Question

2. At 0 O C, ice and water are in equilibrium and ΔH = 6.0 kJ mol -1 for the process H2O (s) ⇌ H2O (l), What will be ΔS and ΔG for the conversion of ice to liquid water.

Answer 1

Answer:

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Explanation:

Since the given process is in equilibrium, ΔG = 0 Putting this value in the relationship, ΔG = ΔH – TΔS, we get 0 = ΔH – TΔS or TΔS = ΔH or ΔS =ΔH/T We are given ΔH = 6.0 kj mol–1 = 6000 J mol–1 and T = 0°C = 273 K ∴ 6000J mol-1/273 = 21.98 J K mol-1

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