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For ideal solution, μA=μA* +RT lnxA

ln function is an increasing function, means that the higher the xA the higher the μA......(1)

Now consider osmosis. From Atkins' Physical Chemistry Ninth Edition, page 173,

osmotic pressure, π, is the pressure that must be applied to the solution to stop the influx of solvent (into solution across a semi-permeable membrane).

Here is my interpretation and thought process:

Had you not exerted π on solution, the μ(solution)< μ(solvent) based on (1) because lnxA is negative for xA<1, so by Second Law, the osmosis is spontaneous and solvent should move from pure solvent (high μ) to solution (low μ). Sounds right.

However, during this process, μ(pure solvent) unchanged (μ is intensive, not affected by amount) but μ(solution) increases due to dilution. Based on (1), for the solution, since xA increases μ(solution) must also increase. Although the process stops when μ(solution)=μ(pure solvent), the effect is that μ increases. μ increases would in turn imply ΔG>0 which means non-spontaneous, but as far as I know osmosis is spontaneous, so where is my misconception?

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  • $\begingroup$ When I do the math, I get $\Delta G<0$ $\endgroup$ – Chet Miller Apr 11 at 12:55
  • $\begingroup$ @ChetMiller How? But I see the μ increases eh $\endgroup$ – The99sLearner Apr 11 at 13:11
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    $\begingroup$ I’m away from my computer now. I’ll write out my full analysis later. The final result is dG/dn = RTln x, where n is the number of moles of solvent in the mixture container and x is its mole fraction in that container. $\endgroup$ – Chet Miller Apr 11 at 13:56
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Let the subscript A refer to the solvent and B refer to the solute, and let the subscript 1 refer to the pure solvent container and 2 refer to the mixture container. Then the free energy of the combined system is given by: $$G=n_{A1}\mu_A^0+n_{A2}\left(\mu_A^0+RT\ln{\left[\frac{n_{A2}}{n_{A2}+n_{B_2}}\right]}\right)+n_{B2}\left(\mu_B^0+RT\ln{\left[\frac{n_{B2}}{n_{A2}+n_{B_2}}\right]}\right)$$If the number of moles of solvent in container 2 increases by $dn_{A2}$ and the number of moles of solvent in container 1 decreases by $dn_{A2}$, the change in the free energy of the combined system changes by dG. If we do the math correctly, we obtain: $$dG=RT\ln{x_{A2}dn_{A2}}\tag{1}$$where $x_{A2}$ is the mole fraction of solvent in the mixture container at any point: $$x_{A2}=\frac{n_{A2}}{n_{A2}+n_{B_2}}$$Eqn. 1 indicates that the change in free energy is negative (as expected).

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  • $\begingroup$ Just to clarify something, when you derive eqn.1, 1) do you assume n of A1 is approximately constant? 2) why is not μ of pure A (that μ ,A,0) included in dG expression? $\endgroup$ – The99sLearner Apr 11 at 22:42
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    $\begingroup$ Of course I don't assume $n_{A1}$ constant. If you differentiate my expression and take $dn_{A1}=-dn_{A2}$, the $\mu_A^0$ terms cancel out. $\endgroup$ – Chet Miller Apr 11 at 23:27
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    $\begingroup$ It would have been much easier to do this problem if I initially had just written $$dG=\mu_{A1}dn_{A1}+\mu_{A2}dn_{A2}+\mu_{B2}dn_{B2}$$since $dn_{B2}=0$. This would have immediately led to the same final result. $\endgroup$ – Chet Miller Apr 11 at 23:32
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To evaluate whether the process is spontaneous you have to take the difference of the chemical potentials $\mu$. Solvent molecules go from the pure solvent (reactant) to the solution (product):

$$\Delta_r G = \mu_\text{product} - \mu_\text{reactant} =$$ $$ = (\mu_\text{solvent}^\circ + R T \ln x_A) - \mu_\text{solvent}^\circ < 0$$ Throughout the process, $\mu$(solvent with solute)< $\mu$(pure solvent). As the process goes on, the solution becomes more and more diluted, so $\mu$(solvent with solute) will continuously increase, but never reach $\mu$(pure solvent). So the transport will go on, but the osmotic pressure will decrease as you get closer to equilibrium

In a different example, if one solution is at 1 mM and the other at 2 mM, solvent will go from the lower solute concentration to the higher solute concentration (or you could say from higher solvent concentration to lower solvent concentration) until the solute concentrations are equal and the osmotic pressure difference is zero.

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    $\begingroup$ 'μ(solvent with solute) will continuously increase', and if μ(pure solvent) does not change, it means your μ difference is positive, so not spontaneous? $\endgroup$ – The99sLearner Apr 11 at 12:23
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    $\begingroup$ $x_{A}$ is strictly less than 1, so $\mu_{\mathrm{product}} < \mu_{\mathrm{reactant}}$. @The99sLearner $\endgroup$ – Zhe Apr 11 at 12:59
  • $\begingroup$ @Zhe Does the inequality mean ΔG>0 because potential increases? $\endgroup$ – The99sLearner Apr 11 at 13:12
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    $\begingroup$ I think there might just be some looseness in terminology and notation here. The point is that you're continually diluting, so the mole faction is getting repeatedly smaller. That means that over time the potential becomes more negative thanks to the log, so the overall free energy change is also negative. $\endgroup$ – Zhe Apr 11 at 13:30
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    $\begingroup$ To evaluate whether the reaction goes forward you don't compare $\Delta G$ over time but compare the difference in Gibbs energy between products and reactants at each given time point. $\endgroup$ – Karsten Theis Apr 11 at 13:32

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