Thermodynamic favorability versus rate — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Distinguishing thermodynamic favorability (spontaneity) from reaction rate, explaining kinetic vs thermodynamic control of competing reactions, relating ΔG and activation energy, and addressing common misconceptions for AP Chemistry exam questions.

You should already know: Gibbs free energy change (ΔG) and the definition of spontaneity. Reaction rate theory and activation energy from collision theory. The Arrhenius relationship between rate and activation energy.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Thermodynamic favorability versus rate?

Thermodynamic favorability (also called spontaneity) describes whether a reaction can proceed without continuous external input of energy, determined by the overall change in Gibbs free energy $\Delta G$ between reactants and products. Reaction rate describes how fast a thermodynamically allowed reaction proceeds to products, determined by the height of the activation energy barrier $E_a$. This topic accounts for 5-10% of Unit 9 (Applications of Thermodynamics) weight per the official AP Chemistry CED, and it appears in both MCQ (often as concept-based questions testing the core distinction) and FRQ (as part of reasoning about industrial, biological, or environmental reaction outcomes). The core misconception the AP exam consistently tests is confusion between these two independent properties: a reaction can be thermodynamically favorable but kinetically immeasurably slow, or thermodynamically unfavorable but kinetically fast when driven by external energy. Favorability tells you only if a reaction can occur, not how fast it will occur.

2. Core Distinction: Favorability vs Rate

Thermodynamic favorability depends only on the difference in free energy between the initial (reactant) and final (product) states of the reaction. The rule for favorability is simple: $$\Delta G=G_{\text{products}}-G_{\text{reactants}}$$ If $\Delta G<0$, the reaction is thermodynamically favorable (spontaneous): products are lower in free energy than reactants, so the reaction does not require continuous external energy to proceed. If $\Delta G>0$, the reaction is thermodynamically unfavorable (non-spontaneous) and requires continuous energy input to proceed. $\Delta G$ gives no information about how fast the reaction will go, only about the net direction the reaction will take when it reaches equilibrium.

Reaction rate, by contrast, depends on the reaction pathway, not just the start and end states. Rate is governed by activation energy $E_a$, the minimum energy colliding reactants need to reach the transition state and break bonds to form products. From the Arrhenius equation, $k=A{e}^{-E_a/RT}$, so rate decreases exponentially as $E_a$ increases. Even if $\Delta G$ is very negative, a large $E_a$ means almost no collisions have enough energy to react, so the reaction can be immeasurably slow at room temperature. The classic example is the conversion of diamond to graphite: $\Delta G=-2.9$ kJ/mol (favorable) but $E_a$ is so large the reaction takes billions of years.

Worked Example

For the reaction $2{\text{H}}_2(g)+{\text{O}}_2(g)\to 2{\text{H}}_2\text{O}(l)$ at 25°C, $\Delta G^{\circ }=-474$ kJ/mol and $E_a\approx 400$ kJ/mol. Which statement correctly describes the reaction?

1. It is thermodynamically unfavorable and kinetically fast
2. It is thermodynamically favorable and kinetically slow at 25°C without a catalyst
3. It is thermodynamically favorable and kinetically fast at all conditions
4. It is thermodynamically unfavorable and kinetically slow

Steps:

1. First, assess thermodynamic favorability: $\Delta G^{\circ }$ is negative, so the reaction is favorable. Eliminate options 1 and 4.
2. Next, relate $E_a$ to rate: typical $E_a$ for a fast reaction at 25°C is < 50 kJ/mol. An activation energy of 400 kJ/mol is extremely high.
3. Without a catalyst to lower $E_a$, very few collisions between H₂ and O₂ have enough energy to overcome the barrier at 25°C.
4. The correct description is option 2: thermodynamically favorable, kinetically slow without a catalyst.

Exam tip: On any AP question asking about rate vs favorability, always check ΔG first for favorability, then Ea for rate. Never assume a negative ΔG means a fast reaction—this is the most common exam trap.

3. Kinetic vs Thermodynamic Control of Competing Reactions

When the same starting materials can form two different products via two competing reaction pathways, we almost always see a difference between the faster-forming product and the more stable product. This leads to two types of reaction control that determine the final product mixture:

• Kinetic control: Reaction conditions do not allow the reaction to reverse, and only the lower activation energy pathway is accessible. This usually occurs at low temperatures, where there is not enough thermal energy to cross higher activation barriers or reverse the initial product formation. The major product is the one that forms faster (lower $E_a$), even if it is less thermodynamically stable.
• Thermodynamic control: Reaction conditions are sufficiently high temperature that all activation barriers are accessible, and all reactions are reversible. This allows the system to reach equilibrium, so the major product is the more thermodynamically stable (lower $\Delta G$) product, regardless of how fast it forms.

AP exam FRQs often ask to predict the major product for different conditions, testing your ability to connect reaction conditions to control type.

Worked Example

Starting from 1 mol of reactant A, two competing reactions occur: $A\to B$ ($\Delta G=-12$ kJ/mol, $E_a=30$ kJ/mol) and $A\to C$ ($\Delta G=-45$ kJ/mol, $E_a=75$ kJ/mol). Predict the major product at low temperature (only barriers < 40 kJ/mol are accessible, no reversal) and at high temperature (all barriers accessible, fully reversible).

