9.5 Free Energy and Equilibrium

Free energy and equilibrium are two ways of describing whether a reaction favors products or reactants. When the standard free energy change, $\Delta G^\circ$, is negative, products are favored and $K$ is greater than 1; when $\Delta G^\circ$ is positive, reactants are favored and $K$ is less than 1. For AP Chemistry, use both the sign of $\Delta G^\circ$ and the size of $K$ to support the same conclusion.

Why This Matters for the AP Chemistry Exam

This topic ties Gibbs free energy from earlier in Unit 9 back to the equilibrium ideas from Unit 7. On the AP Chemistry exam, you may need to justify whether a process is thermodynamically favored using the relationships between K, ΔG°, and T, and explain your reasoning with chemical principles or math. Many questions ask you to reason about the size and sign of K from ΔG° (or the reverse) instead of just plugging into a formula, so being able to estimate K qualitatively is just as important as the calculation.

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Key Takeaways

• "Thermodynamically favored" (ΔG° < 0) means products are favored at equilibrium, so K > 1 under standard conditions.
• The core equations are ΔG° = -RT ln K and K = e^(-ΔG°/RT), with R = 8.314 J·mol⁻¹·K⁻¹ and T in Kelvin.
• A negative ΔG° gives K > 1 (products favored); a positive ΔG° gives K < 1 (reactants favored).
• When ΔG° is near zero, K is close to 1. When ΔG° is much larger or much smaller than RT, K deviates strongly from 1.
• ΔG (no naught) describes free energy at any conditions; ΔG° describes the standard-state case. They connect through Q: ΔG = ΔG° + RT ln Q.
• At equilibrium, ΔG = 0 and Q = K, which is exactly how the ΔG°-K equation is derived.

Two Ways to Define Equilibrium

Unit 7 used a kinetic definition of equilibrium: the point where the forward and reverse reactions happen at the same rate, so the concentrations of products and reactants stop changing. Equilibrium does not mean nothing is happening. Both reactions are still going, they just cancel out.

There is also a thermodynamic definition: equilibrium is the point of minimum free energy. While a reaction moves toward equilibrium on its own, ΔG (no naught symbol) is less than zero, so the reaction releases free energy. Once it reaches equilibrium concentrations, moving further in either direction would require ΔG to be positive, meaning external energy would be needed.

Picture a graph with free energy (G) on the y-axis and the extent of reaction on the x-axis, running from 100% reactants to 100% products.

A few things to keep straight when reading that kind of graph:

• ΔG is not the height of the curve. It is the slope (the rate of change). When ΔG < 0, the curve is going downhill; when ΔG > 0, it is going uphill.
• The far left point is 100% reactants and the far right is 100% products. The difference in height between those two ends reflects the sign of ΔG° for the reaction, calculated as ΔG° = ΣnΔG°f (products) - ΣnΔG°f (reactants).
• The lowest point on the curve is the equilibrium point, where ΔG = 0.

For a reaction where products sit lower than reactants, ΔG° is negative (thermodynamically favored). The system slides downhill toward the minimum, making products, until it hits the equilibrium point. Past that point ΔG > 0, so pushing the reaction further toward products would require adding energy.

For a reaction where products sit higher than reactants, ΔG° is positive (not favored). The equilibrium point lands closer to the reactant side, so there are many more reactants than products at equilibrium. A reaction that is not favored ends up reactant-heavy at equilibrium.

Relationship Between ΔG°, ΔG, and K

ΔG measures free energy change at nonstandard conditions, so you connect ΔG° and ΔG through Q, the reaction quotient:

ΔG = ΔG° + RT ln Q

To find Q, plug non-equilibrium concentrations or pressures into the equilibrium expression (the law of mass action). R is the gas constant (8.314 J·mol⁻¹·K⁻¹), and T is the temperature in Kelvin.

To link ΔG° directly to K, use the two conditions that are true at equilibrium: ΔG = 0 and Q = K. Substituting in:

0 = ΔG° + RT ln K

ΔG° = -RT ln K

Solving for K:

-ΔG° = RT ln K

-ΔG°/RT = ln K

K = e^(-ΔG°/RT)

These equations show a direct relationship between ΔG° and K:

• A negative ΔG° makes -ΔG°/RT positive, so K = e^(positive number), which is greater than 1 (products favored).
• A positive ΔG° makes -ΔG°/RT negative, so K = e^(negative number), which is less than 1 (reactants favored).
• The more negative ΔG° gets, the larger K becomes; the more positive ΔG° gets, the smaller K becomes.

Estimating K from ΔG° Without a Calculator

A lot of questions want reasoning, not just a number. The size of K compared to 1 depends on how ΔG° compares to RT:

• ΔG° near zero means K is close to 1, so products and reactants are present in similar amounts at equilibrium.
• ΔG° much smaller than RT (a large negative value) means K is much greater than 1, so products are strongly favored.
• ΔG° much larger than RT (a large positive value) means K is much less than 1, so reactants strongly outweigh products.

Being able to say "ΔG° is large and negative, so K is very large" without computing the exact value is the kind of qualitative reasoning the exam rewards.

How to Use This on the AP Chemistry Exam

Problem Solving

• Watch your units. ΔG° often comes in kJ·mol⁻¹ while R is in J·mol⁻¹·K⁻¹. Convert kJ to J before using ΔG° = -RT ln K.
• Always use temperature in Kelvin.
• You can find ΔG° first from standard free energies of formation (ΔG° = ΣnΔG°f products - ΣnΔG°f reactants), then plug it into the K equation.

Free Response

• When asked to justify whether a process is favored, connect the sign of ΔG° to the value of K and to which side is favored at equilibrium. Use chemical reasoning or the math relationship, not just a restated definition.
• If you only need to know whether K is greater or less than 1, you do not have to calculate it. The sign of ΔG° is enough.

Common Trap

• Q and K are not the same. Q can be any value at any moment; K is the specific value of Q at equilibrium. ΔG uses Q, while ΔG° uses K.

Common Misconceptions

• Equilibrium does not mean the reaction stopped. The forward and reverse reactions are still running at equal rates.
• ΔG and ΔG° are different. ΔG° applies only to standard conditions; ΔG applies to whatever conditions the system is actually in. Mixing them up breaks the math.
• A negative ΔG° does not mean the reaction goes to completion. It means K > 1, so products are favored, but there is still some reactant left at equilibrium.
• A positive ΔG° does not mean no reaction happens at all. It means K < 1, so the reaction still proceeds a little, just reactant-favored.
• ΔG° tells you the position of equilibrium, not the speed. A favored reaction can still be extremely slow if it is under kinetic control.
• On the free energy vs. extent of reaction graph, ΔG is the slope of the curve, not the height. The lowest point is equilibrium, where the slope (ΔG) is zero.

Related AP Chemistry Guides

• 9.1 Introduction to Entropy
• 9.3 Gibbs Free Energy and Thermodynamic Favorability
• 9.2 Absolute Entropy and Entropy Change
• 9.8 Galvanic (Voltaic) and Electrolytic Cells
• 9.4 Thermodynamic and Kinetic Control
• Unit 9 Review: Thermodynamics and Electrochemistry