Unit 7: Equilibrium

Dynamic Equilibrium and Reversible Reactions

Chemical equilibrium is the state of a reversible reaction in which the forward and reverse reaction rates are equal, so the macroscopic amounts (concentrations or partial pressures) of reactants and products stay constant over time. “Constant” does not mean equal amounts, and it does not mean the reaction stopped. At equilibrium, molecules are still colliding and reacting in both directions; the net change is zero because the two directions occur at the same rate.

A helpful picture is a crowded hallway with two doors: people continuously move through both doors, but once the flow in both directions matches, the number of people on each side stops changing. The motion continues; the net change becomes zero.

Equilibrium is only meaningful for reversible reactions (reactions that can proceed in both directions under the given conditions). To reach equilibrium and stay there, the system must be closed so that reactants/products cannot escape. If gases leak out or a product is continuously removed, the composition cannot settle into a stable equilibrium state.

The equilibrium position describes the relative amounts of reactants and products at equilibrium. Some equilibria contain mostly products, others mostly reactants, and the position depends on the reaction and on temperature. A common misconception is that equilibrium means “50–50”; in reality, equal rates do not require equal concentrations.

Equilibrium is tightly connected to reaction rates over time. If you start with only reactants, the forward rate is initially high and the reverse rate is near zero. As products form, the reverse rate increases until the rates become equal. On a typical concentration-versus-time graph, reactant concentrations decrease and level off, and product concentrations increase and level off. The leveling off indicates no net change, not “no reaction.”

Stoichiometric coefficients also show up in how fast concentrations change before equilibrium. Because coefficients tell you how much of each substance is created or consumed per “reaction event,” they affect relative slopes on early-time graphs. For example, in the Haber process,

N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)

when N_2 and H_2 start at the same concentration, H_2 is consumed three times faster than N_2 (consistent with the coefficient 3), so its initial concentration-time curve is typically steeper downward.

Example: Identifying equilibrium from a graph

If a graph shows concentrations changing at first and then becoming horizontal (constant), the system has reached equilibrium. If the lines are still changing, it has not.

Exam Focus

Typical question patterns include interpreting concentration vs. time graphs to identify when equilibrium is reached and comparing relative equilibrium amounts, explaining in words what “dynamic equilibrium” means (rates equal, not concentrations), and determining whether a described system can reach equilibrium (open vs. closed system). Common mistakes include saying “equilibrium means the reaction stops” (instead, reactions continue and net change is zero), assuming equal concentrations at equilibrium, and ignoring the need for a closed system when discussing equilibrium.

Equilibrium Constants and the Law of Mass Action: Writing K, K_c, and K_p

When a reversible reaction reaches equilibrium, the ratio of product amounts to reactant amounts settles into a predictable value called the equilibrium constant. The equilibrium expression itself is often referred to as the Law of Mass Action.

For a general reaction,

aA + bB \rightleftharpoons cC + dD

the concentration-based equilibrium constant is:

K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}

Here, brackets indicate equilibrium molar concentrations (mol/L), and the exponents come directly from the balanced equation coefficients. The most important idea is that the equilibrium expression is built from the balanced reaction as written.

Equilibrium constants matter because they compactly describe equilibrium position and allow you to predict which side is favored, calculate unknown equilibrium concentrations/pressures, and analyze how a system responds to disturbances (via comparisons of Q vs. K).

What goes into equilibrium expressions (and what does not)

Include gases and aqueous solutes because their concentrations/pressures can change meaningfully as the reaction proceeds. Exclude pure solids and pure liquids because their effective concentrations are essentially constant; those constants are absorbed into the value of K.

For example:

CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)

Only the gas appears:

K_c = [CO_2]

A common mistake is to include CaCO_3 or CaO in the expression; you should not.

K_p for gas-phase equilibria

For reactions involving gases, you can write an equilibrium constant in terms of partial pressures:

K_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}

Relationship between K_p and K_c

When a reaction involves gases, the two constants are related by:

K_p = K_c(RT)^{\Delta n}

Here, \Delta n is moles of gaseous products minus moles of gaseous reactants.

For example:

N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)

\Delta n = 2 - (1+3) = -2

So:

K_p = K_c(RT)^{-2}

This highlights the key concept: if the number of moles of gas changes, concentration-based and pressure-based constants are not numerically the same.

Units and interpreting the size of K

In AP Chemistry, equilibrium constants are typically treated as unitless, even though algebra can suggest units. (A more rigorous thermodynamic approach uses activities.) Also, the magnitude of K indicates which side is favored:

• K greater than 1 means products are favored over reactants.
• K less than 1 means reactants are favored over products.

Notation reference (common variants)

Other important equilibrium constants you may see

Several specialized equilibrium constants follow the same general “products over reactants” structure, with the same rules about omitting pure solids and pure liquids.

• K_{sp} is the solubility product constant (often no denominator because reactants are solids).
• K_a is the acid dissociation constant.
• K_b is the base dissociation constant.
• K_w is the ionization constant for water.

At room temperature,

K_w = 1.0 \times 10^{-14}

Exam Focus

Typical question patterns include writing K_c or K_p for a balanced equation (often including solids/liquids to test whether you omit them), converting between K_c and K_p using \Delta n, and identifying whether a proposed equilibrium expression matches a given reaction. Common mistakes include using coefficients as multipliers instead of exponents, including pure solids/pure liquids in K, and using \Delta n incorrectly (count only gaseous moles).

The Reaction Quotient Q: Predicting Direction Before Equilibrium

The reaction quotient, Q, has the same mathematical form as the equilibrium constant, but it uses current concentrations/pressures at a particular moment, not necessarily equilibrium values. This makes Q useful at any point in time during a reaction to determine how close or far the reaction is from equilibrium and how it will shift.

