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

Cp[H2O(l)] = 75.3 J mol–1 K –1 

Cp[H2O(s)] = 36.8 J mol–1 K –1

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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. 

= (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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