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 [H 2 O(l)] = 75.3 J mol –1 K –1
C p [H 2 O(s)] = 36.8 J mol –1 K –1
Totalenthalpy change involved in the transformation is the sum of thefollowing changes:
(a) Energy change involved in the transformation of 1 mol of water at10°C to 1 mol of water at 0°C.
(b) Energy change involved in the transformation of 1 mol of water at0° to 1 mol of ice at 0°C.
(c) Energy change involved in the transformation of 1 mol of ice at0°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 thetransformation is –7.151 kJ mol–1.