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How do I find the change in Gibbs free energy of the reaction

$$\ce{H2 + \frac{1}{2}O2 -> H2O}$$

when it is at $\pu{75^\circ C}$ and $\pu{1 atm}$? Do I calculate the Gibbs free energy of each species and then do

$$\sum\Delta G{\mathrm{(products)}} - \sum\Delta G{\mathrm{(reactants)}}?$$

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You're using the correct equation, $\Delta G^0_r(T;P)=\sum \nu_i \mu_i^0(T;P)$, where $\Delta G_r^0(T;P)$ is the standard free energy change for the reaction at $T$ and the pressure $P$. The task is to calculate each $\mu_i^0$. Provided the same convention for the standard state pressure $P^0$ and for the zero value of $\mu_i$ is used, then $\Delta G_r^0$ for any reaction will be the same. Here are some ways to calculate $\mu_i^0(T;P)$ for each species to use in the equation and the data required to calculate them (available in tables). In the following, $P^0$ is the standard state pressure used in the tables and subscript $fi$ denotes the reaction forming the species from the elements in their standard states. If the tabular values are available at a different pressure than the one you want ($P$), you must add $RT\ln(P/P^0)$ to the values of $\mu_i$ indicated below.

  1. $\mu_i^0 = \Delta G^0_{fi}(T;P^0)$.
  2. $\mu_i^0 = \Delta H^0_{fi}(T) - T\Delta S^0_{fi}$.
  3. $\mu_i^0 = \Delta H^0_{fi}(298.15) + \int_{298.15}^Tc_{pi}^0(T) dT -Ts_i^0(T)$.
  4. $\mu_i^0 = \Delta H^0_{fi}(T) - Ts_i^0(T)$.
  5. $\mu_i^0 = \Delta H^0_{fi}(298.15) + \int_{298.15}^Tc_{pi}^0(T)dT -T\left[s_i^0(298.15) + \int_{298.15}^T \frac{c^0_{pi}}{T}dT\right]$

There are many other ways. It's an interesting exercise to prove why all the above routes give the same value of $\Delta G^0_r$ for any reaction.

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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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    $\begingroup$ 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. $\endgroup$ – Nathaniel Sep 12 '12 at 13:05
  • $\begingroup$ @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. $\endgroup$ – Leonardo Sep 12 '12 at 13:47
  • $\begingroup$ 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. $\endgroup$ – Nathaniel Sep 12 '12 at 19:35

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