 Gibbs Free Energy Calculations (ΔG = ΔH - TΔS) Chemistry Tutorial

Key Concepts

• Gibbs free energy is also known as free energy or Gibbs energy.
• Gibbs free energy is given the symbol G
• Gibbs free energy (G) usually has the units kilojoules per mole (kJ mol-1)
• The change in Gibbs free energy (ΔG) for a chemical reaction at constant temperature (T) and pressure can be calculated:

ΔG = ΔH - TΔS

ΔG = change in Gibbs free energy for the reaction (kJ mol-1)

ΔH = enthalpy change for the reaction (kJ mol-1)

T = temperature of the reaction (K) (1)

ΔS = change in entropy for the reaction (kJ K-1 mol-1)

• For a reaction in which the reactants and products are present in their standard states (state at 298.15 K and atmospheric pressure):

ΔG° = ΔH° - TΔS°

ΔG° = change in standard Gibbs free energy for the reaction (kJ mol-1)

ΔH° = standard enthalpy change(2) for the reaction (kJ mol-1)

T = temperature of the reaction (K)

ΔS° = change in standard absolute entropy(3) for the reaction (kJ K-1 mol-1)

• A chemical reaction is spontaneous if

(a) entropy change of the entire universe is positive (ΔStotal = +):

ΔStotal > 0

(b) change in Gibbs free energy for the chemical system is negative (ΔGsystem = −):

ΔGsystem < 0

• A chemical reaction is nonspontaneous (not spontaneous) if

(a) entropy change of the entire universe is negative (ΔStotal = −):

ΔStotal < 0

(b) change in Gibbs free energy for the chemical system is positive (ΔGsystem = +):

ΔGsystem > 0

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Calculating ΔG° : ΔG° = ΔH° - TΔS°

For a reaction in which the reactants and products are present in their standard states (state at 298.15 K and atmospheric pressure):

ΔG° = ΔH° - TΔS°

ΔG° = change in standard Gibbs free energy for the reaction (kJ mol-1)

ΔH° = standard enthalpy change for the reaction (kJ mol-1)

T = temperature of the reaction (K)

ΔS° = change in standard absolute entropy for the reaction (kJ K-1 mol-1)

Note that the units used in the calculation of ΔG° must be consistent:

(a) ΔH° values are usually provided in units of kJ mol-1

(b) Temperature (T) may be given in units of Kelvin (K), or in degrees Celsius (°C)

Convert temperature in °C to K :

T(K) = 273.15 + T(°C)

Standard state temperature is 25°C = 298.15 K

(c) ΔS° values are usually provided in units of J K-1 mol-1

Convert J K-1 mol-1 to kJ K-1 mol-1 :

ΔS° (kJ K-1 mol-1) = ΔS° (J K-1 mol-1) ÷ 1000

Let's take, for example, a reaction that we know occurs spontaneously at 298.15 K and atmospheric pressure, the chemical reaction in which hydrogen in its standard state, hydrogen gas (H2(g)) and oxygen in its standard state, oxygen gas (O2(g)), react under standard conditions, 298.15 K and atmospheric pressure, to produce water in its standard state, liquid water (H2O(l)).
You may have experienced this spontaneous chemical reaction yourself by producing some hydrogen gas in the lab then using a heat source like a lit taper to increase the kinetic energy of the gas molecules giving them sufficient energy to successfully collide to produce products (water molecules) in a self-sustaining reaction, that is, a spontaneous reaction that also produces an audible "pop" (which is why it is usually called the "pop test" for hydrogen gas).
The balanced chemical equation for this reaction is given below:

H2(g) + ½O2(g) → H2O(l)

If we are given the following information:

ΔH°(reaction) = -285.8 kJ mol-1

ΔS°(reaction) = -163.5 J K-1 mol-1

We can calculate ΔG°(reaction) because we know the conditions under which the reaction occurs, that is:

T = 298.15 K

However, first we need to convert the entropy change in J K-1 mol-1 to kJ K-1 mol-1:

ΔS°(reaction) = -163.5 J K-1 mol-1 = -163.5 J K-1 mol-1 ÷ 1000 J kJ-1 = -0.1635 kJ K-1 mol-1

Now we can substitute the values for ΔH°(reaction), T (K), and ΔS°(reaction) into the equation for calculating standard Gibbs free energy change for the reaction:

 ΔG°(reaction) = ΔH°(reaction) − TΔS°(reaction) = −285.8 kJ mol-1 − (298.15 K × −0.1635 kJ K-1 mol-1) = −285.8 kJ mol-1 − (−48.75 kJ mol-1) = −285.8 kJ mol-1 + 48.75 kJ mol-1 = −237.1 kJ mol-1

ΔG° for the spontaneous reaction in which hydrogen gas and oxygen gas combine to produce liquid water under standard conditions is −237.1 kJ mol-1.

