Стандартна вільна енергія утворення

We've been studying thermodynamics. We started with internal energy and the first law. At constant pressures, it was easier to calculate enthalpy. We defined it as heat. Next, we looked at entropy. It measured energy dispersal. Energy spread helped us understand the second and third laws. The combination of enthalpy and entropy gave us Gibb's free energy. Gibb's free energy is the energy that's available ("free") for doing work. Its change during a chemical reaction tells us if the spontaneity. Gibb's free energy is always at a fixed temperature and pressure. What if we set the temperature and pressure to standard conditions. We get the standard free energy of formation.

What is the Standard Free Energy of Formation?

The standard free energy of formation has three components.

1. It's the change in free energy from the formation of one mole of a molecule in its standard state.

2. The molecule is produced from its standard state elements.

3. The molecule's formation occurs at standard temperature and pressure (STP).

In the definition, the molecule forms from its elements. But what about pure elements in their standard states? We define their energy to be zero. That creates a baseline for our calculations.

What are the Standard States?

To define the standard free energy of formation, we need to know the standard states. They are listed below.

$$\begin{array}{c|c} State & Standard\,Condition \\ \hline Solid & Pure\,solid \\ Liquid & Pure\,liquid \\ Gas & 1.0\,atm. \\ Solution & 1.0\,molar \end{array}$$

How to Calculate Standard Free Energy-Part 1

To calculate the standard free energy of formation, we need the standard enthalpy of formation and the entropy change at STP. Typically, we can look up the formation enthalpy values. It's excellent when chemists have calculated or measured things for us! For standard state elements, enthalpy is zero. Remember, entropy is temperature-dependent. So, we need the entropy at $298.15 K$. That means standard state elements will have an entropy value!

Once we have the values, we can use the formula: $\Delta G_{f} ^{o}=\Delta H_{f} ^{o}-T\Delta S^{o}$. The $^{o}$ indicates standard conditions. The "f" denotes "of formation."

Let's look at an example. Let's say we have the thermochemical equation for methane formation. It gives us the standard enthalpy of formation.

• $C_{(s)} + 2H_{2(g)} \rightarrow CH_{4(g)}\, \Delta H^{o}_{f}= -74475.2\frac{J}{mol}$.

• Let's say we're given the reactions entropy change. It's $\Delta S^{o}=-80.9\frac{J}{mol\times K}$.

• If we had to calculate it, we would look up the standard entropy values for methane, carbon, and hydrogen. Then we could use the formula $\Delta S^{o}=\Sigma n \Delta S^{o}_{products} - \Sigma n \Delta S^{o}_{reactants}$. Remember n is the mole ratios.

So, the standard free energy of formation is

• $\Delta G_{f} ^{o}=\Delta H_{f} ^{o}-T\Delta S^{o}\rightarrow -74475.2\frac{J}{mol} - 298.15K \times -80.9\frac{J}{mol\times K} = -50354.7\frac{J}{mol}$.

How to Calculate Standard Free Energy-Part 2

Both enthalpy and entropy are state functions. Gibb's free energy depends on both. So, it is also a state function. That means we can apply the method we learned from Hess's law! Physical chemists have tabulated the standard free energy of formation for thousands of molecules. We can use the standard free energy of formation to calculate the standard free energy of chemical reactions.

The Hess's law formula is

$\Delta G^{o}_{rxn}=\Sigma n \Delta G^{o}_{products} - \Sigma n \Delta G^{o}_{reactants}$

Let's look at the combustion of ethane. Its balanced equation is

$C_{2}H_{6(g)}+ 3.5O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_{2}O_{(l)}$.

We can look up the standard free energy of formation for each molecule.

• $C_{2}H_{6(g)}=-31.8\frac{KJ}{mol\times K}$

• $O_{2(g)}=0$

• Oxygen is a gaseous diatomic element. So, it's in its standard state.

• $CO_{2(g)}=-394.4\frac{KJ}{mol\times K}$

• $H_{2}O_{(l)}=-237.14\frac{KJ}{mol\times K}$

Now we can use Hess's law to calculate the standard free energy.

$\Delta G^{0}_{rxn}= (2\times-394.4+3\times-237.14)-(-31.8+3.5\times0)=-1468.4\frac{KJ}{mol}$

(Video) Calculate Free Energy G from Free Energies of Formation

by Old School Chemistry

In this video, Old School Chemistry demonstrates how to calculate a reaction's Gibbs free energy from the standard free energies of formation. She starts by telling you some common mistakes to avoid. Next, she walks you through two example problems. If you need help with free energy of formation calculations, this video is excellent!