Gibbs Free Energy Revision Notes for HSC SSCE Chemistry

Gibbs Free Energy

What is Gibbs free energy?

When studying chemical reactions, we need to consider two important factors: energy changes (enthalpy, $\Delta H°$ ) and disorder changes (entropy, $\Delta S°$ ). Gibbs free energy brings these two factors together into a single value that tells us whether a reaction will actually occur under specific conditions.

The standard Gibbs free energy change ( $\Delta G°$ ) represents the combined effect of the energy drive and entropy drive for a chemical reaction. It provides a definitive answer to the question: "Will this reaction happen?"

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Gibbs free energy is one of the most powerful tools in chemistry because it combines both enthalpy and entropy into one value that directly predicts reaction spontaneity. Understanding this concept is essential for predicting and controlling chemical processes.

The Gibbs free energy equation

The relationship between Gibbs free energy, enthalpy, and entropy is expressed by this fundamental equation:

$\Delta G° = \Delta H° - T\Delta S°$

where:

$\Delta G°$ is the standard Gibbs free energy change (in $\text{kJ mol}^{-1}$ )
$\Delta H°$ is the standard enthalpy change (in $\text{kJ mol}^{-1}$ )
$T$ is the absolute temperature (in $\text{K}$ , Kelvin)
$\Delta S°$ is the standard entropy change (in $\text{J K}^{-1}\text{mol}^{-1}$ )

chatImportant

Critical note about units: The temperature term ( $T$ ) is included to ensure both parts of the equation have the same energy units. You must convert entropy values from joules to kilojoules by multiplying by $10^{-3}$ when enthalpy is given in kilojoules. This is a common source of errors in calculations!

Understanding standard conditions

The degree symbol (°) in $\Delta G°$ indicates that all species must be in their standard states:

Gases: at a pressure of $100.0$ kPa
Solutions: at a concentration of $1.00$ mol L $^{-1}$
Pure substances: in their most stable form at the specified temperature

These standard conditions allow us to compare different reactions fairly.

Predicting reaction direction

The sign of $\Delta G°$ tells us which direction a reaction will proceed:

When ΔG° is negative

Both the energy drive ( $\Delta H°$ negative) and entropy drive ( $\Delta S°$ positive) favour the forward reaction, OR
One favourable drive outweighs an unfavourable one
The reaction proceeds in the forward direction as written

When ΔG° is positive

Both drives oppose the forward reaction ( $\Delta H°$ positive and $\Delta S°$ negative), OR
One unfavourable drive outweighs a favourable one
The reaction proceeds in the reverse direction

chatImportant

Key principle: At any given temperature, if $\Delta G°$ is negative under standard conditions, the reaction will proceed forward; if $\Delta G°$ is positive, the reaction will proceed in reverse.

This is one of the most important principles in thermodynamics - Gibbs free energy change determines whether a reaction will occur.

Real-world applications

Understanding Gibbs free energy and entropy is essential in many industrial processes. For example, iron and steel manufacturers use these concepts to explain why approximately 20% of the carbon monoxide used to reduce iron ore is not consumed. This "wastage" is not due to factory inefficiency but is a consequence of the fundamental chemistry and thermodynamics of the process.

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Thermodynamics - the study of heat, energy, and motion - is a crucial scientific discipline in both physics and chemistry. It helps us understand and optimise countless chemical processes in industries worldwide.

Calculating Gibbs free energy changes

Let's work through a complete example to see how these calculations are performed.

