R2.3.7 Gibbs free energy and equilibrium (Higher Level Only)

The Relationship Between $Δ G^{\circ}$ and $K$

Standard Gibbs Free Energy Change ( $Δ G^{\circ}$ ) and Its Significance

As discussed in Reactivity 1.4, the Gibbs free energy change ( $Δ G$ ) is a thermodynamic quantity that determines whether a reaction is spontaneous.
A spontaneous reaction proceeds without requiring external energy input.
- $Δ G < 0$ : The reaction is spontaneous in the forward direction.
- $Δ G > 0$ : The reaction is non-spontaneous in the forward direction (but spontaneous in reverse).
- $Δ G = 0$ :The system is at equilibrium.

Note

When we discuss $Δ G^{\circ}$ , we refer to the Gibbs free energy change under standard conditions: 1 atm pressure for gases, 1 $m o l d m^{- 3}$ concentration for solutions, and a specified temperature (usually 298 K unless otherwise stated).

But how does $Δ G^{\circ}$ relate to the equilibrium constant ( $K$ )?

The Equation Relating $Δ G^{\circ}$ and $K$

The connection between $Δ G^{\circ}$ and $K$ is expressed mathematically as:

$Δ G^{\circ} = - R T \ln K$

Where:

$Δ G^{\circ}$ : Standard Gibbs free energy change (in ${J mol}^{- 1}$ ).
$R$ : Gas constant ( $8.31 J {mol}^{- 1} K^{- 1}$ ).
$T$ : Absolute temperature in Kelvin.
$K$ : Equilibrium constant (unitless).

This equation links thermodynamics ( $Δ G^{\circ}$ ) to equilibrium ( $K$ ) and provides critical insights into reaction behavior:

When $Δ G^{\circ} < 0$ :
- The reaction is product-favored.
- $K > 1$ , indicating that the equilibrium lies toward the products.
When $Δ G^{\circ} > 0$ :
- The reaction is reactant-favored.
- $K < 1$ , indicating that the equilibrium lies toward the reactants.
When $Δ G^{\circ} = 0$ :
- The system is at equilibrium.
- $K = 1$ , meaning that reactants and products are present in comparable amounts.

Interpreting $Δ G^{\circ}$ and $K$

$Δ G^{\circ} < 0$ (Spontaneous Forward Reaction)

Suppose a reaction has $Δ G^{\circ} = - 25, kJ mol^{-1}$ at 298 K.
The negative value indicates that the forward reaction is spontanous under standard conditions. Using the equation:

$K = e^{- \frac{Δ G^{\circ}}{R T}}$

Substitute the values:

$Δ G^{\circ} = - 25, 000 J {mol}^{- 1}$
$R = 8.31 J {mol}^{- 1} K^{- 1}$
$T = 298 K$

$K = e^{- (- 25, 000) / (8.31 \times 298)}$

$K = e^{10.08} \approx 23000$

Since $K ≫ 1$ , this equilibrium strongly favors the products.

$Δ G^{\circ} > 0$ (Non-Spontaneous Forward Reaction)

Now consider a reaction with $Δ G^{\circ} = + 10 kJ {mol}^{- 1}$
Substituting into the equation:

$K = e^{- 10, 000 / (8.31 \times 298)}$

$K = e^{- 4.03} \approx 0.018$

Here, $K ≪ 1$ , meaning the equilibrium strongly favors the reactants.

$Δ G^{\circ} = 0$ (Equilibrium)

If $Δ G^{\circ} = 0$ , then:

$K = e^{0} = 1$

This indicates that the concentrations of reactants and products are balanced at equilibrium, depending on their stoichiometric coefficients.

Graphs showing relationship between Gibbs free energy and equilibrium constant for spontaneous and non-spontaneous reactions.

Example question

Calculate $Δ G^{\circ}$ for a reaction at 298 K if $K = 50$ .

Solution

$Δ G^{\circ} = - R T \ln K$ Substitute:

$R = 8.31 J {mol}^{- 1} K^{- 1}$
$T = 298 K$
$K = 50$

$Δ G^{\circ} = - (8.31) (298) \ln (50)$

$Δ G^{\circ} = - (8.31) (298) (3.91) \approx - 9700 J {mol}^{- 1} (or - 9.7 kJ {mol}^{- 1})$

Since $Δ G^{\circ} < 0$ , the reaction is spontaneous under standard conditions.

Note

Keep in mind: $Δ G^{\circ}$ applies only to standard conditions. For non-standard conditions, use $Δ G = Δ G^{\circ} + R T \ln Q$ , where $Q$ is the reaction quotient.

Common Mistakes to Avoid

Common Mistake

Confusing $Δ G^{\circ}$ with $Δ G$ : $Δ G^{\circ}$ applies only under standard conditions. For other conditions, use $Δ G = Δ G^{\circ} + R T \ln Q$ .

Common Mistake

Forgetting to convert $Δ G^{\circ}$ to Joules when using the equation $Δ G^{\circ} = - R T \ln K$ . Since $R$ is in Joules, ensure units are consistent.

Reflection and Broader Implications

Self review

Can you explain why a reaction with $K < 1$ is reactant-favored? How does this relate to $Δ G^{\circ} > 0$ ?

Theory of Knowledge

How does the interpretation of $Δ G^{\circ}$ and $K$ connect to real-world applications?
For example, in the Haber process for ammonia synthesis, how do thermodynamics and equilibrium constants guide industrial decisions about temperature and pressure?

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R2.3.7 Gibbs free energy and equilibrium (Higher Level Only)

The Relationship Between ΔG∘ and K

Standard Gibbs Free Energy Change (ΔG∘) and Its Significance

The Equation Relating ΔG∘ and K

Interpreting ΔG∘ and K

ΔG∘<0 (Spontaneous Forward Reaction)

ΔG∘>0 (Non-Spontaneous Forward Reaction)

ΔG∘=0 (Equilibrium)