Standard Gibbs Free Energy of Formation (ΔG_ƒ°) Calculations Chemistry Tutorial

Key Concepts

• Gibbs free energy is given the symbol G
• Standard Gibbs free energy is given the symbol G°
• Standard Gibbs free energy of formation is given the symbol G_ƒ°
• Standard Gibbs free energy of formation is the change in Gibbs free energy when elements in their standard states combine to form a product also in its standard state.
• Standard Gibbs free energy of formation of a compound can be calculated using standard enthalpy of formation (ΔH_ƒ°), absolute standard entropy (ΔS°) and standard temperature (T = 298.15 K):

  ΔG_ƒ° = ΔH_ƒ°(compound) - TΔS°(compound)

• Values of standard Gibbs free energy of formation are tabulated for many compounds.
• The standard Gibbs free energy of formation of an element in its standard state is 0

  ΔG_ƒ°(element) = 0

• Values of standard Gibbs free energy of formation can be used to calculate the change in Gibbs free energy for a chemical reaction or physical change:

  reactants → products

  ΔG°(reaction) = ΣG_ƒ°(products) − ΣG_ƒ°(reactants)

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Definition of ΔG_ƒ°

The standard Gibbs free energy of formation (ΔG_ƒ°) refers to the change in Gibbs free energy when elements in their standard states react to form a new product in its standard state.

Recall that, for species in their standard states at constant temperature and pressure, the change in standard Gibbs free energy (ΔG°) for chemical reaction or physical change can be calculated using the standard enthlpy change (ΔH°), the temperature (T in kelvin), and the standard absolute entropy change (ΔS°) as shown by the equation below:

ΔG° = ΔH° − TΔS°

So, if we want to calculate the standard Gibbs free energy of formation of a species (ΔG_ƒ°), we will need to know the standard enthalpy of formation of the species (ΔH_ƒ°), the temperature (T in kelvin), and the standard absolute entropy change for the formation of the species (ΔS_ƒ°).
Then we can write the equation for this calculation as:

ΔG_ƒ° = ΔH_ƒ° − TΔS_ƒ°

Values of the standard heat of formation (standard enthalpy of formation) are tabulated for many compounds (at 298.15 K and atmospheric pressure).
Values of standard absolute entropy are also tabulated (at 298.15 K and atmospheric pressure).
So we can use those tabulated values to calculate the value of the standard Gibbs free energy of formation of a compound.

Note that the standard Gibbs free energy of formation of an element in its standard is zero (ΔG_ƒ° = 0) because, at 298.15 K and atmospheric pressure, the element in its standard state already exists, no change has to take place.

Consider the formation of ammonium chloride in its standard state (solid), NH₄Cl_(s), from its elements in their standard states at standard temperature and pressure, that is, from nitrogen gas (N_2(g)), hydrogen gas (H_2(g)) and chlorine gas (Cl_2(g)) as shown in the balanced chemical equation below:

We can look up tabulated values for ΔH_ƒ° and S°. These values are given in the table below:

Note that values for ΔH_ƒ° are given in kJ mol^-1 but values of ΔS° are given in J K^-1 mol^-1 so we need to convert entropy in J to kJ by dividing by 1000 J/kJ as shown in the table below:

Now we calculate the change in entropy when the reactants N_2(g), H_2(g) and Cl_2(g) are converted to the product NH₄Cl_(s), that is, we calculate the standard absolute entropy of formation (ΔS_ƒ°) of NH₄Cl_(s) in kJ K^-1 mol^-1 as shown below:

We can now use the standard temperature, 298.2 K (to 4 significant figures), to calculate the standard Gibbs free energy change (ΔG_ƒ°) for the reaction:

Fortunately, values for the standard Gibbs free energy of formation have been tabulated for many compounds!
Tabulated values makes it much easier to use ΔG_ƒ° in calculations, as we shall see in the section below.
When we use tabulated values of ΔG_ƒ° we need to remember that those values refer to the formation of a product in its standard state from its elements which are also in their standard states.

