The Gibbs Free Energy Equation: Predicting Spontaneity ($\Delta G$, $\Delta H$, $\Delta S$)

One of the most common stumbling blocks in physical chemistry is the difference between energy and spontaneity. Just because a reaction releases heat (exothermic) doesn't guarantee it will happen. To truly predict whether a reaction will occur without outside intervention, we need the ultimate arbiter of the chemical universe: Gibbs Free Energy ($\Delta G$).

This guide breaks down the thermodynamics of spontaneity, how to handle the critical unit conversions that trip up most students, and how to link free energy to the equilibrium constant ($K$).

What is Gibbs Free Energy and Why It Matters ($\Delta G$ vs $\Delta G^\circ$)

Gibbs Free Energy ($G$) represents the maximum amount of reversible work that may be performed by a thermodynamic system at a constant temperature and pressure. In simpler terms, it tells us the "net driving force" of a reaction.

When analyzing a reaction, we look at the change in free energy ($\Delta G$). The sign of $\Delta G$ is the definitive test for spontaneity:

• $\Delta G < 0$ (Negative): The process is spontaneous (exergonic). It releases free energy and moves forward toward products.
• $\Delta G > 0$ (Positive): The process is non-spontaneous (endergonic). It requires an input of energy to proceed. Ideally, the reverse reaction is spontaneous.
• $\Delta G = 0$ (Zero): The system is at equilibrium. There is no net change.

The Difference Between $\Delta G$ and $\Delta G^\circ$

Students often conflate these two terms. It is crucial to distinguish them:

• $\Delta G^\circ$ (Standard Free Energy Change): This is a fixed value for a specific reaction under standard conditions (1 atm, 1 M concentrations, usually 298 K). It tells you about the position of equilibrium (whether products or reactants are favored ultimately).
• $\Delta G$ (Non-Standard Free Energy Change): This is the instantaneous free energy change based on the current conditions and concentrations. It tells you which direction the reaction will shift right now to reach equilibrium.

Illustrative Example:
Consider a reaction where $\Delta G^\circ = -15 \text{ kJ/mol}$ but under current conditions $\Delta G = +25 \text{ kJ/mol}$.

• Since $\Delta G^\circ$ is negative, the reaction favors products at equilibrium ($K > 1$).
• However, since $\Delta G$ is positive, the reaction is currently non-spontaneous in the forward direction. This implies the system has "overshot" equilibrium (Reaction Quotient $Q > K$) and will shift in reverse to stabilize.

The $\Delta G = \Delta H - T\Delta S$ Equation: Calculating Spontaneity

To calculate Gibbs Free Energy, we combine the two driving forces of the universe: Enthalpy ($\Delta H$), the heat energy, and Entropy ($\Delta S$), the disorder.

Analyzing Driving Forces

A reaction is "driven" by negative enthalpy (exothermic) and positive entropy (more disorder). We can predict spontaneity based on signs:

1. $\Delta H(-)$ and $\Delta S(+)$: Always spontaneous at any temperature (both terms contribute to a negative $\Delta G$).
2. $\Delta H(+)$ and $\Delta S(-)$: Never spontaneous (both terms contribute to a positive $\Delta G$).
3. $\Delta H(-)$ and $\Delta S(-)$: Spontaneous only at low temperatures (enthalpy driven).
4. $\Delta H(+)$ and $\Delta S(+)$: Spontaneous only at high temperatures (entropy driven).

Step-by-Step Case Study

Problem: Determine $\Delta G^\circ_{\text{rxn}}$ for the following reaction at 298 K.

Given: $\Delta H^\circ = +1.9 \text{ kJ}$ and $\Delta S^\circ = -109.6 \text{ J/K}$.

Solution:

1. Convert Units: Convert $\Delta S$ to kJ to match $\Delta H$. $$ -109.6 \text{ J/K} = -0.1096 \text{ kJ/K} $$
2. Apply the Equation: $$ \Delta G^\circ = 1.9 \text{ kJ} - (298 \text{ K})(-0.1096 \text{ kJ/K}) $$
3. Calculate: $$ \Delta G^\circ = 1.9 - (-32.66) $$ $$ \Delta G^\circ = +34.56 \text{ kJ} $$
4. Conclusion: Because $\Delta G^\circ$ is positive, the reaction is non-spontaneous under standard conditions.

The Relationship Between $\Delta G^\circ$ and the Equilibrium Constant ($K$)

Thermodynamics and Equilibrium are directly linked. If $\Delta G^\circ$ is negative, the reaction runs forward, creating more products than reactants, resulting in a large $K$.

Where:

• $R$ = 8.314 J/mol·K (Gas constant)
• $T$ = Temperature in Kelvin
• $K$ = Equilibrium constant ($K_{eq}, K_{sp}, K_a$, etc.)

Predicting K from $\Delta G^\circ$

• 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 $K = 1$.

Example: Calculating $\Delta G$ from $K_{sp}$

Problem: The $K_{sp}$ for the dissolution of NaCl is 37.7 at $24.85^\circ\text{C}$. Calculate $\Delta G^\circ$.

1. Prepare Temperature: $T = 24.85 + 273.15 = 298 \text{ K}$.
2. Identify Variables: $R = 8.314 \text{ J/mol K}$, $K = 37.7$.
3. Substitute: $$ \Delta G^\circ = -(8.314)(298) \ln(37.7) $$ $$ \Delta G^\circ = -(2477.57)(3.63) $$ $$ \Delta G^\circ \approx -8993 \text{ J/mol} $$
4. Final Unit Conversion: $$ \Delta G^\circ = -8.99 \text{ kJ/mol} $$

Since $\Delta G^\circ$ is negative, the dissolution of salt is spontaneous.

Calculating the Equilibrium Temperature ($T_{eq}$)

Sometimes a reaction is non-spontaneous at room temperature but becomes spontaneous if we heat it up (or cool it down). This occurs when $\Delta H$ and $\Delta S$ have the same sign.

The "crossover temperature" is the point where the system is exactly at equilibrium ($\Delta G = 0$).

Practice Problem: Finding the Temperature Threshold

Reaction: $C(s) + H_2O(g) \rightarrow CO(g) + H_2(g)$

Given: $\Delta H^\circ = 131.3 \text{ kJ/mol}$, $\Delta S^\circ = 133.6 \text{ J/K}\cdot\text{mol}$.

Analysis: Both values are positive. This is an endothermic reaction that increases disorder. It will become spontaneous at high temperatures where the entropy term ($T\Delta S$) is large enough to overcome the enthalpy cost.

Calculation:

1. Convert Units: $\Delta H = 131,300 \text{ J/mol}$.
2. Solve for T: $$ T = \frac{131,300 \text{ J/mol}}{133.6 \text{ J/K}\cdot\text{mol}} $$ $$ T \approx 982.8 \text{ K} $$

Answer: The reaction is spontaneous at temperatures greater than 983 K.

Key Takeaways: Thermodynamics Summary

• Spontaneity: Determined solely by the sign of $\Delta G$ (negative = spontaneous).
• Watch Your Units: Always convert $\Delta S$ (J) to kJ or $\Delta H$ (kJ) to J before calculating.
• Equilibrium Connection: A large $K$ corresponds to a very negative $\Delta G^\circ$.
• Temperature Dependence: If $\Delta H$ and $\Delta S$ have the same sign, temperature determines spontaneity. Use $T = \Delta H / \Delta S$ to find the tipping point.