Because free energy, *G* is a state function, we can calculate the free energy change of a reaction, Δ*G*^{o}_{rxn} by subtracting the free energies of the reactants of the reaction from the free energies of the products of the reaction.

Remember, a state function depends only on the initial and final values and not on the path that leads to these states. A classic example of a state function is attitude. Regardless of the path the hikers have taken, they both are at the same altitude:

The standard free energy change of the reaction is calculated by **subtracting the standard free energies of formation of the reactants ***multiplied by their stoichiometric coefficients* from the **standard free energies of formation of the products** *multiplied by their stoichiometric coefficients*:

** **

** **

** **

The ** standard free energy of formation **of a compound, Δ

*G*

^{o}

_{f}is

*the free-energy change that occurs when*

*1 mole of the compound is synthesized from its elements in their standard states*. This is similar to the definition of standard enthalpies of formation that we used to calculate the enthalpy of a reaction.

- The
and products are given in textbooks under the appendix for thermodynamics data.*standard free energy of formation* - Like the standard enthalpy of formation, Δ
*H*^{o}_{f}, Δ*G*^{o}_{f}for elements is**equal to zero at 25 °C.**

**For example**, given the values of Δ*G*^{o}_{f}, calculate Δ*G°*_{rxn} for the following reaction at 25°C:

2C_{2}H_{6}(*g*) + 7O_{2}(*g*) → 4CO_{2}(*g*) + 6H_{2}O(*l*)

**Solution:** To calculate 𝚫*G*°_{rxn}, we need to subtract the standard free energies of formations of the reactants from the standard free energies of formations of products, both multiplied by their stoichiometric coefficients:

Δ*G*°_{rxn} = Σ*n*_{p}Δ*G*^{o}_{f} (products) – Σ*n*_{r}Δ*G*^{o}_{f }(reactants)

*where n _{p} and n_{r} are the molar coefficients of the products and reactants in the balanced chemical equation*

Remember, the standard free energies of formations of elements or their molecules in standard states are equal to zero.

From an appendix for thermodynamic data, we find that:

*ΔG*_{f}**° **(kJ/mol): for C_{2}H_{6}(*g*) = -46.2, for O_{2}(*g*) = 0, for CO_{2}(*g*) = -394.4, for H_{2}O(*l*) = -237.2

We can now enter the numbers to determine the heat of our reaction:

*ΔG°*_{rxn} = [6 x *G*_{f}° (H_{2}O) + 4 x *G*_{f}° (CO_{2})] – [2 x *G*_{f}° (C_{2}H_{6}) + 7 x *G*_{f}° (O_{2})]

*ΔG*°_{rxn} = [6 x (-237.2 kJ) + 4 x (-394.4)] – [2 x (-32.89 kJ) + 7 x 0] = **-2935 kJ/mol**

As already mentioned, another way of determining the standard free energy change of a reaction is using the Gibbs free energy equation.

Which equation you need to use, depends largely on which one you *can* use. In some problems, it will be mentioned which one to use, in some, you will need to go based on the data given.

For example, *this equation, ΔG° _{rxn} = Σn_{p}ΔG^{o}_{f} (products) – Σn_{r}ΔG^{o}_{f }(reactants), is not suitable for calculating the free energy charge of reactions at higher temperatures because the values of *Δ

*G*

^{o}

_{f }are usually given for 25

^{o}C.

Instead, we **use the Gibbs free energy formula** correlating Δ*G*° with Δ*H*° and Δ*S*°:

**Δ G° = ΔH° – TΔS°**

You may ask how come we can use the tabulated values of Δ*H*°_{f }and *S*° but not *G*^{o}_{f.}

The reason is that although the enthalpies and entropies are affected by temperature, the difference between their values is very little compared to the temperature change. As a result, the values of Δ*H*° and Δ*S*° do not change much with temperature. However, **Δ G° is more sensitive to temperature** which we can see by the presence of T in the equation Δ

*G*° = Δ

*H*° – TΔ

*S*°.

**Check Also**

- Standard Entropy Change (𝚫
*S*^{o}_{rxn}) of a Reaction - The Gibbs Free Energy
- The Effect of 𝚫H, 𝚫S, and T on 𝚫G – Spontaneity
- Entropy and State Change
- Entropy Changes in the Surroundings
- Gibbs Free Energy and Hess’s Law
- Gibbs Free Energy Under Nonstandard Conditions
- Gibbs Free Energy and Equilibrium Constant
**Entropy, Enthalpy, and Gibbs Free Energy Practice Problems**

#### Practice

Using the data for standard free energies of formations in the attached appendix, calculate the standard free energy of each reaction at 25 ^{o}C.

**a) **6Cl_{2}(*g*) + 2Fe_{2}O_{3}(*s*) → 4FeCl_{3}(*s*) + 3O_{2}(*g*)

**b) **2CH_{3}OH(*g*) + H_{2}(*g*) → C_{2}H_{6}(*g*) + 2H_{2}O(*g*)

**c)** Fe_{2}O_{3}(*s*) + 3CO(*g*) → 2Fe(*s*) + 3CO_{2}(*g*)