Gibbs Free Energy and Spontaneity of Reaction

 

HSC Chemistry Syllabus

 

• Explain reaction spontaneity using terminology, including

– Gibbs Free Energy

– Entropy

Entropy and Gibbs Free Energy

This video will discuss the concept of Entropy and also explain how the combination of entropy and enthalpy values can be utilised along with the Gibbs Free Energy equation to predict the spontaneity of a reaction. 

 

Enthalpy (H) and Entropy (S)

The change in enthalpy (`\Delta H`) and change in entropy (`\Delta S`) of a chemical reaction determine its spontaneity. These two thermodynamic quantities enthalpy and entropy are explained separately.

• Enthalpy change: Negative enthalpy change (`\Delta H < 0`, exothermic reaction) is enthalpically favourable

• Entropy change: Positive entropy change (`\Delta S > 0`) is entropically favourable

 

Gibbs Free Energy: Combining Entropy and Enthalpy

Gibbs free energy (G) is a thermodynamic potential that can be used to predict whether a chemical reaction will occur spontaneously under constant temperature and pressure. It accounts for both the contributions of enthalpy change and entropy change towards the favourability or potential of a reaction to occur.

In chemistry, spontaneous means a process or reaction can proceed on its own without a continuous input of external energy. A non-spontaneous reaction is one that requires a continuous supply of external energy to happen.

Gibbs Free energy combines enthalpy (H) and entropy (S). The change in Gibbs free energy (ΔG) of a process is used to predict its spontaneity, which is represented by the equation:

 

$$\Delta G = \Delta H – T \Delta S$$

 

`\Delta G` represents Gibbs Free Energy change, `\Delta H` is enthalpy change, `T` is temperature in Kelvin, and `\Delta S` is entropy change. 

The sign of ΔG tells us whether a reaction is spontaneous or not:

• If ΔG is negative (< 0): The reaction is spontaneous (or exergonic). It can proceed without the continuous input of external energy.

• If ΔG is positive (> 0): The reaction is non-spontaneous (or endergonic). It requires a continuous input of energy to occur.

• If ΔG is zero (= 0): The reaction is at equilibrium. The rates of the forward and reverse reactions are equal. This will be discussed in greater detail in Year 12 HSC Chemistry.

 

Case 1: The Perfect Scenario

• ΔH is negative (-): The reaction is exothermic (releases heat), which is a favourable outcome.

• ΔS is positive (+): The reaction becomes more disordered, which is also favourable.

Since both factors driving the reaction are favourable, it doesn't matter what the temperature is. The equation works out to $\Delta G=(\text{negative})-(\text{positive})\times (\text{positive})$, which will always be negative, and thus the reaction is always spontaneous. 

Case 2: The Impossible Scenario

• ΔH is positive (+): The reaction is endothermic (absorbs heat), which is unfavourable.

• ΔS is negative (-): The reaction becomes more ordered, which is also unfavorable.

Since both factors work against the reaction, it can never happen on its own. The equation is $\Delta G=(\text{positive})-(\text{positive})\times (\text{negative})$, which simplifies to a positive plus a positive. The result will always be positive and thus the reaction is never spontaneous.

Case 3: Spontaneous Only When Cold

• ΔH is negative (-): The reaction is exothermic (favourable).

• ΔS is negative (-): The reaction becomes more ordered (unfavourable).

Here we have a conflict. The favourable heat release is competing against the unfavourable decrease in disorder. The temperature is the tie-breaker. The equation is $\Delta G=(\text{negative})-T(\text{negative})$, which is $\Delta G=(\text{negative})+T(\text{positive})$.

• At low temperatures, the "$T\Delta S$" term is small, so the negative $\Delta H$ wins and $\Delta G$ is negative.

• At high temperatures, the "$T\Delta S$" term becomes very large, overpowering the $\Delta H$ and making $\Delta G$ positive.

This combination of ΔH and ΔS means that the reaction is spontaneous only at low temperatures.

Case 4: Spontaneous Only When Hot

• ΔH is positive (+): The reaction is endothermic (unfavourable).

• ΔS is positive (+): The reaction becomes more disordered (favourable).

This is another conflict. The unfavourable heat absorption competes with the favourable increase in disorder. Again, temperature decides the outcome. The equation is $\Delta G=(\text{positive})-T(\text{positive})$.

• At low temperatures, the "$T\Delta S$" term is small and cannot overcome the positive $\Delta H$, so $\Delta G$ remains positive.

• At high temperatures, the "$T\Delta S$" term becomes a large negative value, overpowering the positive $\Delta H$ and making $\Delta G$ negative.

Therefore, the reaction is spontaneous only at high temperatures. 

Practice Questions

Example 1

The formation of water from hydrogen and oxygen gas is spontaneous under certain temperature conditions. Use the following information for this question.

∆H = –286 kJ mol^–1

∆S = –164 J K^–1 mol^–1

(a) Calculate the Gibbs free energy at 298 K.

(b) Is this reaction spontaneous at 298 K?

(c) Determine the temperature condition required for this reaction to be spontaneous. 

Example 2

The enthalpy and entropy values of the complete combustion of ethyne gas (C₂H₂) are shown.

∆H = –1300 kJ mol^–1

∆S = –216 J K^–1 mol^–1

(a) Determine whether the reaction is spontaneous at 50ºC.

(b) Determine temperature requirement for the reaction to be spontaneous.

 

 

RETURN TO MODULE 4: DRIVERS OF REACTION