Introduction to equilibrium — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Dynamic equilibrium definition, the law of mass action, writing $K_c$ and $K_p$ equilibrium constant expressions, calculation of the reaction quotient $Q$, and comparing $Q$ and $K$ to predict reaction direction.

You should already know: Reversible vs irreversible reaction classification. Balanced chemical equation stoichiometry. Partial pressure and concentration units for gases and solutions.

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 Introduction to equilibrium?

Introduction to equilibrium is the core foundational topic for Unit 7 (Equilibrium) in the AP Chemistry CED, which accounts for 7-9% of the total AP exam score. This topic introduces the unique behavior of reversible reactions, which do not go to completion (unlike the irreversible reactions you studied for limiting reactant calculations) but instead reach a steady state where reactant and product concentrations no longer change over time. Equilibrium can be established for any closed system, including phase changes (like liquid-vapor equilibrium for water in a sealed container) and chemical reactions (like the reversible synthesis of ammonia). On the AP exam, content from this topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections: MCQ typically tests conceptual understanding of dynamic equilibrium and expression writing, while FRQ integrates this content with calculations and reasoning about reaction direction.

2. Dynamic Equilibrium

Dynamic equilibrium is the steady state reached by a reversible reaction in a closed system at constant temperature where the rate of the forward reaction equals the rate of the reverse reaction. The key detail here is that the equilibrium is dynamic, not static: reactions do not stop once equilibrium is reached. Instead, forward and reverse processes occur at equal rates, so there is no net change in the concentration of any reactant or product over time. This means observable properties (color, pressure, concentration) remain constant, even at dynamic equilibrium. For example, if you seal liquid bromine in a flask, after some time the brown color of bromine vapor stops changing intensity: the rate of evaporation of liquid bromine equals the rate of condensation of gaseous bromine, so the concentration of bromine vapor is constant, but both processes continue. The origin of the equilibrium constant comes directly from the equality of forward and reverse rates: for a general reaction $aA+bB\rightleftharpoons cC+dD$, the rate law for the forward reaction is $rat{e}_f=k_f[A{]}^a[B{]}^b$, and the rate law for the reverse reaction is $rat{e}_r=k_r[C{]}^c[D{]}^d$. At equilibrium, $rat{e}_f=rat{e}_r$, so rearranging gives $\frac{k_f}{k_r}=\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}$, where the left side is the equilibrium constant $K$.

Worked Example

A student places solid iodine into a closed, temperature-controlled flask. After 30 minutes, the intensity of purple color from gaseous iodine stops changing. (a) Explain why the color stops changing even though solid iodine is still evaporating, (b) Is the system at dynamic equilibrium? Justify your answer.

1. Recognize that constant color intensity means the concentration of gaseous iodine ($I_2(g)$) is no longer changing, but this does not mean all processes have stopped.
2. The reversible process here is $I_2(s)\rightleftharpoons I_2(g)$. As solid iodine evaporates, the concentration of $I_2(g)$ increases, which increases the rate of the reverse process (deposition of $I_2(g)$ back to solid).
3. Eventually, the rate of evaporation equals the rate of deposition. Because the rates are equal, the concentration of $I_2(g)$ does not change, so the color intensity remains constant.
4. Since both forward and reverse processes continue to occur at equal rates (rather than stopping), the system is at dynamic equilibrium.

Exam tip: On AP MCQ, any answer that claims "at equilibrium, all reaction stops" or "concentrations of reactants equal concentrations of products" is always wrong. Remember: equal rates, not equal concentrations, and reactions never stop at equilibrium.

3. The Equilibrium Constant and Law of Mass Action

The law of mass action is the empirical rule that defines the form of the equilibrium constant expression for any balanced reversible reaction. For the general balanced reaction: $$aA+bB\rightleftharpoons cC+dD$$ the equilibrium constant $K$ is defined as the ratio of product concentrations (raised to their stoichiometric coefficients) over reactant concentrations (raised to their stoichiometric coefficients): $$K_c=\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}$$ The subscript $c$ indicates that $K_c$ uses molar concentrations (in mol/L) for all species. For gas-phase reactions, we can also write $K_p$, which uses partial pressures of gases (typically in atm) instead of concentrations, with the same ratio of products to reactants. A critical rule for writing any equilibrium expression is: pure solids, pure liquids, and solvents in dilute solutions do not appear in the expression. This is because their concentrations are constant at constant temperature, so they are incorporated into the numerical value of $K$ and do not need to be included explicitly. For reactions with gaseous species, the relationship between $K_p$ and $K_c$ is: $$K_p=K_c(RT{)}^{\Delta n}$$ where $\Delta n=\text{moles of gaseous products}-\text{moles of gaseous reactants}$, $R=0.0821\text{Lcdotpatm/molcdotpK}$, and $T$ is absolute temperature in Kelvin.

Worked Example

Write the $K_c$ and $K_p$ expressions for the reversible decomposition of solid calcium carbonate to solid calcium oxide and carbon dioxide gas: $CaC{O}_3(s)\rightleftharpoons CaO(s)+C{O}_2(g)$.

