Gibbs Free Energy ($\Delta G$), Spontaneity, and the Equilibrium Constant (K)

Why do some chemical reactions happen instantly while others require a massive input of energy? Why does ice melt at room temperature even though melting absorbs heat? The answer lies in the "ultimate arbiter" of chemical fate: Gibbs Free Energy ($\Delta G$).

For students mastering thermodynamics, understanding the interplay between Enthalpy ($\Delta H$), Entropy ($\Delta S$), and the Equilibrium Constant ($K$) is non-negotiable. Whether you are calculating cell potentials or predicting reaction spontaneity, these concepts are the foundation.

In this guide, we will break down the thermodynamics of equilibrium, moving from standard definitions to complex calculations involving $K$ and $E^{\circ}_{cell}$.

Calculating Standard State Thermodynamics ($\Delta G^{\circ}, \Delta H^{\circ}, \Delta S^{\circ}$)

Before predicting if a reaction will happen, you must quantify the energy changes involved. Chemists use "standard state" values (denoted by the degree symbol $\circ$, usually at $25^{\circ}\text{C}$ or $298\text{K}$, $1\text{ atm}$, and $1\text{ M}$ concentrations) to establish a baseline.

The most reliable method to find these values is using Hess’s Law principles and tabulated standard formation data ($f$). The universal formula for any state function ($X$) is:

Practice Problem: The Formation of HBr

Let's determine the standard Gibbs Free Energy change ($\Delta G^{\circ}$) for the reaction of hydrogen and bromine:

$$ 2H_2(g) + 2Br_2(l) \rightarrow 4HBr(g) $$

Given Data at 298 K:

• $\Delta G^{\circ}_f [H_2(g)] = 0 \text{ kJ/mol}$ (Element in standard state)
• $\Delta G^{\circ}_f [Br_2(l)] = 0 \text{ kJ/mol}$ (Element in standard state)
• $\Delta G^{\circ}_f [HBr(g)] = -36.2 \text{ kJ/mol}$

Step-by-Step Calculation:

1. Sum of Products: $4 \text{ mol} \times (-36.2 \text{ kJ/mol}) = -144.8 \text{ kJ}$
2. Sum of Reactants: $2(0) + 2(0) = 0 \text{ kJ}$
3. Apply Formula: $$ \Delta G^{\circ} = (-144.8) - (0) = -144.8 \text{ kJ} $$

Since $\Delta G^{\circ}$ is negative, this reaction is spontaneous under standard conditions.

The $\Delta G = \Delta H - T\Delta S$ Equation (Predicting Spontaneity)

Calculating $\Delta G$ from formation data is useful, but it doesn't explain why a reaction is spontaneous. To understand the driving forces, we use the Gibbs-Helmholtz equation:

• $\Delta H$ (Enthalpy): The heat energy. Exothermic ($\Delta H < 0$) favors spontaneity.
• $\Delta S$ (Entropy): The disorder. An increase in disorder ($\Delta S > 0$) favors spontaneity.
• $T$ (Temperature): The amplifier for entropy. As $T$ rises, the $T\Delta S$ term dominates.

Case Study 1: When Forces Oppose Each Other

Consider the decomposition of dinitrogen pentoxide at $25^{\circ}\text{C}$:

$$ 2N_2O_5(g) \rightleftharpoons 4NO_2(g) + O_2(g) $$

Thermodynamic Analysis:

• Enthalpy ($\Delta H^{\circ} = +106.0 \text{ kJ}$): The reaction is endothermic (requires heat). This opposes spontaneity.
• Entropy ($\Delta S^{\circ} = +453.0 \text{ J/K}$): We are going from 2 moles of gas to 5 moles of gas. Disorder increases massively. This favors spontaneity.

Who wins?

We plug the values into the Gibbs equation (ensuring units match—convert Entropy to kJ!):

Result: Despite being endothermic, the reaction is spontaneous ($\Delta G < 0$) because the entropy increase is significant enough to overcome the energy cost at this temperature. This perfectly illustrates how $\Delta S$ drives reactions that $\Delta H$ would otherwise forbid.

