1.

Calculate the enthalpy change on freezing of 1.0 mole of water at 10.0^(@)C to ice at -10.0^(@)C [Delta_("fus")H=6.03kJmol^(-1) at 0^(@)C] C_(p)[H_(2)O(l)]=75.3Jmol^(-1)K^(-1) C_(p)[H_(2)O(s)]=36.8Jmol^(-1)K^(-1)

Answer»

`-753Jmol^(-1)`
`-368Jmol^(-1)`
`-7.151kJmol^(-1)`
`-6.03kJmol^(-1)`

SOLUTION :Enthalpy change for the conversion of 1 mole of LIQUID water at `10^(@)C` into 1 mole of liquid water at `0^(@)`C,
`DeltaH_(1)=C_(p)[H_(2)O(l)]xxDeltaT`
`=-75.3"J "mol^(-1)K^(-1)xx10K=-753J" "mol^(-1)`
Enthalpy of fusion,
`DeltaH_(2)=DeltaH_("freezing")=-DeltaH_("fusion")=-6.03"kJ "mol^(-1)`
Enthalpy change for the conversion of 1 mole of ice at `0^(@)C`
to 1 mole of ice at `10^(@)C`,
`DeltaH_(3)=C_(p)[H_(2)O(s)]xxDeltaT=-36.8J" "mol^(-1)K^(-1)xx10K`
`=-368"J "mol^(-1)`
`DeltaH_("total")=-(0.753+6.03+0.368)=-7.151" kJ "mol^(-1)`


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