Let us now consider a reaction that we know is NOT spontaneous, such as the decomposition of liquid water to produce hydrogen gas and oxygen gas.
You may have set an experiment up in the laboratory in which you pass an electric current through water and separately collect and test for hydrogen gas (H2(g)) and oxygen gas (O2(g)).
This reaction is said to be nonspontaneous because if you turn off the electricity the chemical reaction stops occurring. If this reaction were spontaneous it would continue going after you turn off the electricity.

The balanced chemical equation for this nonspontaneous decomposition of water under standard conditions (298.15 K and atmospheric pressure) is geven below:

H2O(l) → H2(g) + ½O2(g)

If we are given the following information we can calculate ΔG° for this reaction:

Substance ΔHƒ° (kJ mol-1) ΔS° (J K-1 mol-1)
H2(g) 0 130.6
O2(g) 0 205.1
H2O(l) −285.8 69.9

First, calculate the standard enthalpy change for the reaction, ΔH°(reaction), in kJ mol-1 using the standard heat of formation (ΔHƒ°) for each reactant and product as shown below:

 ΔH°(reaction) = ΣΔHƒ°(products) − ΣΔHƒ°(reactants) = [ΔHƒ°(H2(g)) + ½ × ΔHƒ°(O2(g))] − [ΔHƒ°(H2O(l))] = [0 + (½ × 0) ] − [−285.8] = 0 + 285.8 ΔH°(reaction) = +285.8 kJ mol-1

Next, calculate the standard absolute entropy change for the reaction, ΔS°(reaction), in J K-1 mol-1 using the tabulated standard absolute entropies of each reactant and product as shown below:

 ΔS°(reaction) = ΣΔS°(products) − ΣΔS°(reactants) = [ΔS°(H2(g)) + ½ × ΔS°(O2(g))] − [ΔS°(H2O(l))] = [(1 × 130.6) + ( ½ × 205.1)] − [1 × 69.9] = [130.6 + 102.55] − [69.9] = 233.15 − 69.9 = + 163.25 J K-1 mol-1

After that, convert ΔS°(reaction) from J K-1 mol-1 to kJ K-1 mol-1 so that the units are consistent with those for the ΔH°(reaction) term as shown below:

 ΔS°(reaction) = + 163.25J K-1 mol-1 ÷ 1000 J kJ-1 = + 0.1633 kJ K-1 mol-1

Finally, calculate ΔG°(reaction) in kJ mol-1 at standard temperature (T = 298.15 K) and pressure by substituting these values for ΔH°(reaction), T (reaction in K) and ΔS°(reaction) into the equation for the change in Gibbs free energy of the chemical system as shown below:

 ΔG°(reaction) = ΔH°(reaction) − TΔS°(reaction) = +285.8 − (298.15 × 0.1633) = +285.8 − 48.69 = +237.1 kJ mol-1

ΔG° for the nonspontaneous reaction in which liquid water decomposes under standard conditions to produce hydrogen gas and oxygen gas is +237.1 kJ mol-1.

If a chemical reaction is spontaneous, then ΔGreaction is negative, ΔG < 0

If a chemical reaction is nonspontaneous, then ΔGreaction is positive, ΔG > 0

But why is ΔG negative for a spontaneous reaction and positive for a nonspontaneous reaction?

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ΔG° and Spontaneity of Chemical Reactions

Recall the definition of Gibbs free energy in which the change in Gibbs free energy for a chemical system at constant temperature and pressure is defined as:

ΔGsystem = −TΔStotal

where ΔStotal is the entropy of the chemical system plus the entropy of its entire surroundings, the entropy of the entire universe.

ΔStotal = ΔSsystem + ΔSsurroundings

In the section above we calculated the change in standard Gibbs free energy change (ΔG°) for the spontaneous reaction between hydrogen gas and oxygen gas to produce liquid water at 298.15 K and atmospheric pressure:

H2(g) + ½O2(g) → H2O(l)     ΔG° = −237.1 kJ mol-1

We note that this reaction is exothermic (ΔH°system = −285.8 kJ mol-1), and that the entropy of the chemical system decreases (ΔS°system = -163.5 J K-1 mol-1).

