Approximate Values of Activity Coefficients from LLE Data for a Partially Miscible Mixture

When n-octanol and water are mixed, two liquid phases form. The water-rich phase is essentially pure water containing only $0.7 \times 10^{-4}$ mole fraction n-octanol, while the octanol-rich phase contains approximately 0.26 mole fraction water. Estimate approximately the activity coefficient of n-octanol in water and water in n-octanol.

Question

Approximate Values of Activity Coefficients from LLE Data for a Partially Miscible Mixture

When n-octanol and water are mixed, two liquid phases form. The water-rich phase is essentially pure water containing only [latex]0.7 imes 10^{-4}[/latex] mole fraction n-octanol, while the octanol-rich phase contains approximately 0.26 mole fraction water. Estimate approximately the activity coefficient of n-octanol in water and water in n-octanol.

Accepted Answer

The basic equation for the solution of this problem is Eq. 11.2-2:

[latex]x_{ i }^{ I } \gamma_{ i }^{ I }\left(\underline{x}^{ I } ight)=x_{ i }^{ II } \gamma_{ i }^{ II }\left(\underline{x}^{ II } ight)[/latex] (11.2-2)

Since the water-rich phase (which we designate with superscriptW) is essentially pure water, so that [latex]x_{ W }^{ W } \cong 1 ext { and } \gamma_{ W }^{ W } \cong 1[/latex], we have for water

[latex]1=x_{ W }^{ O } \gamma_{ W }^{ O }[/latex]

where the superscript O indicates the octanol-rich phase. Consequently, for water in the octanol phase,

[latex]\gamma_{ W }^{ O }=\frac{1}{x_{ W }^{ O }}=\frac{1}{0.26}=3.846[/latex]

For octanol we have

[latex]x_{ O }^{ W } \gamma_{ O }^{ W }=0.7 imes 10^{-4} \gamma_{ O }^{ W }=x_{ O }^{ O } \gamma_{ O }^{ O }[/latex]

Since the octanol-rich phase is not pure octanol, we do not know the value of [latex]\gamma_{ O }^{ O }[/latex]. However, we see from Fig. 11.2-4 (for the system isobutane-furfural) that in the composition range of liquidliquid equilibrium we do not introduce a serious error by assuming that the product [latex]x_{ O }^{ O } \gamma_{ O }^{ O }[/latex] is approximately equal to unity. Therefore, as an estimate, we have

[latex]0.7 imes 10^{-4} \gamma_{ O }^{ W } \approx 1 \quad ext { or } \quad \gamma_{ O }^{ W } \approx \frac{1}{0.7 imes 10^{-4}}=1.43 imes 10^{4}=14300[/latex]

Comment

1. Since the concentration of n-octanol in water is so low, the value of [latex]\gamma_{ O }^{ W }[/latex] above is essentially the value at infinite dilution.

2. While the value of the infinite-dilution activity coefficient of n-octanol in water above is large, in fact much larger activity coefficient values will be seen to occur in the next chapter. Consequently, although an activity coefficient represents a correction to ideal solution behavior, it can be a very large correction.

3. The value of the product [latex]x_{ O }^{ O } \gamma_{ O }^{ O }[/latex] cannot be greater than unity; otherwise a pure octanol phase would form. (Can you explain why this is so?) Therefore, the value of [latex]x_{ O }^{ O } \gamma_{ O }^{ O }[/latex] is likely to be somewhat less than unity, so our estimate above for the activity coefficient of octanol in water is too high, but probably only slightly so.

Question 11.2.4: Approximate Values of Activity Coefficients from LLE Data fo...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions