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How do I find the change in Gibbs free energy of the reaction $\ce{H_2 + \frac{1}{2}O_2 -> H_2O}$ when it is at $\ce{75}$°$\ce{C}$ and $\ce{1 atm}$. Do I calculate the Gibbs free energy of each species and then do $\sum\Delta G{(products)}$ - $\sum\Delta G{(reactants)}$

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2 Answers

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You may notice, that at $\ce{75}$°$\ce{C}$ this reaction is heterogenous (water in a liquid phase). So change of free Gibbs energy might be calculated from equilibrium constant:

$\Delta G=-RT\ln{K_p}=-RT\ln{\frac{1}{P_{H_2}\sqrt{P_{O_2}}}}=RT\ln{(P_{H_2}\sqrt{P_{O_2}})}$

Where $P_{H_2}$ and $P_{O_2}$ are partial pressures, and can be calculated from Dalton's law. Also, if you use fugacity coefficients - you'll get more clear results.

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Warning: this is only a way to estimate and not necessarily good practice.

I never actually did one like this, but I believe that one must use the equation $\Delta G$° $= \Delta H$°$ - T\Delta S$° with the assumptions that the information you will need to look up for $\Delta H$° and $\Delta S$° are temperature independent. This will only allow you to estimate $\Delta G$° and remember to convert to kelvin when plugging in the temp.

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This isn't quite right. You'd think that $\Delta G^\circ$ would equal $\Delta H^\circ - T \Delta S^\circ$, but in fact $\Delta S^\circ$ is usually defined on a scale with a zero point defined by the third law, whereas $\Delta H^\circ$ and $\Delta G^\circ$ are defined to be zero for a pure substance at standard conditions. These incompatible scales mean that equation doesn't hold. Luckily, though, the values of $\Delta G^\circ$ are usually tabulated, so you can just look them up. – Nathaniel Sep 12 '12 at 13:05
@Nathaniel You are probably very correct, I only was offering a way to estimate, unfortunately I am not an engineer so my answer should be more clearly stated as an estimate, not a good answer. I wonder though, how far off the estimate would be from the real answer. – Leonardo Sep 12 '12 at 13:47
Well, you can always give it a try. A good textbook will have all three values ($\Delta G^\circ$, $\Delta H^\circ$ and $\Delta S^\circ$) for various compounds in a table at the back, so you can try calculating $\Delta H^\circ - T\Delta S^\circ$ and see what you get. My guess is that it will usually be very far away from $\Delta G^\circ$ though - they're on different scales, so it's a bit like adding miles to kilometres. – Nathaniel Sep 12 '12 at 19:35

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