Now we can substitute the value for ΔG° and T into the equation above which defines ΔGsystem to determine the total entropy change:

 ΔGsystem = −TΔStotal −237.1 kJ mol-1 = −298.15 K × ΔStotal Divide both sides by −298.15 K −237.1 kJ mol-1−298.15 K = −298.15 K × ΔStotal−298.15 K +0.785 kJ K-1 mol-1 = ΔStotal

Note that while the entropy of the chemical system has decreased, the total entropy (entropy of system + entropy of surroundings) has increased (ΔStotal is positive).
Recall that the second law of thermodynamics states that the total entropy, or the entropy of the entire universe, is constantly increasing.
We infer that a chemical reaction occurs spontaneously if it induces an increase in the total entropy of the entire universe, that is ΔStotal > 0 (ΔStotal is positive, ΔStotal = +)

Now, the temperature of any chemical system will always be positive when measured in kelvin, T > 0 K (that is, T = +)(4)

So, for a spontaneous chemical reaction, the total entropy change is positive (ΔStotal = +) and the temperature in kelvin is also positive (T = +), then the Gibbs free energy change for the spontaneous chemical reaction is:

ΔGsystem = −TΔStotal = −(+K × +kJ K-1 mol-1) = −kJ mol-1

ΔGsystem < 0

The third law of thermodynamics states that the entropy of the entire universe is constantly increasing, ΔStotal > 0 (ΔStotal) is positive.
It is not possible for there to be a spontaneous decrease in total entropy, ΔStotal cannot be negative.
IF we consider the possibility that ΔStotal was negative (−) we can see what happens to the sign of ΔGsystem:

ΔGsystem = −TΔStotal = −(+K × −kJ K-1 mol-1) = +kJ mol-1

ΔGsystem > 0

If ΔGsystem > 0 then the reaction is NOT spontaneous, it is nonspontaneous.(5)
This means that, in order for the reaction to proceed at all, energy must be constantly supplied.
If this energy is removed then the reaction cannot continue to proceed, it is not self-sustaining and therefore not spontaneous.

A chemical reaction will proceed spontaneously if

• the total entropy change of the entire universe is positive (ΔStotal > 0)
(Note that the entropy change for the chemical reaction, ΔSreaction, may be either positive or negative)
• ΔGreaction is negative (ΔGreaction < 0)

and, a chemical reaction will NOT proceed spontaneously if it results in:

• the total entropy change of the entire universe being negative (ΔStotal < 0)
(Note that the entropy change for the chemical reaction, ΔSreaction, may be either positive or negative)
• ΔGreaction being positive (ΔGreaction > 0)

A chemical reaction that does NOT proceed spontaneously is referred to as a nonspontaneous chemical reaction.

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Worked Example of Gibbs Free Energy Calculation

Question:

The combustion of acetylene gas, C2H2(g), as shown in the balanced chemical equation below

C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(l)

has been used historically to provide artifical light for lamps used by miners, cyclists, motorists and in homes. The same chemical reaction is used in the oxy-acetylene torches employed today to cut through steel. (6)

Determine the value of ΔG° in kJ mol-1 for this reaction under standard conditions given the following data:

Substance ΔHƒ° (kJ mol-1) ΔS° (J K-1 mol-1)
C2H2(g) +226.8 200.8
O2(g) 0 205.1
CO2(g) −393.5 213.6
H2O(l) −285.8 69.9

Solution:

(Based on the StoPGoPS approach to problem solving.)

STOP STOP! State the Question.
What is the question asking you to do?
Determine the value of ΔG° in kJ mol-1

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:

C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(l)

(b) Values of ΔHƒ°, ΔS°

Substance ΔHƒ° (kJ mol-1) ΔS° (J K-1 mol-1)
C2H2(g) +226.8 200.8
O2(g) 0 205.1
CO2(g) −393.5 213.6
H2O(l) −285.8 69.9

(c) Temperature must be standard temperature: T = 298.15 K
(since we are given data for standard enthalpy of formation and standard absolute entropy, and need to calculate standard Gibbs free energy change)

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

ΔG°reaction = ΔH°reaction − TΔS°reaction

where:

ΔH°reaction = ΣΔHƒ°(products) - ΣΔHƒ°(reactants)

T = 298.15 K

ΔS°reaction = ΣΔS°(products) - ΣΔS°(reactants)

So

ΔG°reaction = ΔH°reaction − (298.15 × ΔS°reaction)

GO GO with the Game Plan

(i) Calculate ΔH°reaction in kJ mol-1:

 ΔH°reaction = ΣΔHƒ°(products) − ΣΔHƒ°(reactants) = [2 × ΔHƒ°(CO2(g)) + 1 × ΔHƒ°(H2O(l))] − [1 × ΔHƒ°(C2H2(g)) + 5/2 × ΔHƒ°(O2(g))] = [(2 × −393.5) + (1 × −285.8)] − [(1 × 226.8) + (5/2 × 0)] = [−787 + −285.8] − 226.8 = -1072.8 − 226.8 ΔH°reaction = −1299.6 kJ mol-1