Case Study 2: The Entropy of the Surroundings

Sometimes, exam questions test your understanding of the "Universe." A reaction is spontaneous only if the total entropy of the universe increases.

$$ \Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr} $$

Consider the reaction $2CO_2 + N_2 \rightarrow 2CO + 2NO$ at $25^{\circ}\text{C}$.

• $\Delta H^{\circ} = +748.6 \text{ kJ}$ (Strongly Endothermic)
• $\Delta S^{\circ}_{sys} = +197.8 \text{ J/K}$ (Positive Entropy Change)

While the system gains entropy, the surroundings lose a massive amount of entropy because the system absorbs heat from them.
Calculation: $\Delta S_{surr} = -\Delta H_{sys} / T = -748600 \text{ J} / 298 \text{ K} \approx -2512 \text{ J/K}$.

Because the loss in the surroundings ($-2512$) outweighs the gain in the system ($+198$), the $\Delta S_{univ}$ is negative. Consequently, $\Delta G > 0$ and the reaction is non-spontaneous.

The Link: $\Delta G^{\circ}$ and the Equilibrium Constant (K)

Standard Gibbs Free Energy tells us the potential of a reaction to proceed from standard states to equilibrium. The position of that equilibrium is represented by $K$. The bridge between these two worlds is:

Where:

• $R$ = $8.314 \text{ J}/(\text{mol} \cdot \text{ K})$ (Gas Constant)
• $T$ = Temperature in Kelvin
• $K$ = Equilibrium constant ($K_c$ or $K_p$)

Quick Interpretation Rules:

• If $\Delta G^{\circ} < 0$, then $\ln K > 0$, so $K > 1$ (Products Favored).
• If $\Delta G^{\circ} > 0$, then $\ln K < 0$, so $K < 1$ (Reactants Favored).
• If $\Delta G^{\circ} = 0$, then $\ln K = 0$, so $K = 1$ (At Equilibrium).

Example: Calculating K from Thermodynamic Data

Let's find the equilibrium constant ($K$) for the reduction of tin oxide at $298 \text{ K}$:

$$ SnO_2(s) + 2CO(g) \rightarrow Sn(s) + 2CO_2(g) $$

Step 1: Calculate $\Delta G^{\circ}$

Using tabulated $\Delta G^{\circ}_f$ values:

• Products: $1(0) + 2(-394.4) = -788.8 \text{ kJ}$
• Reactants: $1(-520.5) + 2(-137.2) = -794.9 \text{ kJ}$
• $\Delta G^{\circ} = -788.8 - (-794.9) = +6.1 \text{ kJ} = +6100 \text{ J}$

Step 2: Rearrange to solve for K

To find $K$, take the exponent ($e$) of both sides:

Since $\Delta G^{\circ}$ was slightly positive, $K$ is small (less than 1), meaning reactants are slightly favored at equilibrium.

Equilibrium in Electrochemical Cells

In electrochemistry, the "driving force" of a reaction is measured in Volts rather than Joules. However, they describe the same reality. A positive cell potential ($E^{\circ}_{cell}$) indicates a spontaneous reaction.

The relationship connects thermodynamics to electricity:

Combining this with our equilibrium equation, we get the Nernst-related equilibrium expression:

Where $n$ is the moles of electrons transferred and $F$ is Faraday's constant ($96,485 \text{ C/mol}$).

Key Takeaway:

• Spontaneous: $\Delta G^{\circ} (-)$, $K (>1)$, $E^{\circ}_{cell} (+)$
• Non-Spontaneous: $\Delta G^{\circ} (+)$, $K (<1)$, $E^{\circ}_{cell} (-)$

Summary: Mastering Spontaneity

To ace your exams, organize your thoughts using this logic flow:

1. Check the signs: Look at $\Delta H$ and $\Delta S$. Do they agree or fight?
2. Check the units: $\Delta G$ and $\Delta H$ are usually in kJ, while $\Delta S$ and $R$ are in J. Always convert before subtracting!
3. Check the magnitude: Spontaneity ($\Delta G^{\circ}$) determines the position of equilibrium ($K$). The more negative the $\Delta G^{\circ}$, the larger the $K$.