Calculate the entropy change of the system when 2.00 mol of water freezes to form ice at 0 °C, given that ΔfusH°(H2O) = +6.02 kJ mol–1.
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I assume the system is at 1 bar pressure, so that 0°C is the equilibrium melting point of water.
At the melting point, the solid and liquid phases are in equilibrium, so the Gibbs free energy change for the reaction:
H2O(l) -> H2O(s)
is zero.
The enthalpy change for the above reaction is simply the negative of the enthalpy of fusion, so we have that:
ΔG = 0 = ΔH - T*ΔS = (-6,020 J/mol) - (273.15K)*ΔS
ΔS = -(6,020 J/mol)/(273.15K) = -22.04 J/(K*mol)
In this case, we have two moles of water freezing, so the entropy change of the system is twice the per-mole value, or -44.08 J/K
At the melting point, the solid and liquid phases are in equilibrium, so the Gibbs free energy change for the reaction:
H2O(l) -> H2O(s)
is zero.
The enthalpy change for the above reaction is simply the negative of the enthalpy of fusion, so we have that:
ΔG = 0 = ΔH - T*ΔS = (-6,020 J/mol) - (273.15K)*ΔS
ΔS = -(6,020 J/mol)/(273.15K) = -22.04 J/(K*mol)
In this case, we have two moles of water freezing, so the entropy change of the system is twice the per-mole value, or -44.08 J/K