Steps:

1. First, identify the kinetically favored product: this is the product with lower $E_a$. B has $E_a=30$ kJ/mol < 75 kJ/mol for C, so B forms faster and is kinetically favored.
2. Identify the thermodynamically favored product: this is the product with more negative $\Delta G$. C has $\Delta G=-45$ kJ/mol < -12 kJ/mol for B, so C is more stable and thermodynamically favored.
3. At low temperature: only barriers < 40 kJ/mol are accessible, so only B can form, and no reversal occurs. Major product = B, under kinetic control.
4. At high temperature: all barriers are accessible, and reactions are reversible. The system reaches equilibrium, so the more stable C is the major product, under thermodynamic control.

Exam tip: On FRQ, always explicitly connect control type to conditions: kinetic control = low temperature, irreversibility, only low Ea barriers accessible. Do not just state "low T gives kinetic product" without this reasoning to earn full points.

4. Catalyst Effects on Favorability vs Rate

A common misconception tested repeatedly on the AP exam is what catalysts change versus what they do not. A catalyst works by providing an alternative reaction mechanism (pathway) from reactants to products, with a lower activation energy than the uncatalyzed pathway. Lower $E_a$ increases the rate constant $k$, so the reaction proceeds faster for both the forward and reverse reactions equally.

Because a catalyst does not change the chemical identity or free energy of the starting reactants or final products, it does not change the overall $\Delta G$ for the reaction. This means a catalyst cannot change the thermodynamic favorability of a reaction: it cannot make a non-spontaneous ($\Delta G>0$) reaction become spontaneous, and it does not change the equilibrium constant or the final product yield at equilibrium. It only makes the reaction reach equilibrium faster.

Worked Example

The decomposition of hydrogen peroxide is $2{\text{H}}_2{\text{O}}_2(aq)\to 2{\text{H}}_2\text{O}(l)+{\text{O}}_2(g)$, with $\Delta G^{\circ }=-234$ kJ/mol at 25°C. The uncatalyzed activation energy is 71 kJ/mol, and the enzyme catalase lowers $E_a$ to 8 kJ/mol. Which statement is correct after adding catalase? A) ΔG becomes more negative, and the rate increases B) The rate increases, and ΔG remains unchanged C) ΔG becomes positive, and the rate decreases D) The rate remains unchanged, and ΔG remains unchanged

Steps:

1. Recall the core rule for catalysts: catalysts change the reaction pathway (lower $E_a$) but do not change the free energy of reactants or products.
2. Therefore, $\Delta G$ for the reaction stays at -234 kJ/mol, so the reaction remains thermodynamically favorable. Eliminate options A and C.
3. Lowering $E_a$ from 71 kJ/mol to 8 kJ/mol means far more colliding H₂O₂ molecules have enough energy to react, so the reaction rate increases dramatically. Eliminate D.
4. The correct answer is B.

Exam tip: Any multiple-choice option that claims a catalyst changes ΔG, spontaneity, or the equilibrium constant K is automatically wrong. Only rate and activation energy are changed.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Claiming a thermodynamically favorable reaction will always occur at an observable rate. Why: Students associate the everyday definition of "spontaneous" (happens immediately) with the chemical definition, so they confuse thermodynamic possibility with kinetic feasibility. Correct move: Always explicitly separate the two properties: state that ΔG < 0 means the reaction is possible, but rate depends on Ea, so it may be too slow to observe.
• Wrong move: Claiming a catalyst changes the spontaneity (ΔG) of a reaction. Why: Students know catalysts speed up reactions, so they incorrectly extrapolate that catalysts can make a non-spontaneous reaction spontaneous. Correct move: Remember catalysts only change the reaction pathway (lower Ea) not the free energy of reactants or products, so ΔG and spontaneity are always unchanged.
• Wrong move: Predicting the thermodynamically favorable product as the major product under all conditions. Why: Students forget competing reactions can be under kinetic control if equilibrium cannot be reached. Correct move: Always check whether the reaction is reversible and whether the activation barrier for the thermodynamic product is accessible before predicting the major product.
• Wrong move: Stating a non-thermodynamically favorable reaction can never occur. Why: Students confuse "will not occur on its own" with "can never occur". Correct move: Recall non-spontaneous reactions can occur if driven by external free energy input (e.g. electrolysis) or coupled to a favorable reaction; thermodynamic favorability only means no continuous input is needed, not that it can never happen.
• Wrong move: Assigning kinetic control to high temperature and thermodynamic control to low temperature. Why: Students reverse the relationship between temperature and control type. Correct move: Memorize that low temperature = only low Ea barriers accessible, no reversal = kinetic control; high temperature = all barriers accessible, reversibility = thermodynamic control.
• Wrong move: Using activation energy to determine the sign of ΔG and spontaneity. Why: Students mix up kinetic and thermodynamic properties, using Ea (kinetic) to infer favorability. Correct move: Only use the overall free energy difference between reactants and products to determine thermodynamic favorability; Ea is only used for rate, never for spontaneity.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

