The enthalpy change on freezing of 1 mol of water at 5 ° C to ice at - 5 ° C is: (Given Δ 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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In order to calculate the enthalpy change for $H_{2} O$ at $5 ° C$ to ice at $- 5 ° C,$ we need to calculate the enthalpy change of all the transformation involved in the process.

(a) Energy change of $1 mol, H_{2} O (l),$ at $5 ° C \to 1 mol, H_{2} O (l), 0 ° C$

(b) Energy change of $1 mol, H_{2} O (l)$ at $0 ° C \to 1 mol, H_{2} O (s) (ice), 0 ° C$

(c) Energy change of $1 mol$ , Ice $(s)$ , at $0 ° C \to 1 mol,$ Ice $(s), - 5 ° C$

Total $ΔH$

$= C_{P} [H_{2} O (l)] ΔT + ΔH$ freezing $+$ $C_{P} [H_{2} O (s)] ΔT$

$= (75.3 J {mol}^{- 1} K^{- 1}) (- 5) K + (- 6 \times 10^{3} J {mol}^{- 1}) + (36.8 J {mol}^{- 1} K^{- 1}) (- 5) K$

$ΔH = - 6.56 kJ {mol}^{- 1}$ (exothermic process)

So, $ΔH = 6.56 kJ {mol}^{- 1}$ .

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