 # Gibbs Free Energy and Spontaneity of Reactions Chemistry Tutorial

## Key Concepts

• Gibbs free energy is also known as:

⚛ free energy

⚛ Gibbs free enthalpy

⚛ free enthalpy

• Gibbs free energy (free energy) is given the symbol G
• Gibbs free energy (free energy), G, of a chemical system is defined as:

G = H -TS

H = enthalpy
T = temperature
S = entropy

• ΔG represents the change in Gibbs free energy for a chemical system at constant temperature and pressure

ΔG = ΔH -TΔS

ΔH = enthalpy change of the system
T = temperature of the system
ΔS = entropy change of the system

• For a chemical system at constant temperature and pressure, the reaction is

⚛ spontaneous if ΔG < 0

(the sign of ΔG is negative)

⚛ nonspontaneous (not spontaneous) if ΔG > 0

(the sign of ΔG is positive)

⚛ at equilibrium if ΔG = 0

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## Defining Gibbs Free Energy, G

Whether or not a chemical reaction or a physical change is spontaneous depends on the total entropy change of the entire system.
The total entropy change of the entire system includes both the entropy of the chemical system under investigation AND also the entropy of the surroundings.
That is:

 total entropy change = entropy change change for the chemical system + entropy change of the surroundings ΔStotal = ΔSchemical system + ΔSsurroundings

While it is often possible to estimate the entropy change for a chemical system, ΔSchemical system, it is almost impossible to determine the entropy change for the surroundings, ΔSsurroundings, given that the surroundings are actually the entire universe!
The problem of estimating the entropy change for the entire universe can be overcome by restricting ourselves to a consideration of chemical processes that occur only at constant temperature and constant pressure.

At constant temperature, the entropy change of the surroundings (ΔSsurroundings) depends on the:

• amount of heat energy absorbed by the surroundings when there is a change in the chemical system being investigated
• temperature (T) at which heat is transferred to the surroundings from the chemical system being investigated

We can write an equation to represent this entropy change in the surroundings at constant temperature as shown below:

 ΔSsurroundings = heat absorbed by surroundings T

At constant pressure, the amount of heat absorbed by the surroundings equals -q where q equals the amount of heat absorbed by the chemical system.

qsystem = heat absorbed by chemical system

-qsystem = heat absorbed by surroundings (that is, heat is transferred from surroundings to be absorbed by the chemical system)

Therefore, we can write:

 ΔSsurroundings = heat absorbed by surroundings T ΔSsurroundings = -qsystem T

Note that if the chemical system:

• gains heat (absorbs heat), q is a positive number, so -q is a negative number
• loses heat (releases heat), q is a negative number, so -q is a positive number

At constant pressure, q equals the enthalpy change for the system:

qsystem = ΔHsystem

At constant pressure, the amount of heat absorbed by the surroundings = -qsystem
So, the amount of heat absorbed by the surroundings = -ΔHsystem

At constant temperature AND pressure, we can write the following equation to represent the change in entropy for the surroundings:

 ΔSsurroundings = -ΔHsystem T

Note that if the chemical process is:

• exothermic, ΔHsystem is negative, -ΔHsystem is positive, and ΔSsurroundings is positive, entropy of the surroundings increase
• endothermic, ΔHsystem is positive, -ΔHsystem is negative, and ΔSsurroundings is negative, entropy of the surroundings decrease

Now we can substitute our expression for ΔSsurroundings into our original equation for determining the change in total entropy:

 change in total entropy = change in entropy for the chemical system + change in entropy of the surroundings ΔStotal = ΔSchemical system + ΔSsurroundings ΔStotal = ΔSchemical system + -ΔHchemical system T ΔStotal = ΔSchemical system - ΔHchemical system T

Multiply throughout by -T

 -T × ΔStotal = -T × ΔSchemical system - -T × ΔHchemical system T -TΔStotal = -TΔSchemical system - -ΔHchemical system -TΔStotal = -TΔSchemical system + ΔHchemical system

Which can be rearranged in give:

-TΔStotal = ΔHchemical system - TΔSchemical system

We define the Gibbs free energy of a chemical system, or free energy of a chemical system, Gsystem, as:

Gsystem = Hsystem - TSsystem

where:

Hsystem = enthalpy of the chemical system

Tsystem = temperature of the chemical system

Ssystem = entropy of the chemical system

Gsystem, Hsystem, and Ssystem depend only on the state of a system.
If there is a change of state, say from state 1 to state 2, then:

Gstate 2 - Gstate 1 = Hstate 2 -Hstate 1 - (Tstate 2Sstate 2 - Tstate 1Sstate 1)

That is:

ΔGsystem = ΔHsystem - (Tstate 2Sstate 2 - Tstate 1Sstate 1)

At constant temperature, Tsystem = Tstate 2 = Tstate 1
So we can write the equation for the change in Gibbs free energy of the system (Gsystem) as:

ΔGsystem = ΔHsystem - (TsystemSstate 2 - TsystemSstate 1)

ΔGsystem = ΔHsystem - Tsystem(Sstate 2 - Sstate 1)

ΔGsystem = ΔHsystem - TsystemΔSsystem

Compare this with our equation for determining the total entropy change (ΔStotal) for a chemical process occurring at constant temperature and pressure:

-TΔStotal = ΔHchemical system - -TΔSchemical system

The change in Gibbs free energy for the system (ΔGsystem) can be equated with the total entropy change multiplied by -temperature:

ΔGsystem = -TΔStotal

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## Using Gibbs Free Energy (G) to Determine if a Chemical Reaction is Spontaneous or Nonspontaneous

According to the Second Law of Thermodynamics, the entropy of the universe increases, that is, the total entropy (ΔStotal) increases.

