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NCERT Class XI Chemistry Thermodynamics Solutions

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Question : 10
Total: 22
Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at –10.0°C. ΔfusH = 6.03 kJmol–1 at 0°C, CP[H2O(l)] = 75.3 Jmol–1K–1, CP[H2O(s)] = 36.8 Jmol–1K–1.
Solution:  
The enthalpy change on freezing from 10°C to –10°C may be expressed as
Liquid (10°C)
ΔH1
→
 Liquid (0°C)
ΔH2
→
 Solid (0°C)
ΔH3
→
 Solid (–10°C)
ΔH1 = nCp[H2O(l)] × ΔT = 1 × 75.3 Jmol–1K–1 × (–10)
= –753 Jmol–1 = – 0.753 kJmol–1
ΔH2 = n(–ΔfusH) = –1 × 6.03 = –6.03 kJmol–1
ΔH3 = nCP[H2O(s)] = –1 × 36.8 × 10 = –368 Jmol–1 = –0.368 kJmol–1
ΔH = ΔH1+ΔH2+ΔH3 = – 0.753 – 6.03 – 0.368 kJmol–1 = –7.151 kJmol–1
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