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Calcualte the enthalpy change on freezin...

Calcualte the enthalpy change on freezing of 1.0 mole of water at `10.0^(@)C` to ice at `-10^(@)`C. `Delta_(fs)H=6.03 kJ mol^(-1)` at `0^(@)C`.
`C_(p)[H_(2)O(l)] = 75.3 J mol^(-1) K^(-1), C_(P)[H_(2)O(s)] = 36.8 Jmol^(-1)K^(-1)`

Text Solution

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The enthalpy change on freezing 1.0 mol of water from `10^(@)C` to `-10^(@)C` may be expressed as :
1 mol water `(10^(@)C) overset(DeltaH_(1))to 1` mol water `(0^(@)C) overset(DeltaH_(2)) to 1` mol ice `(0^(@)C) overset(DeltaH_(3)) to` 1 mol ice `(-10^(@)C)`
`DeltaH_(1) = nC_(p)[H_(2)O(l)] xx DeltaT = 1 xx 75.3 J "mol"^(-1) K^(-1) xx (0 - 10)K`
` = -753 J "mol"^(-1) = -0.753 kJ "mol"^(-1)`
`DeltaH_(2) = n(Delta_(freezing)H) = n(-Delta_(fus)H)(because Delta_(freezing)H = -Delta_(fus)H)`
`= 1 xx (-6.03) = -6.03 kJ "mol"^(-1)`
`DeltaH_(3) = nC_(p)[H_(2)O(s)] DeltaT = 1 xx 36.8 J "mol"^(-1) K^(-1) xx (-10-0)K`
` = -368 J "mol"^(-1) = -0.368 kJ "mol"^(-1)`
`DeltaH = DeltaH_(1) + DeltaH_(2) + DeltaH_(3)`
` = -0.753 kJ "mol"^(-1) - 6.03 kJ "mol"^(-1) - 0.368 kJ "mol"^(-1)`
` = -7.151 kJ "mol"^(-1)`.
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Calculate the enthalpy change on freezing 1.0 mol of water at 10.0^(@)C to ice at - 10.0 ^(@)C , Delta_(fus) H= 6.03 kJ mol^(-1) mol^(-1) at 0^(@)C C_(p) [H_(2)O(l)]= 75.3J mol^(-1) K^(-1) C_(p)[ H_(2)O(s)] = 36.8 J mol^(-1) K^(-1)

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Knowledge Check

  • 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)

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    B
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  • The enthalpy change of freezing of 1 mol of water at 5^(@)C to ice at -5^(@)C is (Given Delta_(fus)H=6kJ"mol"^(-1) at 0^(@)C , (C_(p)(H_(2)O,s)=36.8J"mol"^(-1)K^(-1))

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