0
$\begingroup$

I am a high school student reading about the second law of thermodynamics, and one of the equations given was the free energy change in dilution.

$$\Delta G_{\text{dil}} = -RT \log(C_1/C_2)$$

This implies that as you make a solution more dilute, the free energy decreases. However this equation does not take into account the entropy change in breaking a solid crystal, which I would expect to be highly positive as the crystal is a highly symmetric arrangement. Also it assumes ideal solution.

If we take into account the entropy change in breaking up the solid crystal, we see that the free energy increases while diluting, and also increases in breaking up the solid , that is, concentrating. So I would expect a minima in the free energy vs concentration curve. My guess is it might be the saturation point?

Is my reasoning correct? If it is, then is it possible to calculate the minima?

$\endgroup$
4
  • 1
    $\begingroup$ I think that the use of dilution in the description of the equation implies that you start with a solution, and then dilute, rather than starting from a solid. Starting from a solid, you would need to calculate the energy change for dissolving the solid first. Once the solid was completely dissolved, then you could use this equation for diluting it further. $\endgroup$ – Sol Jul 31 '20 at 4:46
  • $\begingroup$ Okay, but even without the equation my logic still holds, $\endgroup$ – Manit Agarwal-El psy congroo Jul 31 '20 at 5:36
  • 2
    $\begingroup$ If entropy in dilution is positive, then according to $\Delta G = \Delta H - T\Delta S$, the free energy decreases, does it not? $\endgroup$ – Safdar Faisal Jul 31 '20 at 5:40
  • $\begingroup$ @ManitAgarwal-Elpsycongroo: Sol is correct. You can see this for yourself by substituting numbers into your expression. C1 is the initial concentration, and C2 is the final concentration. Suppose C1 =20 and C2 = 10 (this would be a two-fold dilution; the final concentration, C2, is half the initial concentration, C1). Then $\Delta G = -RT \ln (C1/C2) = -RT \ln (20/10) = –RT \ln 2 = -0.693 RT$. And since $RT>0$ , $\Delta G < 0$. I suspect you got confused because you were thinking more dilute means a higher C, when in fact it's the opposite. $\endgroup$ – theorist Aug 2 '20 at 4:04
1
$\begingroup$

For Gibbs Free Energy, the standard equation is ΔG=ΔH−TΔS. The overall change in Gibbs Free Energy will depend, at least in part, on the enthalpy of dissolution for the specific substance. As for entropy, increasing entropy by breaking up the crystal decreases Gibbs Free Energy. The saturation point is where ΔG for the dissolution equals zero: where ΔH = TΔS. As ΔH and ΔS vary depending on the particulars of the solute and solvent, the saturation point also varies. The equation you cite is specifically for when you have two solutions, so the dissolution of the solid is not described by it. Since the saturation point depends on the enthalpy of dissolution, you can't use the log equation to find it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.