Find ΔH and q if 2.000 mol of supercooled liquid water at −15.00°C freezes irreversibly at a constant pressure of 1.000 atm and a temperature of −15.00°C. Assume that CP,m of liquid water is constant and equal to 75.48 JK^−1 mol^−1, and that CP,m of ice is constant and equal to 37.15 JK^−1 mol^−1.

Question

Find ΔH and q if 2.000 mol of supercooled liquid water at −15.00°C freezes irreversibly at a constant pressure of 1.000 atm and a temperature of −15.00°C. Assume that [latex]C_{P,m}[/latex] of liquid water is constant and equal to [latex]75.48 J K^{−1} mol^{−1}[/latex], and that [latex]C_{P,m}[/latex] of ice is constant and equal to [latex]37.15 J K^{−1} mol^{−1}[/latex].

Accepted Answer

We assume that the supercooled water is metastable and treat it as though it were at equilibrium. We calculate ΔH along the reversible path shown in Figure 2.9. Step 1 is the reversible heating of the supercooled liquid to 0.00°C, the equilibrium freezing temperature. Step 2 is the reversible freezing of the system at 0.00°C, and step 3 is the reversible cooling of the solid to −15.00°C. The enthalpy change for the irreversible process is equal to the enthalpy change of this reversible process.

[latex]\Delta H_{1}=\int_{258.15K}^{273.15K}{C_{P}(1)dT} ={C_{P}(1)\Delta T}[/latex]

[latex]=(2.000mol)(75.48JK^{-1}mol^{-1})(15.00K)=2264J[/latex]

The enthalpy change of step 2 was calculated in an earlier example:

[latex]\Delta H_{2}=-12020J[/latex]

The enthalpy change of step 3 is similar to that of step 1.

[latex]\Delta H_{3}=\int_{273.15K}^{258.15K}{C_{P}(s)dT} ={C_{P}(s)\Delta T}[/latex]

[latex]=(2.000mol)(37.15JK^{-1}mol^{-1})(15.00K)=1114J[/latex]

The total enthalpy change is

[latex]\Delta H=\Delta H_{1}+\Delta H_{2}+\Delta H_{3}=-10870J=-10.87kJ[/latex]

Because the process is at constant pressure, q=ΔH.

Question 2.28: Find ΔH and q if 2.000 mol of supercooled liquid water at −1...

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