Using ΔG_ƒ° to Calculate Standard Gibbs Free Energy Change of a Chemical Reaction (ΔG°)

The value of Gibbs free energy is dependent on temperature.
Values of standard Gibbs free energy are tabulated for a specific temperature, usually 298.15 K.
An example of a table of values for standard Gibbs free energy of formation is given below:

These tabulated values give us the change in Gibbs free energy when product in its standard state is formed from its elements, also in their standard states.

From the table above we see that the standard Gibbs free energy of formation of solid ammonium chloride, NH₄Cl_(s) is −202.7 kJ mol^-1.
In the section above we calculated the standard Gibbs free energy of formation of NH₄Cl_(s) to be −202.9 kJ mol^-1.
Note that both values are consistent since there is uncertainty in the value of the digit after the decimal point.

For example, from the table we see that the standard Gibbs free energy of formation of liquid water is −237.2 kJ mol^-1.
This tells us that when 1 mole of liquid water at 298.2 K and atmospheric pressure is formed from its elements in their standard states, that is, from hydrogen gas (H_2(g)) and oxygen gas (O_2(g)), the change in standard Gibbs free energy is 237.2 kJ and the reaction is spontaneous (because the sign of ΔG is negative).
The balanced chemical equation for the formation of 1 mole of liquid water from its elements in their standard states is shown below:

H_2(g) + ½O_2(g) → H₂O_(l)     ΔG_ƒ° = −237.2 kJ mol^-1

This chemical equation tell us that, under standard conditions, if we react (combust) 1 mole of hydrogen gas with ½ mole oxygen gas then 1 mole of liquid water will form in a spontaneous (or self-sustaining) chemical reaction. The difference between the enthalpy change for the reaction (ΔH°) and the dissipation of the energy within the chemical system (TΔS°) is 237.2 kJ for every mole of liquid water produced. 237.2 kJ of energy per mole of liquid water formed is available to do work on the surroundings, that is, by using this chemical reaction we achieve 237.2 kJ of energy that is "free" to do work.⁽¹⁾

What if I decompose liquid water into its elements: hydrogen gas (H_2(g)) and oxygen gas (O_2(g))?
What is the change in standard Gibbs free energy (ΔG°) for this decomposition reaction?

H₂O_(l) → H_2(g) + ½O_2(g)

The magnitude of the change in Gibbs free energy will be the same, but, the reaction will no longer be spontaneous, that means we will have to constantly do work on the chemical system by supplying energy to keep the reaction going, so the sign of ΔG° will now be positive!

H₂O_(l) → H_2(g) + ½O_2(g)     ΔG° = −ΔG_ƒ° = +237.2 kJ mol^-1

We can generalise and say that if we reverse the chemical reaction we must also reverse the sign of ΔG_ƒ°

If I combust one mole of hydrogen gas in ½ mole of oxygen gas to form 1 mole of liquid water at 298.2 K, then the Gibbs free energy of formation is −237.2 kJ mol^-1.

H_2(g) + ½O_2(g) → H₂O_(l)     ΔG_ƒ° = −237.2 kJ mol^-1

But what if I form 10 moles of liquid water instead of 1 mole? Surely 10 times more "free" energy will be released to do work on the surroundings?

10 × H_2(g) + 10 × ½O_2(g) → 10 × H₂O_(l)     ΔG° = 10 × ΔG_ƒ°(H₂O_(l)) = 10 × −237.2 kJ mol^-1

That's right!

10H_2(g) + 5O_2(g) → 10H₂O_(l)     ΔG° = −2372 kJ mol^-1

We can use what we have learnt above to calculate ΔG° for a chemical reaction using just the values of standard Gibbs free energy of formation (ΔG_ƒ°) for each reactant and product in the reaction.

