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The enthalpy change on freezing of 1 mol of water at `5^(@)C` to ice at `-5^(@)C` is: (Given `Delta_("fus") H = 6 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 J"mol"^(-1) K^(-1))` |
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Answer» Total `DeltaH` `=(1 "mol water at" 10^(@)C rarr 1 "mol of water at" 0^(@)C)+(1 "mol water at" 0^(@)C rarr 1 "mol ice at" 0^(@)C)+(1 "mol ice at" 0^(@)C rarr 1 "mol ice at" -10^(@)C)` `=C_(P)[H_(2)O(l)]xxDeltaT+(DeltaH_("freezing"))/T_(f)+C_(P)[H_(2)O(s)]xxDeltaT` `=(75.3 J K^(-1) mol^(-1))(10K)+(-6.03 k J mol^(-1))+(36.8 J K^(-1) mol^(-1))(-10K)` `=753 J mol^(-1)-6.03 kJ mol^(-1)-368 J mol^(-1)` `=0.753 kJ mol^(-1)-6.03 kJ mol^(-1)-0.368 kJ mol^(-1)` `=-5.645 kJ mol^(-1)` |
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