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The enthalpy change on freezing of 1 mol...

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))`

Text Solution

Verified by Experts

`DeltaH = C_p . DeltaT `(at constant pressure)
(i) `DeltaH` involved in changing the temperature of 1 mole of liquid water from `-10^@C " to " 0^@C`
`=C_p [ H_2O(l)] xx DeltaT`
`=75.3 xx 10 = 753 J = 0.753 kJ`
(ii) `DeltaH` involved in changing 1 mole of liquid water at `0^@C`
`=-Delta_(fus)H = - 6.03 kJ`
(iii) `DeltaH` involved in changing 1 mole of ice at `0^@C` (273 K) to water at `10^@C` (283K)
` = C_p [ H_2O(s)] xx DeltaT`
` = 36.8 xx (273 -283) = -368 J`
`=-0.368 kJ`.
`:.` Total enthalpy change in the entire process
`= 0.753 + (-6.03) + (-0.368)`
` =-5.645 kJ mol^(-1)`
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