## 5G 21

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

LBacker_2E
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### 5G 21

I am confused on why the components that make up the delta G*r are different for a, b, and c. For a only the product is considered (H2O, g), for b the product (CO2, g) is subtracted from one of the reactants (CO, g), and for part C, the reactant (CaCO3, s) is subtracted from the sum of the products (CaO, s, and CO2, g). Can someone please explain how you determine this?

Sebastian Lee 1L
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### Re: 5G 21

In this problem, you are solving for the change in gibbs free energy of the reaction by adding/subtracting standard gibbs free energy of formation of the products and reactants. The equation is $\Delta G_{rxn}^{\circ} = \sum \Delta G^{\circ}_{f}(products)-\sum \Delta G^{\circ}_{f}(reactants)$

They found the standard delta G of the products and subtracted it from the standard delta G of the reactants. In the case of a), the delta G of the reactants is 0 since they are in the most stable states (H2 and O2) so the equation of products - reactants is just products since reactants is 0. In part b), the delta G of O2 is 0 and in part c), none of the species are in their most stable form.