-285.8 kJThe molar enthalpy of vaporization of water at 1 atm and 25^(@)C is 44 kJ. The standard enthalpy of formation of 1 mole of water vapour at 25^(@)C is

Answer»

`-241.8` KJ
241.8 kJ
329.8 kJ
`-329.8` kJ

Solution :`H_(2)+(1)/(2)O_(2)rarrH_(2)O_((l)),DeltaH=-285.8 kJ`
`H_(2)O_((l))rarrH_(2)O_((G)),DeltaH=44 kg`
`THEREFORE H_(2)+(1)/(2)O_(2)rarrH_(2)O_((g)),DeltaH^(@)=-241.8 kJ`.


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