Consider a chemical reaction in which ammonia gas reacts with hydrogen chloride gas to produce solid ammonium chloride (at 298.2 K and atmospheric pressure) as shown in the balanced chemical equation below:

NH_3(g) + HCl_(g) → NH₄Cl_(s)

Using the table of values of standard Gibbs free energy of formation (ΔG_ƒ°) above we find that:

ΔG_ƒ°(NH_3(g)) = −16.5 kJ mol^-1

ΔG_ƒ°(HCl_(g)) = −95.3 kJ mol^-1

ΔG_ƒ°(NH₄Cl_(s)) = −202.7 kJ mol^-1

So we could write a chemical equation to represent the formation of each compound as shown below:

½N_2(g) + ³/₂H_2(g) → NH_3(g)     ΔG_ƒ°(NH_3(g)) = −16.5 kJ mol^-1

½H_2(g) + ½Cl_2(g) → HCl_(g)     ΔG_ƒ°(HCl_(g)) = −95.3 kJ mol^-1

½N_2(g) + 2H_2(g) + ½Cl_2(g) → NH₄Cl_(s)     ΔG_ƒ°(NH₄Cl_(s)) = −202.7 kJ mol^-1

Now, notice that we are interested in the formation of solid ammonium chloride, that is:

NH_3(g) + HCl_(g) → NH₄Cl_(s)

So we need to "decompose" the reactants, ammonia (NH_3(g)) and hydrogen chloride (HCl_(g)), that is, we reverse the "formation" reactions AND reverse the sign of ΔG° as shown below:

NH_3(g) → ½N_2(g) + ³/₂H_2(g)     ΔG°(NH_3(g)) = −ΔG_ƒ°(NH_3(g)) = +16.5 kJ mol^-1

HCl_(g) → ½H_2(g) + ½Cl_2(g)     ΔG°(HCl_(g)) = −ΔG_ƒ°(HCl_(g)) = +95.3 kJ mol^-1

Note that we do not need to reverse the reaction for the formation of ammonium chloride:

½N_2(g) + 2H_2(g) + ½Cl_2(g) → NH₄Cl_(s)     ΔG°(NH₄Cl_(s)) = ΔG_ƒ°(NH₄Cl_(s)) = −202.7 kJ mol^-1

Now we can just add together the chemical equations and the values of ΔG° as shown below:

And perhaps now you can see an even quicker way to calculate the change in standard Gibbs free energy for a reaction (ΔG°) using standard Gibbs free energy of formation (ΔG_ƒ°) values?
First we reversed the equations, and sign of ΔG_ƒ°, for each reactant (ΔG° = −ΔG_ƒ°).
Next we added that result to the Gibbs free energy of formation of the product.
We can express this as a generalised form of mathematical equation as shown below:

ΔG°(reaction) = −ΣΔG_ƒ°(reactants) + ΣΔG_ƒ°(products)

We usually write this equation in a slightly different, but equivalent, way as shown below:

ΔG°(reaction) = ΣΔG_ƒ°(products) − ΣΔG_ƒ°(reactants)

So for the reaction:

NH_3(g) + HCl_(g) → NH₄Cl_(s)

for which:

ΔG_ƒ°(NH_3(g)) = −16.5 kJ mol^-1

ΔG_ƒ°(HCl_(g)) = −95.3 kJ mol^-1

ΔG_ƒ°(NH₄Cl_(s)) = −202.7 kJ mol^-1

we can calculate the value of ΔG°(reaction) as followings:

We can use the Gibbs free energy of formation value (ΔG_ƒ°) for each reactant and product in a chemical reaction to calculate the overall change in standard Gibbs free energy for that chemical reaction (ΔG°_reaction) using the following relationship:

ΔG°(reaction) = ΣΔG_ƒ°(products) - ΣΔG_ƒ°(reactants)

Worked Example : Calculating Standard Gibbs Free Energy Change for a Reaction Using Standard Gibbs Free Energy of Formation Data

Question:

Calculate the standard Gibbs free energy of combustion of acetylene, C₂H_2(g), to form carbon dioxide gas, CO_2(g), and liquid water, H₂O_(l), at 25°C, according to the balanced chemical equation below:

C₂H_2(g) + ⁵/₂O_2(g) → 2CO_2(g) + H₂O_(l)

The standard Gibbs free energy of formation of some substances at 298.15 K are given below:

Substance	ΔGƒ° (kJ mol-1)
C2H2(g)	209.2
CO(g)	−137.2
CO2(g)	−394.4
H2O(l)	−237.2
H2O(g)	−228.6

Solution:

(Based on the StoPGoPS approach to problem solving.)