The solubility of $O_2(g)$ in water at $25^o \text{C}$ and $\text{1 bar}$ pressure of $O_2(g)$ above the solution is $0.00126$ $mol$ $O_2(g)/kg$ $water$. Find $\Delta_fG_{298.15}^o(O_2(g),sln)$.

I kinda have a clue on how to solve this, but there's just one variable I'm stuck on figuring out. I'm not sure how to get the molal-scale activity coefficient for O2 gas in solution. The answer is $\Delta_fG_{298.15}^o(O_2(g),sln) = \text{16.8 kJ/mol}$... now how would I get there?

Here's a derivation I went through to show what I have tried. The chemical potential for $O_2(g)$ relative to some reference standard state is for a real solution on the molal-scale is:

$\mu_{O_2(g)} = \mu_{m,O_2(g)}^o + RTln(\gamma_{m,O_2(g)}\frac{m_{O_2(g)}}{m^o})$

By subtracting the standard molar Gibbs energy of reactants in their elemental state, $G_{elem}^o$, and knowing that the chemical potential of a pure substance is equal to the chemical potential of that substance in a saturated solution, $\mu_{O_2(g)} = \mu_{O_2(g)}^*$, and we should get:

$\mu_{O_2(g)}^* - G_{elem}^o = \mu_{m,O_2(g)}^o - G_{elem}^o + RTln(\gamma_{m,O_2(g)}\frac{m_{O_2(g)}}{m^o})$

$\Delta_fG_T^o(O_2(g)) = \Delta_fG_T^o(O_2(g),sln) + RTln(\gamma_{m,O_2(g)}\frac{m_{O_2(g)}}{m^o})$

where $\Delta_fG_{298.15}^o(O_2(g))$ is $\text{0 kJ/mol}$ since oxygen gas is in its elemental state, and $\Delta_fG_{298.15}^o(O_2(g),sln)$ is what I am supposed to solve for. So, the final equation I think I'm supposed to use is:

$\Delta_fG_T^o(O_2(g),sln) = -RTln(\gamma_{m,O_2(g)}\frac{m_{O_2(g)}}{m^o})$

But I don't have $\gamma_{m,O_2(g)}$... so how could I do this? This is due next Tuesday, but it would be nice to have it addressed asap.

  • $\begingroup$ The textbook I am using is Physical Chemistry, 6th ed., by Ira N. Levine, if that helps. This is problem 10.55a at the end of the chapter. $\endgroup$ – timaeus222 Nov 23 '16 at 1:42
  • $\begingroup$ Isn't the chemical potential of the oxygen in the liquid solution the same as that of the oxygen in the gas phase (at equilibrium)? $\endgroup$ – Chet Miller Nov 23 '16 at 12:37
  • $\begingroup$ Yeah, it should be; that's what I said when I wrote $\mu_{O_2(g)} = \mu_{O_2(g)}^{*}$, right? $\endgroup$ – timaeus222 Nov 23 '16 at 18:48
  • $\begingroup$ So the free energy of formation of the oxygen in the solution should also be zero, right? $\endgroup$ – Chet Miller Nov 24 '16 at 1:07
  • $\begingroup$ Well, no, because oxygen doesn't naturally exist in solution. It is supposed to be $\text{16.8 kJ/mol}$. However, as you would expect, the free energy of formation of the oxygen in its natural, gaseous state is 0 though. $\endgroup$ – timaeus222 Nov 24 '16 at 2:17

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