 # Standard Gibbs Free Energy of Formation (ΔGƒ°) 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 Gƒ°
• 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 (ΔHƒ°), absolute standard entropy (ΔS°) and standard temperature (T = 298.15 K):

ΔGƒ° = ΔHƒ°(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

ΔGƒ°(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) = ΣGƒ°(products) − ΣGƒ°(reactants)

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## Definition of ΔGƒ°

The standard Gibbs free energy of formation (ΔGƒ°) 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 (ΔGƒ°), we will need to know the standard enthalpy of formation of the species (ΔHƒ°), the temperature (T in kelvin), and the standard absolute entropy change for the formation of the species (ΔSƒ°).
Then we can write the equation for this calculation as:

ΔGƒ° = ΔHƒ° − TΔSƒ°

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 (ΔGƒ° = 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 (N2(g)), hydrogen gas (H2(g)) and chlorine gas (Cl2(g)) as shown in the balanced chemical equation below:

 reactants → product ½N2(g) + 2H2(g) + ½Cl2(g) → NH4Cl(s) elements (in their standard states) → compound(in its standard state)

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

at 298.2 K and 1 atmosphere pressure
Substance ΔHƒ°
(kJ mol-1)

(J K-1 mol-1)
N2(g) 0 191.5
H2(g) 0 130.6
Cl2(g) 0 223.0
NH4Cl(s) −314.4 94.6

Note that values for ΔHƒ° 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:

at 298.2 K and 1 atmosphere pressure
Substance ΔHƒ°
(kJ mol-1)

(J K-1 mol-1)
S° ÷ 1000
(kJ K-1 mol-1)
N2(g) 0 191.5 0.1915
H2(g) 0 130.6 0.1306
Cl2(g) 0 223.0 0.2230
NH4Cl(s) −314.4 94.6 0.0946

Now we calculate the change in entropy when the reactants N2(g), H2(g) and Cl2(g) are converted to the product NH4Cl(s), that is, we calculate the standard absolute entropy of formation (ΔSƒ°) of NH4Cl(s) in kJ K-1 mol-1 as shown below:

 ΔSƒ° = ΣS°(products) − ΣS°(reactants) = {1 × S°(NH4Cl(s))} − {[½ × S°(N2(g))] + [2 × S°(H2(g))] + [½ × S°(Cl2(g))]} = {1 × 0.0946} − {[½ × 0.1915] + [2 × 0.1306] + [½ × 0.2230]} = 0.0946 − {0.09575 + 0.2612 + 0.1115} = 0.0946 − 0.46845 = −0.3739 kJ K-1 mol-1

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

 ΔGƒ° = ΔHƒ° − TΔSƒ° = −314.4 − (298.2 × −0.3739) = −314.4 − (−111.5) = −314.4 + 111.5 = −202.9 kJ mol-1

Fortunately, values for the standard Gibbs free energy of formation have been tabulated for many compounds!
Tabulated values makes it much easier to use ΔGƒ° in calculations, as we shall see in the section below.
When we use tabulated values of ΔGƒ° 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.

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## Using ΔGƒ° 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:

Standard Free Energies of Formation at 298.2 K
Solid
Substances
ΔGƒ°
(kJ mol-1)
Liquid
Substances
ΔGƒ°
(kJ mol-1)
Gaseous
Substances
ΔGƒ°
(kJ mol-1)
NH4Cl(s) −202.7   CCl4(l) −68.7   C2H2(g) 209.2
K2O(s) −318.8   H2O2(l) −120.4   O3(g) 163.2
NaNO3(s) −365.9   CH3OH(l) −166.3   C2H4(g) 68.1
Na2O(s) −376.6   C2H5OH(l) −174.8   NH3(g) −16.5
NaCl(s) −384.0   H2O(l) −237.2   C2H6(g) −32.9
LiCl(s) −383.7   H2SO4(l) −690.1   H2S(g) −33.6
KCl(s) −408.3         CH4(g) −50.8
Li2O(s) −560.2         HCl(g) −95.3
MgO(s) −569.6         CO(g) −137.2
MgCl2(s) −592.3         H2O(g) −228.6
CaO(s) −604.2         SO2(g) −300.2
CaCl2(s) −748.1         SO3(g) −371.1
Al2O3(s) −1576         CO2(g) −394.4