The conversion of ozone to oxygen in the lower atmosphere is $2{\text{O}}_3(g)\to 3{\text{O}}_2(g)$, $\Delta G^{\circ }=-326$ kJ/mol at 25°C. At 25°C, the reaction is extremely slow, with a half-life of multiple years. Which of the following correctly explains this observation? A) The reaction is thermodynamically unfavorable, so it proceeds slowly B) The activation energy for the reaction is very high, so very few collisions have enough energy to react at 25°C C) ΔG is negative, so the reaction should be fast, meaning the equilibrium constant is very small D) The enthalpy change for the reaction is positive, leading to slow reaction rate

Worked Solution: First, eliminate A because ΔG° is negative, so the reaction is thermodynamically favorable. Eliminate C: a negative ΔG° means the equilibrium constant K is large, not small, and ΔG sign does not determine rate. Eliminate D: this reaction is exothermic (ΔH < 0), and enthalpy sign does not control reaction rate anyway. The slow rate is caused by a high activation energy, which means only a tiny fraction of ozone collisions have enough energy to overcome the barrier at 25°C. The correct answer is B.

Question 2 (Free Response)

Acid-catalyzed isomerization of glucose produces two products from the same starting material, with the following data at 1 atm:

• Glucose → Fructose: $\Delta G^{\circ }=-2.5$ kJ/mol, $E_a=48$ kJ/mol
• Glucose → Sorbose: $\Delta G^{\circ }=-12.1$ kJ/mol, $E_a=92$ kJ/mol

(a) Identify which product is kinetically favored and which is thermodynamically favored. Justify your answer. (b) Predict the major product after running the reaction at 0°C, where the maximum activation energy accessible is 60 kJ/mol, and reactions do not reverse. What type of reaction control is this? (c) A chemist allows the reaction to run for 48 hours at 100°C, where all activation barriers are accessible and both reactions are fully reversible. Predict the major product, and explain why it is different from the product at 0°C.

Worked Solution: (a) Fructose is kinetically favored, because it has a lower activation energy (48 kJ/mol < 92 kJ/mol) so it forms faster. Sorbose is thermodynamically favored, because it has a more negative ΔG° (-12.1 kJ/mol < -2.5 kJ/mol), so it has lower free energy and is more stable. (b) At 0°C, only activation barriers below 60 kJ/mol are accessible. Fructose's Ea (48 kJ/mol) is below this threshold, while sorbose's Ea (92 kJ/mol) is above. No reversal of fructose formation occurs, so the major product is fructose. This is kinetic control. (c) At 100°C, all activation barriers are accessible, and all reactions are reversible. The system can reach equilibrium, which favors the more thermodynamically stable product. Therefore the major product is sorbose. The difference arises because high temperature provides enough energy to cross the higher activation barrier for sorbose, and reversibility allows glucose and fructose to convert to the more stable sorbose over time.

Question 3 (Application / Real-World Style)

Iron metal rusts in the presence of oxygen and water via the reaction $4\text{Fe}(s)+3{\text{O}}_2(g)+2{\text{H}}_2\text{O}(l)\to 2{\text{Fe}}_2{\text{O}}_3\cdot {\text{H}}_2\text{O}(s)$, $\Delta G^{\circ }=-1504$ kJ/mol at 25°C. At 25°C, a block of pure iron placed in dry air (10% relative humidity) loses only 0.1% of its mass to rusting per year. Explain this observation using the concepts of thermodynamic favorability and rate, and describe the effect of increasing humidity on the rate of rusting.

Worked Solution: The very negative ΔG° confirms that rusting is highly thermodynamically favorable at 25°C, so it is spontaneous overall. However, the reaction rate is extremely slow in dry air because water acts as a catalyst for the rusting redox reaction, lowering the activation energy. In dry air, very little water is available, so the activation energy for rusting remains very high, leading to a very slow rate even though the reaction is thermodynamically favorable. Increasing humidity increases the amount of water available, which lowers the activation energy for the reaction. This increases the reaction rate, leading to much faster mass loss from rusting. This matches the real-world observation that iron rusts much faster in humid environments than in dry climates, even though thermodynamic favorability is unchanged.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundation for understanding how thermodynamics and kinetics interact to determine real-world reaction outcomes, which you will apply next to the study of kinetic versus thermodynamic control in organic synthesis and electrochemical cell potential analysis, where you will need to distinguish between thermodynamically allowed redox reactions and kinetically slow reactions that do not produce usable current even with a positive cell potential. Without mastering the distinction between favorability and rate, you will struggle to explain observations like why some thermodynamically allowed reactions do not occur at observable rates, or why organic reactions yield different major products at different temperatures. This topic also connects to the overarching AP Chemistry theme of energy changes in chemical systems, linking thermodynamics and kinetics into a coherent framework for predicting chemical behavior.

Gibbs free energy and spontaneity, Arrhenius equation and activation energy, Cell potential and thermodynamics