A waste liquor consists of a 0.5 molal solution of CaCl2 in water. Calculate the minimum work, per mole of CaCl2 , required to separate the liquor into anhydrous CaCl2 and pure water at atmospheric temperature and pressure. The mean ionic activity coefficient of 0.5 molal CaCl2 is 0.448.

Question

A waste liquor consists of a 0.5 molal solution of [latex]CaCl_2 [/latex] in water. Calculate the minimum work, per mole of [latex]CaCl_2 [/latex], required to separate the liquor into anhydrous [latex]CaCl_2 [/latex] and pure water at atmospheric temperature and pressure. The mean ionic activity coefficient of 0.5 molal [latex]CaCl_2 [/latex] is 0.448.

Accepted Answer

The minimum work is that required when the separation is conducted reversibly; that is, [latex]w=-\Delta G[/latex] for the process. Further, 0.5 molal [latex]CaCl_2 [/latex] comprises 0.5 moles of [latex]CaCl_2 [/latex] and 1000 g of [latex]H_2O[/latex] or 0.5 moles of [latex]CaCl_2 [/latex] and 1000/18 = 55.55 moles of water. Thus, 1 mole of [latex]CaCl_2 [/latex] exists in 111.1 moles of water, and the mole fraction of water is 111.1/112.1 = 0.991.

1. Move 1 mole of dissolved [latex]CaCl_2 [/latex] from a concentration of 0.5 molar to the 1 molal standard state.

[latex]\Delta G_{(1)}=RT\ln \frac{a_{CaCl_{2(m=1)}}}{a_{CaCl_{2(m=0.5)}}} [/latex]

[latex]=-RT\ln 4(\gamma_{\pm} m_{CaCl_2})^3[/latex]

[latex]=-8.3144 imes 298\ln \left[4 imes (0.448 imes 0.5)^3 ight]=7686 [/latex] J

2. Transfer the 1 mole of [latex]Ca^{2+}[/latex] ions from the 1 molal standard state to solid Ca at 298 K. From Table 14.1, [latex]\varepsilon ^{\circ ,Ca}=-2.87 [/latex] volts. Therefore, for the reaction

[latex]Ca^{2+}_{(m)}+2e^{-}=Ca_{(s)}[/latex]

[latex]\Delta G_{(2)}=-(2 imes 96,487 imes- 2.87)=553,835[/latex] J

3. Transfer the 2 moles of [latex]Cl^{2-}[/latex] ions from the 1 molal standard state to [latex]Cl_2[/latex] gas at 1 atm pressure at 298 K. From Table 14.1, [latex]\varepsilon ^{\circ ,Cl}=1.3595 [/latex] volts. Therefore, for the reaction

[latex]2Cl^{-}_{(m)} =Cl_{2(g)}+2e^{-}[/latex]

[latex]\Delta G_{(3)}=-(2 imes 96,487 imes- 1.3595)=262,348[/latex] J

4. Transfer 111.1 moles of [latex]H_2O[/latex] from a mole fraction of 0.991 to a mole fraction of 1.0. Assuming Raoultian behavior,

[latex]\Delta G_{(4)}=-n_{H_2O}RT\ln X_{H_2O}[/latex]

[latex]=-111.1 imes 8.3144 imes 298\ln 0.991=2486[/latex] J

5. Allow the 1 mole of solid Ca to react with the mole of gaseous [latex]Cl_2[/latex] to form 1 mole of solid [latex]CaCl_2[/latex] at 298 K. For the reaction

[latex]Ca_{(s)}+Cl_{2(g)}=CaCl_{2(s)}[/latex]

[latex]\Delta G_{(5)}=\Delta G^{\circ }_{298K}=-752,100[/latex] J

Thus, the change in the Gibbs free energy for the separation process is

[latex]\Delta G_{(1)}+\Delta G_{(2)}+\Delta G_{(3)}+\Delta G_{(4)}+\Delta G_{(5)}=74,255[/latex] J

which is the minimum amount of work required per mole of [latex]CaCl_2[/latex] separated.

Question 14.14.2: A waste liquor consists of a 0.5 molal solution of CaCl2 in ...

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