1. First, identify all species and exclude pure solids per the rule: both $CaC{O}_3$ (reactant) and $CaO$ (product) are pure solids, so they are omitted from both expressions.
2. For $K_c$, the only species with a variable concentration is gaseous $C{O}_2$, so the expression becomes $K_c=\frac{[C{O}_2]}{1}=[C{O}_2]$ (the denominator is 1 because all reactants are pure solids).
3. For $K_p$, the only species with a variable partial pressure is gaseous $C{O}_2$, so the expression becomes $K_p=P_{C{O}_2}$.
4. Verify that the product is in the numerator, matching the direction of the balanced reaction, and no pure solids are included: the expressions are correct.

Exam tip: Always check for pure solids and liquids before writing a K or Q expression on FRQ. If you include them, your expression will be marked wrong, which is an easy point to avoid losing.

4. The Reaction Quotient and Predicting Reaction Direction

The reaction quotient ($Q$) is a value calculated from non-equilibrium concentrations (or partial pressures) of reactants and products at any point in a reaction before equilibrium is reached. $Q$ has the exact same form as the equilibrium constant $K$: the only difference is that $K$ uses equilibrium concentrations, while $Q$ uses current non-equilibrium concentrations. By comparing $Q$ to $K$, we can predict which direction the reaction will proceed to reach equilibrium:

• If $Q<K$: The numerator (product terms) is too small, and the denominator (reactant terms) is too large. The reaction will proceed forward (toward products) to increase $Q$ until it equals $K$.
• If $Q>K$: The numerator (product terms) is too large. The reaction will proceed in reverse (toward reactants) to decrease $Q$ until it equals $K$.
• If $Q=K$: The system is already at equilibrium, so no net change occurs.

This comparison is one of the most frequently tested skills on the AP exam for this topic, and it forms the basis for Le Châtelier's principle later in the unit.

Worked Example

For the reaction $N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g)$, $K_c=0.50$ at 400 K. A student mixes the gases at 400 K with the following initial concentrations: $[N_2]=0.10$ M, $[H_2]=0.10$ M, $[N{H}_3]=0.050$ M. Predict which direction the reaction will shift to reach equilibrium.

1. Write the expression for $Q_c$, which matches the form of $K_c$: $Q_c=\frac{[N{H}_3{]}^2}{[N_2][H_2{]}^3}$.
2. Substitute the initial non-equilibrium concentrations into the expression: $Q_c=\frac{(0.050{)}^2}{(0.10)(0.10{)}^3}=\frac{0.0025}{0.0001}=25$.
3. Compare the calculated $Q_c$ to the given $K_c$: $Q_c=25>K_c=0.50$.
4. Since $Q>K$, the reaction will proceed in the reverse direction (toward reactants) to reach equilibrium.

Exam tip: A simple mnemonic to avoid mixing up direction: if $Q<K$, you need more of the top (products) to increase Q to K. If $Q>K$, you need more of the bottom (reactants) to decrease Q to K. This avoids misremembering the rule.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Including pure solids or pure liquids in the equilibrium constant expression, with their concentration values. Why: Students are used to including all species in stoichiometry and rate law problems, so they forget the rule that constant concentrations do not appear. Correct move: Before writing any K or Q expression, cross out all pure solids, pure liquids, and dilute solvents, only include aqueous or gaseous species with variable concentrations.
• Wrong move: Claiming that at equilibrium, the concentration of reactants equals the concentration of products. Why: Students confuse equal rates of forward and reverse reaction with equal concentrations of reactants and products. Correct move: Always remember: at equilibrium, rates are equal, concentrations are constant (not necessarily equal). Only the ratio of concentrations (raised to stoichiometric coefficients) equals K.
• Wrong move: Flipping the ratio, putting reactants in the numerator and products in the denominator for K or Q. Why: Students mix up the direction of the reaction or mis-memorize the ratio order. Correct move: Always write the balanced reaction first, then put all species on the right (products) of the equilibrium arrow in the numerator, and all species on the left (reactants) in the denominator.
• Wrong move: Forgetting to raise concentrations/partial pressures to their stoichiometric coefficients in K or Q. Why: Students rush through calculations and skip checking exponents. Correct move: After writing the expression, check each term's exponent against the stoichiometric coefficient in the balanced reaction before plugging in any numbers.
• Wrong move: Predicting a forward shift when $Q>K$. Why: Students reverse the direction rule because they misremember which value corresponds to which shift. Correct move: Always ask "do I need to increase or decrease Q to reach K?" If Q < K, increase Q by making more products; if Q > K, decrease Q by making more reactants.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

For the reversible reaction $2HI(g)\rightleftharpoons H_2(g)+I_2(g)$, $K_c=0.020$ at a certain temperature. A reaction mixture has the following concentrations: $[HI]=0.25$ M, $[H_2]=0.15$ M, $[I_2]=0.10$ M. Which of the following statements is correct? A) The system is at equilibrium, so no shift occurs. B) The reaction will shift forward to produce more $H_2$ and $I_2$. C) The reaction will shift in reverse to produce more $HI$. D) The reaction will shift forward initially, then shift reverse to reach equilibrium.