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 (H2(g)) and oxygen gas (O2(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:

H2(g) + ½O2(g) → H2O(l)     ΔGƒ° = −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 (H2(g)) and oxygen gas (O2(g))?
What is the change in standard Gibbs free energy (ΔG°) for this decomposition reaction?

H2O(l) → H2(g) + ½O2(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) → H2(g) + ½O2(g)     ΔG° = −ΔGƒ° = +237.2 kJ mol-1

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

 Formation of Compound Decomposition of Compound elements in standard states → compound in standard state ΔGƒ° = x kJ mol-1 compound in standard state → elements in standard states ΔG° = −ΔGƒ° = −x kJ mol-1

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.

H2(g) + ½O2(g) → H2O(l)     ΔGƒ° = −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 × H2(g) + 10 × ½O2(g) → 10 × H2O(l)     ΔG° = 10 × ΔGƒ°(H2O(l)) = 10 × −237.2 kJ mol-1

That's right!

10H2(g) + 5O2(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 (ΔGƒ°) 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:

NH3(g) + HCl(g) → NH4Cl(s)

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

ΔGƒ°(NH3(g)) = −16.5 kJ mol-1

ΔGƒ°(HCl(g)) = −95.3 kJ mol-1

ΔGƒ°(NH4Cl(s)) = −202.7 kJ mol-1

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

½N2(g) + 3/2H2(g) → NH3(g)     ΔGƒ°(NH3(g)) = −16.5 kJ mol-1

½H2(g) + ½Cl2(g) → HCl(g)     ΔGƒ°(HCl(g)) = −95.3 kJ mol-1

½N2(g) + 2H2(g) + ½Cl2(g) → NH4Cl(s)     ΔGƒ°(NH4Cl(s)) = −202.7 kJ mol-1

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

NH3(g) + HCl(g) → NH4Cl(s)

So we need to "decompose" the reactants, ammonia (NH3(g)) and hydrogen chloride (HCl(g)), that is, we reverse the "formation" reactions AND reverse the sign of ΔG° as shown below:

NH3(g) → ½N2(g) + 3/2H2(g)     ΔG°(NH3(g)) = −ΔGƒ°(NH3(g)) = +16.5 kJ mol-1

HCl(g) → ½H2(g) + ½Cl2(g)     ΔG°(HCl(g)) = −ΔGƒ°(HCl(g)) = +95.3 kJ mol-1

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

½N2(g) + 2H2(g) + ½Cl2(g) → NH4Cl(s)     ΔG°(NH4Cl(s)) = ΔGƒ°(NH4Cl(s)) = −202.7 kJ mol-1

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

 reactants → products ΔG° (kJ mol-1) NH3(g) → ½N2(g) + 3/2H2(g) +16.5 HCl(g) → ½H2(g) + ½Cl2(g) +95.3 ½N2(g) + 2H2(g) + ½Cl2(g) → NH4Cl(s) −202.7 NH3(g) + HCl(g) → NH4Cl(s) −90.9

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 (ΔGƒ°) values?
First we reversed the equations, and sign of ΔGƒ°, for each reactant (ΔG° = −ΔGƒ°).
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) = −ΣΔGƒ°(reactants) + ΣΔGƒ°(products)

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

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

So for the reaction:

NH3(g) + HCl(g) → NH4Cl(s)

for which:

ΔGƒ°(NH3(g)) = −16.5 kJ mol-1

ΔGƒ°(HCl(g)) = −95.3 kJ mol-1

ΔGƒ°(NH4Cl(s)) = −202.7 kJ mol-1

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

 ΔG°(reaction) = ΣΔGƒ°(products) − ΣΔGƒ°(reactants) = ΔGƒ°(NH4Cl(s)) − {ΔGƒ°(NH3(g)) + ΔGƒ°(HCl(g))} = −202.7 − {−16.5 + −95.3} = −202.7 − {−111.8} = −202.7 + 111.8 = −90.9 kJ mol-1

We can use the Gibbs free energy of formation value (ΔGƒ°) 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) = ΣΔGƒ°(products) - ΣΔGƒ°(reactants)

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## 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, C2H2(g), to form carbon dioxide gas, CO2(g), and liquid water, H2O(l), at 25°C, according to the balanced chemical equation below:

C2H2(g) + 5/2O2(g) → 2CO2(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.)

STOP STOP! State the Question.
What is the question asking you to do?
Calculate change in standard Gibbs free energy for the reaction at 298.15 K (25°C)

ΔG°(reaction) = ? kJ mol-1

PAUSE PAUSE to Prepare a Game Plan
(1) What information (data) have you been given in the question?

(a) Balanced chemical equation for the reaction:

C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(l)

(b) Relevant ΔGƒ° values at 25°C (298.15 K):

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

(2) What is the relationship between what you know and what you need to find out?

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

GO GO with the Game Plan

 ΔG°(reaction) = ΣΔGƒ°(products) − ΣΔGƒ°(reactants) = {[2 × Gƒ°(CO2(g))] + [1 × Gƒ°(H2O(l))]} − {[1 × ΔGƒ°(C2H2(g))] + [5/2 × Gƒ°(O2(g))]} Note that ΔGƒ°(O2(g)) = 0 (an element in its standard state) Substitute the values for ΔGƒ° into the equation and solve for ΔG°(reaction) ΔG°(reaction) = {[2 × −394.4] + [1 × −237.2]} − {[1 × 209.2] + [5/2 × 0]} = {−788.8 −237.2} − {209.2 + 0} = −1026 − 209.2 = −1235.2 kJ mol-1

PAUSE PAUSE to Ponder Plausibility

Yes, we have calculated ΔG° for the reaction.

(a) The combustion of a fuel like acetylene must be a spontaneous process, that is, ΔG will be negative.
Our value of ΔG°(reaction) is −1235.2 kJ mol-1, that is, our value is negative so it is plausible.

(b) We could check our value by using known tabulated values of for the heat of combustion of acetylene (ΔH° = −1300 kJ mol-1) and standard absolute entropy to calculate ΔG° at 298.2 K as given below:

Substance S° (J K-1 mol-1) S° (kJ K-1 mol-1)
C2H(g) 200.8 0.2008
O2(g) 205.1 0.2051
CO2(g) 213.6 0.2136
H2O(l) 69.9 0.0699

ΔS°(reaction) = ΣS°(products) − ΣS°(reactants)
ΔS°(reaction) = {(2 × 0.2136) + (1 × 0.0699)} − {(1 × 0.2008 ) + (5/2 × 0.2051)}
ΔS°(reaction) = 0.4971 − 0.7136 = −0.2165 kJ K-1 mol-1

At 298.2 K :
ΔG°(reaction) = ΔH° −(298.2 × ΔS°)
ΔG°(reaction) = −1300 − (298.2 × −0.2165)
ΔG°(reaction) = −1300 + 64.56 = −1235.4 kJ mol-1

Because this result agrees with the result we got when we calculated ΔG°(reaction) using standard Gibbs free energy of formation values, we are reasonably confident that our answer of −1235.2 kJ mol-1 is correct.

STOP STOP! State the Solution
ΔG°(reaction) = −1235.2 kJ mol-1

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Footnotes:

(1) Actually, this is the amount of energy available to do non-PV work since the system must be maintained at constant pressure.
Refer to the definition of Gibbs Free Energy.