(ii) Calculate ΔS°reaction in J mol-1:

 ΔS°reaction = ΣΔS°(products) − ΣΔS°(reactants) = [2 × ΔS°(CO2(g)) + 1 × ΔS°(H2O(l))] − [1 × ΔS°(C2H2(g)) + 5/2 × ΔS°(O2(g))] = [ (2 × 213.6) + (1 × 69.9)] − [(1 × 200.8) + (5/2 × 205.1)] = [427.2 + 69.9] − [200.8 + 512.75] = 497.1 − 713.55 = −216.45 J K-1 mol-1 Then convert J K-1 mol-1 to kJ K-1 mol-1 ΔS°reaction = −216.45 J K-1 mol-1 ÷ 1000 J kJ-1 ΔS°reaction = −0.21645 kJ K-1 mol-1

(iii) Calculate ΔG°(reaction):

ΔG°reaction = ΔH°reaction − (T × ΔS°reaction)

ΔH°reaction = −1299.6 kJ mol-1

Treaction = 298.15 K

ΔS°reaction = −0.21645 kJ K-1 mol-1

ΔG°reaction = −1299.6 − (298.15 × −0.21645)

ΔG°reaction = −1299.6 − (−64.53)

ΔG°reaction = −1299.6 + 64.53

ΔG°reaction = −1235.1 kJ mol-1

PAUSE PAUSE to Ponder Plausibility

Yes, we calculated of ΔG° for the reaction.

(a) Check that the answer has the right sign:
The reaction must be spontaneous because this reaction was used to produce artificial light, that is, once started the combustion of acetylene gas is self-sustaining.
For a spontaneous reaction ΔGreaction < 0
−1235.1 < 0
Therefore the predicted sign of ΔG° agrees with our negative answer.

(b) Work backwards to check your calculations:
Assume ΔG° = −1235.1 kJ mol-1, T = 298.15 K and ΔS° = −0.21645 kJ K-1 mol-1 then calculate ΔH°(reaction) to see if it agrees with the value we calculated in the question:
ΔG° = ΔH° -TΔS°
−1235.1 = ΔH° -(298.15 × -0.21645)
−1235.1 = ΔH° + 64.53
−1235.1 - 64.53 = ΔH° = -1299.6 kJ mol-1 which is the same as that calculated in the question.
Note that to be useful as a fuel, acetylene must release lots of energy spontaneously, and a release of 1299.6 kJ per mole of acetylene consumed sounds quite a sensible amount when compared to other hydrocarbon fuels.

We are therefore reasonably confident that our answer is plausible.

STOP STOP! State the Solution
ΔG°(reaction) = −1235.1 kJ mol-1

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Footnotes

(1) Kelvin (K) is the S.I. unit of temperature. If temperature is given in other units such as °C or °F you will need to convert this temperature to units of kelvin (K).

(2) The standard enthalpy change for a reaction is also referred to as the standard heat of reaction.

(3) Standard absolute entropy is also referred to as standard third law entropy.
Note that standard absolute entropies (third law entropies) are often tabulated in units of J K-1 mol-1. You will need to convert this to kJ K-1 mol-1 before you can calculate ΔG° for the reaction.

(4) The third law of thermodynamics can be generalised as saying that it is impossible to reach absolute zero, 0 K, in a finite number of steps. It is therefore inpossible for a temperature to below 0 K, that is, negative temperatures in units of kelvin are impossible.
Therefore the temperature of any chemical system must be greater than 0 K (temperature in kelvin is always positive)
Note that 0 K = −273.15°C
So it is very important that we express temperatures in kelvin (K)!

(5) If you are finding this concept of "spontaneous" and "nonspontaneous" a bit tricky, try thinking about a rechargable battery.
When you connect the battery to your phone, it spontaneously generates electricity to run the phone. No energy is required to keep the battery's chemical reactions going to produce electricity, indeed if you don't turn off the battery the chemical reaction generating the electricity will continue spontaneously until it has used up all the reactants, even if you never use the phone!
Clearly the products of the chemical reaction in your battery do not start re-making the reactants all by themselves, they require an input of energy which you supply by connecting your battery to the mains electricity during recharge.
The difference here is that if you remove the battery from the charger while it is only 50% charged, the reaction does not continue spontaneously to increase the concentration of reactants and hence increase the charge on your battery. In fact, quite the opposite is true, the reverse reaction that consumes reactants and generates electricity will begin, reducing the charge in your battery to below 50%.
The discharging of the battery to make products and electricity is spontaneous, a self-sustainging chemical reaction.
The recharging of the battery is a nonspontaneous process, you must continually supply the energy required in order for the chemical reaction to keep going.