Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at –10.0°C. Δ fus H = 6.03 kJ mol –1 at 0°C.

C p [H2O(l)] = 75.3 J mol–1K–1

C p [H2O(s)] = 36.8 J mol–1K–1

Total enthalpy change involved in the transformation is the sum of the following changes:

(a) Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C.

(b) Energy change involved in the transformation of 1 mol of water at 0° to 1 mol of ice at 0°C.

(c) Energy change involved in the transformation of 1 mol of ice at 0°C to 1 mol of ice at –10°C.

Total ΔH = Cp [H2O (l)] ΔT + ΔHfreezing + Cp[H2O (s)] ΔT

= (75.3 J mol–1 K–1) (0 – 10)K + (–6.03 × 103 J mol–1) + (36.8 J mol–1 K–1) (–10 – 0)K

= –753 J mol–1 – 6030 J mol–1 – 368 J mol–1

= –7151 J mol–1

= –7.151 kJ mol–1

Hence, the enthalpy change involved in the transformation is –7.151 kJ mol–1.

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