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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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Total `DeltaH` = (1 mol of water at `10^(@)crarr` 1 mol 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 mole ice at `-10^(@)c`)
`DeltaH=C_(P)[(H_(2)O)_(l)]DeltaT+Deltaunderset("freezing")(H)+C_(P)[(H_(2)O)_(s)]DeltaT`
`=75.3JK^(-1)mol^(-1)(0-10)K+(-6.03KJmol^(-1))+(36.8JK^(-1)mol^(-1))(-10K)`
`=-753Jmol^(-1)-6.03KJmol^(-1)-368Jmol^(-1)`
`=-0.753KJmol^(-1)-6.03KJmol^(-1)-0.368KJmol^(-1)`
`=-7.151KJmol^(-1)`
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