How do I find change in Gibbs free energy of water formation?

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 


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




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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