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Worked Example: Predicting Calcium Hydroxide Decomposition

Calculate $\Delta H°$ , $\Delta S°$ , and $\Delta G°$ for the decomposition of calcium hydroxide at 298 K, then predict whether the reaction will occur under standard conditions:

$\text{Ca(OH)}_2\text{(s)} \rightarrow \text{CaO(s)} + \text{H}_2\text{O(g)}$

Step-by-step solution

Calculation	Method and notes
Step 1: Calculate ΔH°
$\Delta H° = \Delta H_f°(\text{CaO}) + \Delta H_f°(\text{H}_2\text{O}) - \Delta H_f°(\text{Ca(OH)}_2)$	Use the enthalpy of formation equation
$= -636 + (-242) - (-987)$	Substitute values from data tables
$= +109 \text{ kJ mol}^{-1}$	Be careful with negative signs

Step 2: Calculate ΔS°
$\Delta S° = S°(\text{CaO}) + S°(\text{H}_2\text{O}) - S°(\text{Ca(OH)}_2)$	Use the entropy equation
$= 38 + 189 - 83$	Substitute values from entropy tables
$= 144 \text{ J K}^{-1}\text{mol}^{-1}$	Check units carefully

Step 3: Calculate ΔG°
$\Delta G° = \Delta H° - T\Delta S°$	Use the Gibbs free energy equation
$= 109 - 298 \times 144 \times 10^{-3}$	Convert entropy to kJ by multiplying by $10^{-3}$
$= 109 - 42.9$
$= +66 \text{ kJ mol}^{-1}$	Positive value obtained

Step 4: Interpret the result
Because $\Delta G°$ is positive, the reaction does not proceed as written under standard conditions at 298 K	Use the prediction principle

Exam tip: Always check that your units are consistent. When $\Delta H°$ is in kJ mol $^{-1}$ , you must convert $\Delta S°$ from J K $^{-1}$ mol $^{-1}$ to kJ K $^{-1}$ mol $^{-1}$ by multiplying by $10^{-3}$ .

Spontaneity revisited: four types of reactions

Previously, we defined a spontaneous reaction as one that occurs as written. However, the situation is more nuanced when we consider temperature effects. We can identify four distinct types of reactions:

Type 1: Spontaneous at room temperature

These reactions occur immediately when reactants are mixed, with no external energy input needed.

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Examples of Type 1 reactions:

Copper reacting with silver nitrate solution
Magnesium reacting with hydrochloric acid: $\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}$

Type 2: Require initial activation

These reactions need a "starting prod" such as a spark, match, or brief heating to begin. Once started, they continue without further external energy input because they are exothermic - the heat released maintains the elevated temperature.

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Examples of Type 2 reactions:

Burning magnesium
Combustion of petrol (octane) or natural gas (methane): $\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(g)}$

Type 3: Require continuous heating

These reactions only occur while we maintain an elevated temperature throughout the process.

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Example of Type 3 reaction:

Decomposition of calcium carbonate (limestone) at 1500 K: $\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}$

Type 4: Non-spontaneous even at high temperatures

These reactions do not occur even when heated to very high temperatures.

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Example of Type 4 reaction:

Decomposition of water at 2000 K: $2\text{H}_2\text{O(g)} \rightarrow 2\text{H}_2\text{(g)} + \text{O}_2\text{(g)}$

Classification summary:

Type 1 reactions are clearly spontaneous
Type 4 reactions are non-spontaneous
Type 2 reactions are considered spontaneous (just very slow at room temperature)
Type 3 reactions have intermediate behaviour

Remember!

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Key Points to Remember:

Gibbs free energy ( $\Delta G°$ ) combines enthalpy and entropy to predict whether a reaction will occur: $\Delta G° = \Delta H° - T\Delta S°$
A negative $\Delta G°$ means the reaction proceeds forward under standard conditions; a positive $\Delta G°$ means it proceeds in reverse
Always check your units when calculating $\Delta G°$ - convert entropy from J to kJ by multiplying by $10^{-3}$ when enthalpy is in kJ
Standard conditions mean gases at 100.0 kPa and solutions at 1.00 mol L $^{-1}$ - the ° symbol reminds you of this
Spontaneity has different levels - some reactions occur immediately at room temperature, others need activation or continuous heating, and some won't occur at all even when heated

Gibbs Free Energy (HSC SSCE Chemistry): Revision Notes