R1.4.2 Gibbs free energy (ΔG) (Higher Level Only)

Gibbs Free Energy and Reaction Spontaneity

Imagine striking a match. Instantly, the chemicals in the match head react with oxygen, producing heat and light—a spontaneous reaction.
Now, think about splitting water into hydrogen and oxygen.
This requires a significant energy input.
Why do some reactions happen on their own, while others need help?

The answer lies in Gibbs free energy ( $Δ G^{\circ}$ ), a concept that combines energy (enthalpy), disorder (entropy), and temperature to predict whether a reaction will occur spontaneously.

Gibbs Free Energy: The Key to Spontaneity

Definition

Gibbs free energy

Gibbs free energy ( $Δ G$ ) is a thermodynamic property that determines whether a chemical reaction is spontaneous under constant pressure and temperature.

A reaction is spontaneous if it proceeds toward completion or equilibrium without requiring external energy.
The relationship between Gibbs free energy and other thermodynamic properties is expressed by the equation:

$Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$

Where:

$Δ G^{\circ}$ : Standard Gibbs free energy change ( $k J m o l^{- 1}$ )
$Δ H^{\circ}$ : Standard enthalpy change ( $k J m o l^{- 1}$ )
$T$ : Temperature in Kelvin (K)
$Δ S^{\circ}$ : Standard entropy change ( $J K^{- 1} m o l^{- 1}$ )

Let’s break this down:

Enthalpy ( $Δ H^{\circ}$ ):
- Represents the heat energy absorbed or released during a reaction.
- Exothermic reactions ( $Δ H^{\circ} < 0$ ) release heat, while endothermic reactions ( $Δ H^{\circ} > 0$ ) absorb heat.
Entropy ( $Δ S^{\circ}$ ):
- Measures the disorder or randomness in a system.
- Reactions that increase disorder ( $Δ S^{\circ} > 0$ ) are more likely to be spontaneous.
Temperature ( $T$ ):
- Higher temperatures amplify the influence of entropy on Gibbs free energy.

Hint

Remember: Temperature must always be in Kelvin. To convert from °C to K, add 273.15.

When Is a Reaction Spontaneous?

The sign of $Δ G$ determines whether a reaction is spontaneous:

$Δ G < 0$ : The reaction is spontaneous.
$Δ G = 0$ : The reaction is at equilibrium.
$Δ G > 0$ : The reaction is non-spontaneous.

The interplay between $Δ H^{\circ}$ , $Δ S^{\circ}$ , and $T$ influences $Δ G^{\circ}$ . Table 1 summarizes these relationships:

Tip

Use this table as a quick guide to predict reaction spontaneity before performing detailed calculations!

Calculating Gibbs Free Energy

To calculate $Δ G^{\circ}$ , you need thermodynamic data for $Δ H^{\circ}$ and $Δ S^{\circ}$ , which are available in the IB Chemistry data booklet.

Example

Combustion of Propane ( $C_{3} H_{8}$ )

Consider the combustion of propane:
$C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 {CO}_{2} (g) + 4 H_{2} O (g)$

Given data:

$Δ H^{\circ} = - 2045 {kJ mol}^{- 1}$
$Δ S^{\circ} = + 103 {J K}^{- 1} {mol}^{- 1}$
Temperature: $T = 298 K$
Step 1: Convert $Δ S^{\circ}$ to $k J K^{- 1} m o l^{- 1}$ :
$Δ S^{\circ} = \frac{103}{1000} = 0.103 /, {kJ K}^{- 1} {mol}^{- 1}$
Step 2: Substitute values into the Gibbs free energy equation:
$Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$
$Δ G^{\circ} = - 2045 - (298 \times 0.103)$
$Δ G^{\circ} = - 2045 - 30.7 = - 2075.7 {kJ mol}^{- 1}$
Step 3: Interpret the result:
Since $Δ G^{\circ} < 0$ , the reaction is spontaneous under standard conditions.

Hint

Always check your units! Entropy values in the IB data booklet are often in ${J K}^{- 1} {mol}^{- 1}$ , so convert them to ${kJ K}^{- 1} {mol}^{- 1}$ by dividing by 1000.

Temperature and Spontaneity: When Does $Δ G = 0$ ?

For some reactions, spontaneity depends on temperature. At the temperature where $Δ G = 0$ , the reaction is at the boundary of becoming spontaneous. Rearranging the Gibbs free energy equation gives:

$T = \frac{Δ H^{\circ}}{Δ S^{\circ}}$

Example

Dissociation of Ammonium Chloride

The decomposition of ammonium chloride ( ${NH}_{4} Cl$ ) is:
${NH}_{4} Cl (s) \to {NH}_{3} (g) + HCl (g)$

Given data:

$Δ H^{\circ} = + 176 {kJ mol}^{- 1}$
$Δ S^{\circ} = + 285 {J K}^{- 1} {mol}^{- 1}$
Step 1: Convert $Δ S^{\circ}$ to ${kJ K}^{- 1} {mol}^{- 1}$ :
$Δ S^{\circ} = \frac{285}{1000} = 0.285 {kJ K}^{- 1} {mol}^{- 1}$
Step 2: Calculate the temperature:
$T = \frac{Δ H^{\circ}}{Δ S^{\circ}}$
$T = \frac{176}{0.285} = 617.5 K$
Step 3: Interpret the result:
The reaction becomes spontaneous above 617.5 K.

Using Thermodynamic Data to Calculate $Δ G^{\circ}$

To calculate $Δ G^{\circ}$ for a reaction using data booklet values:

Find $Δ H^{\circ}$ and $Δ S^{\circ}$ for each species in the reaction.
Use the equations:
$Δ H^{\circ} = \sum Δ H_{products}^{\circ} - \sum Δ H_{reactants}^{\circ}$
$Δ S^{\circ} = \sum S_{products}^{\circ} - \sum S_{reactants}^{\circ}$
Substitute into $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$ .

Common Mistake

Many students forget to convert entropy from ${J K}^{- 1} {mol}^{- 1}$ to ${kJ K}^{- 1} {mol}^{- 1}$ or temperature from °C to K. Always double-check your units!

Reflection and Broader Implications

Self review

What happens to $Δ G^{\circ}$ if both $Δ H^{\circ}$ and $Δ S^{\circ}$ are negative?
Under what conditions would the reaction be spontaneous?

Theory of Knowledge

How does the concept of spontaneity in chemistry connect to broader ideas about energy efficiency and sustainability?
Can we always prioritize spontaneity over energy input?

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