Worked Solution: First, write the expression for $Q_c$ matching the form of $K_c$: $Q_c=\frac{[H_2][I_2]}{[HI{]}^2}$. Substitute the given concentrations: $Q_c=\frac{(0.15)(0.10)}{(0.25{)}^2}=\frac{0.015}{0.0625}=0.24$. Compare $Q_c=0.24$ to $K_c=0.020$: $Q_c>K_c$, so the reaction has more product than it would at equilibrium, meaning it must shift reverse to make more reactant (HI). Eliminate A (Q ≠ K), B (Q < K would be required for a forward shift), and D (no secondary shift occurs once direction is determined). The correct answer is C.

Question 2 (Free Response)

Nitrogen dioxide gas dimerizes to form dinitrogen tetroxide gas in a closed container at 25°C. The balanced reaction is $2N{O}_2(g)\rightleftharpoons N_2{O}_4(g)$. (a) Write the $K_c$ and $K_p$ expressions for this reaction. (b) The initial partial pressures of the gases are: $P_{N{O}_2}=0.50$ atm, $P_{N_2{O}_4}=0.25$ atm. $K_p=8.8$ at 25°C. Calculate $Q_p$ and predict the direction of shift to reach equilibrium. (c) A student claims that "after the system reaches equilibrium, no more $N{O}_2$ is converted to $N_2{O}_4$". Is the student correct? Justify your answer.

Worked Solution: (a) All species are gaseous, so all are included in the expressions: $$K_c=\frac{[N_2{O}_4]}{[N{O}_2{]}^2}{K}_p=\frac{P_{N_2{O}_4}}{(P_{N{O}_2}{)}^2}$$ (b) Substitute initial partial pressures into the $Q_p$ expression: $$Q_p=\frac{0.25}{(0.50{)}^2}=\frac{0.25}{0.25}=1.0$$ Since $Q_p=1.0<K_p=8.8$, the reaction proceeds forward (toward products, forming more $N_2{O}_4$) to reach equilibrium. (c) The student is incorrect. At equilibrium, the system is dynamic: the rate of conversion of $N{O}_2$ to $N_2{O}_4$ (forward reaction) equals the rate of conversion of $N_2{O}_4$ back to $N{O}_2$ (reverse reaction). Conversion still occurs continuously, but there is no net change in concentrations, so the student's claim is false.

Question 3 (Application / Real-World Style)

Hemoglobin (Hb) carries oxygen in human blood via the reversible reaction: $Hb(aq)+4O_2(aq)\rightleftharpoons Hb(O_2{)}_4(aq)$. At body temperature (37°C), $K_c=2.8\times 10^2$ M⁻⁴. An unacclimatized climber at high altitude has initial blood concentrations: $[Hb]=1.6\times 10^{-6}$ M, $[O_2]=6.1\times 10^{-5}$ M, $[Hb(O_2{)}_4]=1.3\times 10^{-11}$ M. Calculate $Q_c$ and determine the direction of shift, then explain what this means for oxygen loading at high altitude.

Worked Solution: Write the $Q_c$ expression: $Q_c=\frac{[Hb(O_2{)}_4]}{[Hb][O_2{]}^4}$. Substitute values: $$Q_c=\frac{1.3\times 10^{-11}}{(1.6\times 10^{-6})(6.1\times 10^{-5}{)}^4}\approx 5.8\times 10^{11}$$ Compare $Q_c$ to $K_c$: $Q_c=5.8\times 10^{11}\gg K_c=2.8\times 10^2$, so the system shifts reverse toward reactants. In context, this means that at the low oxygen concentration found at high altitude, hemoglobin cannot bind enough oxygen to sustain normal activity, leading to the oxygen deprivation (hypoxia) experienced by unacclimatized climbers.

7. Quick Reference Cheatsheet

8. What's Next

This introduction to equilibrium is the foundational prerequisite for all subsequent topics in Unit 7. Next, you will learn how to manipulate equilibrium constant values for reversed, multiplied, or combined reactions, then calculate equilibrium concentrations from initial concentrations using the ICE table method. You will also apply the $Q$ vs $K$ comparison you learned here to Le Châtelier's principle, which predicts how equilibrium shifts in response to changes in concentration, temperature, and pressure. Without mastering the basics of writing K expressions, calculating Q, and predicting direction from Q vs K, none of these more advanced topics will make sense, and you will lose easy points on both MCQ and FRQ. This topic also feeds into the study of acid-base equilibrium and solubility equilibrium later in the course, which together account for nearly 15% of the total AP exam score. Manipulating equilibrium constants Equilibrium calculations with ICE tables Le Châtelier's principle Acid-base dissociation constants