Topic 9.5 Notes – Free Energy and Equilibrium

1. What Thermodynamically Favored Means

A process is thermodynamically favored when

$$\Delta G^\circ < 0$$

That statement is about equilibrium position under standard conditions, not about speed.

• Thermodynamics → where equilibrium lies (products or reactants favored)
• Kinetics → how fast equilibrium is reached

Students mix those up constantly. A reaction can be thermodynamically favored and still painfully slow.

Connecting ΔG° and K

Under standard conditions:

• ΔG° < 0 → products favored → K > 1
• ΔG° > 0 → reactants favored → K < 1
• ΔG° = 0 → neither favored → K = 1

At equilibrium (no matter the starting amounts):

• ΔG = 0
• Forward rate = reverse rate
• The system is at minimum free energy

Important distinction:

• ΔG° is under standard conditions.
• ΔG is under current conditions.
• Only ΔG° connects directly to K.

That connection comes from one equation.

2. The Mathematical Relationship Between ΔG° and K

The two equations you must know cold:

$$\Delta G^\circ = -RT \ln K$$

$$K = e^{-\Delta G^\circ/RT}$$

Where:

• $R = 8.314 \text{ J/mol}\\cdot\\text{K}$
• $T$ is in Kelvin
• ln is natural log

These equations are two versions of the same idea.

What the Signs Tell You

Look at the exponent in $K = e^{-\Delta G^\circ/RT}$.

If ΔG° is negative:

• $-\Delta G^\circ$ is positive
• Exponent is positive
• $K = e^{\text{positive}}$
• K > 1 → products favored

If ΔG° is positive:

• $-\Delta G^\circ$ is negative
• $K = e^{\text{negative}}$
• K < 1 → reactants favored

If ΔG° = 0:

• Exponent = 0
• $K = e^{0} = 1$

On multiple-choice questions, they often give you only the sign of ΔG° and ask about K. No calculator needed if you understand the logic above.

3. Estimating K from the Size of ΔG°

The key comparison is between ΔG° and RT.

At room temperature (298 K):

$$RT \approx 2.5 \text{ kJ/mol}$$

That number helps you judge scale.

When ΔG° Is Close to Zero

If ΔG° ≈ 0:

• $-\Delta G^\circ/RT ≈ 0$
• $\ln K ≈ 0$
• K ≈ 1

That means appreciable amounts of both reactants and products at equilibrium.

When |ΔG°| Is Much Larger Than RT

Suppose ΔG° = −40 kJ/mol at 298 K.

• Compare 40 kJ to RT ≈ 2.5 kJ
• 40 is much larger than 2.5
• $-\Delta G^\circ/RT$ is a large positive number
• K is huge

Strongly product-favored.

If ΔG° = +40 kJ/mol, same reasoning gives K ≪ 1, strongly reactant-favored.

On free-response questions, they often want qualitative reasoning like:
“Because ΔG° is large and negative relative to RT, K is much greater than 1.”
That comparison language earns points.

4. Free Energy and the Position of Equilibrium

Equilibrium is the point of minimum free energy.

Here’s the idea visually. The three panels show different signs of ΔG° and where the minimum in G occurs along the reaction progress:

In each graph:

• The system moves in the direction that lowers G.
• It stops at the lowest point of the curve.
• At that point, ΔG = 0 and the system is at equilibrium.

Two big cases:

ΔG° < 0

• Products have lower standard free energy.
• The minimum lies closer to the products side.
• K > 1

ΔG° > 0

• Reactants have lower standard free energy.
• The minimum lies closer to the reactants side.
• K < 1

If ΔG° = 0, the curve is symmetric and K = 1.

That’s the thermodynamic definition of equilibrium. The system sits where G is lowest.