For a spontaneous change going from state 1 to state 2 the entropy of state 2 (Sstate 2) must be greater than the entropy of state 1 (Sstate 1)

Sstate 2 > Sstate 1

So the change in total entropy (ΔStotal) going from state 1 to state 2 spontaneously must be positive:

ΔStotal = Sstate 2 - Sstate 1

ΔStotal > 0

For a chemical process occurring at constant temperature and pressure, this means that the change in the Gibbs free energy of the system (ΔGsystem) must be negative for a spontaneous reaction:

ΔGsystem = -TΔStotal

ΔStotal is positive for a spontaneous reaction

ΔGsystem = negative

ΔGsystem < 0

A process in which the total entropy decreases (ΔStotal < 0) would be nonspontaneous.

At constant temperature and pressure, the change in Gibbs free energy for this nonspontaneous process would be positive:

ΔGsystem = -TΔStotal

ΔStotal is negative for a nonspontaneous reaction

ΔGsystem = positive

ΔGsystem > 0

For a system at equilibrium there is no change in total entropy (ΔStotal = 0)

At constant temperature and pressure, the change in Gibbs free energy for this process at equilibrium would be 0:

ΔGsystem = -TΔStotal

ΔStotal = 0 for a system at equilibrium

ΔGsystem = -T × 0

ΔGsystem = 0

The table below summarises the values for the change in Gibbs free energy for spontaneous reactions, nonspontaneous reactions and systems at equilibrium at constant temperature and pressure:

ΔG
(constant T, P)
Change
ΔG < 0
(ΔG negative)
spontaneous
ΔG = 0 equilibrium
(no net change)
ΔG > 0
(ΔG positive)
nonspontaneous

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## Worked Example of Gibbs Free Energy and Spontaneity of Reactions

Question:

Iso-octane (2,2,4-trimethylpentane), C8H18, is a component of petrol (gasoline), a fuel commonly used in internal combustion engines.

The complete combustion of iso-octane in an internal combustion engine is described by the balanced chemical equation given below:

C8H18(g) + 12½O2(g) → 8CO2(g) + 9H2O(g)

At 25°C (298 K) and 1 atmosphere pressure (101.3 kPa), the combustion of 1 mole of iso-octane releases 5109 kJ heat energy.

Predict the sign of the change in Gibbs free energy (ΔG) for this reaction.

Solution:

(Based on the StoPGoPS approach to problem solving.)

 STOP STOP! State the Question. What is the question asking you to do? Predict whether:     (i) ΔG is negative, ΔG < 0     or     (ii) ΔG is positive, ΔG > 0 PAUSE PAUSE to Prepare a Game Plan (1) What information (data) have you been given in the question? (i) iso-octane is a fuel (ii) C8H18(g) + 12½O2(g) → 8CO2(g) + 9H2O(g) (iii) reaction is exothermic: ΔH = -5109 kJ mol-1 (298 K and 101.3 kPa) (2) What is the relationship between what you know and what you need to find out? (i) ΔG is negative (ΔG < 0) if reaction is spontaneous (ii) ΔG is positive (ΔG > 0) if reaction is nonspontaneous (i) ΔG is 0 (ΔG = 0) if reaction is at equilibrium GO GO with the Game Plan Decide if reaction is spontaneous and hence ΔG is negative or nonspontaneous in which case ΔG is positive (i) iso-octane is a fuel. A substance can only be used as a fuel if the combustion reaction proceeds spontaneously. (ii) For a spontaneous reaction, ΔG for the reaction is negative (ΔG < 0). For the combustion of iso-octane at 25°C and 101.3 kPa, ΔG is negative. PAUSE PAUSE to Ponder Plausibility Is your answer plausible? (i) Have you answered the question that was asked? Yes, we predicted the sign of ΔG for the reaction. (ii) Check that your answer is plausible: Reactions proceed spontaneously to minimise enthalpy (by releasing heat) and maximise entropy (increase distribution of energy or randomness). The combustion of iso-octane: ⚛ is exothermic (ΔH is negative) so enthalpy is minimised ⚛ increases entropy of the system (13½ moles of gas of left hand side of equation, 17 moles of gas of right hand side) Combustion of iso-octane is predicted to proceed spontaneously therefore the sign of ΔG is negative. Since this is the same as the answer we got above, we are reasonably confident that our answer is plausible. STOP STOP! State the Solution ΔG is negative

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