For:

aA + bB \rightleftharpoons cC + dD

Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}

For gases:

Q_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}

You compare Q to K at the same temperature:

• If Q < K, there are too few products relative to equilibrium, so the reaction proceeds forward (toward products).
• If Q > K, there are too many products, so the reaction proceeds reverse (toward reactants).
• If Q = K, the system is at equilibrium.

Example 1: Using Q to predict direction

For:

N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)

Suppose:

• [N_2] = 0.50
• [H_2] = 0.50
• [NH_3] = 0.20

Compute:

Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3}

Q_c = \frac{(0.20)^2}{(0.50)(0.50)^3}

Q_c = \frac{0.040}{(0.50)(0.125)}

Q_c = \frac{0.040}{0.0625}

Q_c = 0.64

If (at the same temperature) K_c = 6.4, then Q_c < K_c, so the reaction proceeds forward to form more NH_3.

Example 2: Connecting Q to Le Châtelier reasoning

If you suddenly add product, Q instantly increases because the numerator increases. If that makes Q > K, the system must shift left to reduce products and restore Q back to K. Thinking “a stress bumps Q, then the reaction shifts to fix it” is often more reliable than memorizing shift rules.

Exam Focus

Typical question patterns include calculating Q from given concentrations/pressures and comparing to K to predict direction, describing what happens to Q immediately after a disturbance (before shifting), and identifying whether a system is at equilibrium from given concentrations. Common mistakes include using equilibrium concentrations when asked to compute Q, forgetting to omit solids and liquids in Q, and comparing Q_c to K_p instead of matching forms.

Interpreting and Manipulating K: Size, Reversals, and Reaction Algebra

Equilibrium constants encode the balance point for a specific reaction at a specific temperature, so they are more than just calculator inputs.

A large K means products are favored at equilibrium (equilibrium lies to the right), a small K means reactants are favored, and K near 1 means appreciable amounts of both sides are present. “Favored” does not mean “100%.”

K depends only on temperature

For a given balanced reaction, K changes only when temperature changes. Changing initial concentrations, adding catalysts, or changing volume/pressure can change the equilibrium position (the actual equilibrium concentrations), but not the numerical value of K.

Manipulating equilibrium equations (Hess’s-law style)

Because K is tied to the reaction as written, rewriting reactions changes K in predictable ways.

Reversing a reaction

If:

A \rightleftharpoons B

then:

K = \frac{[B]}{[A]}

For the reverse:

B \rightleftharpoons A

K_{rev} = \frac{1}{K}

Multiplying coefficients by a factor

If you multiply an entire reaction by a factor n, the new constant becomes:

K_{new} = K^n

For example, “multiply by 2” means “square the constant.”

Adding reactions

If you add two reactions to obtain an overall reaction, you multiply their constants:

K_{overall} = K_1K_2

Worked example: building an overall K

Suppose:

A \rightleftharpoons B

has K_1 = 4.0 and:

B \rightleftharpoons C

has K_2 = 0.50. Adding them gives:

A \rightleftharpoons C

So:

K_{overall} = (4.0)(0.50) = 2.0

If you instead need:

C \rightleftharpoons A

then:

K = \frac{1}{2.0} = 0.50

Temperature and equilibrium (conceptual)

Changing temperature changes K and shifts equilibrium in a direction tied to whether heat is a reactant or product.

• For an endothermic forward reaction (heat acts like a reactant), increasing temperature favors products and increases K.
• For an exothermic forward reaction (heat acts like a product), increasing temperature favors reactants and decreases K.

A key real example is the Haber process, which is exothermic with enthalpy change:

\Delta H = -92.6 \text{ kJ/mol}

So heat is on the product side:

N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) + energy

If temperature goes up, the equilibrium shifts left (toward reactants) to consume the added heat. If temperature goes down, the equilibrium shifts right (toward products) to produce heat.

Exam Focus

Typical question patterns include finding a new K after reversing, multiplying, or adding reactions; using the magnitude of K to justify which side is favored; and predicting how K changes with temperature for endothermic vs. exothermic reactions. Common mistakes include claiming catalysts change K, forgetting to invert K when reversing a reaction, and multiplying K by a coefficient factor instead of raising K to a power.

Equilibrium Calculations with ICE Tables: Finding Unknown Concentrations or Pressures

Most equilibrium calculations connect equilibrium concentrations/pressures to K, but you often do not know those equilibrium values directly. An ICE table organizes the problem by tracking Initial values, the Change as the system moves toward equilibrium, and the Equilibrium values that must satisfy the equilibrium expression.

A reliable workflow is: write the balanced reaction, write the correct equilibrium expression (omitting solids/liquids), set up ICE using stoichiometry-based variables, substitute into the K expression, solve (approximation or quadratic), and check that results are physically reasonable (no negative concentrations and a direction consistent with Q vs. K).

Example 1: Finding equilibrium concentrations (quadratic)

Consider:

N_2O_4(g) \rightleftharpoons 2NO_2(g)

Given:

K_c = 0.50

Start with:

• [N_2O_4]_{initial} = 1.00
• [NO_2]_{initial} = 0.00

Let x be the amount of N_2O_4 that dissociates. Then at equilibrium:

• [N_2O_4] = 1.00 - x
• [NO_2] = 2x

Use:

K_c = \frac{[NO_2]^2}{[N_2O_4]}

Substitute:

0.50 = \frac{(2x)^2}{1.00 - x}

0.50 = \frac{4x^2}{1.00 - x}

0.50(1.00 - x) = 4x^2

0.50 - 0.50x = 4x^2

4x^2 + 0.50x - 0.50 = 0

Divide by 0.50:

8x^2 + x - 1 = 0

Quadratic formula:

x = \frac{-1 \pm \sqrt{1 - 4(8)(-1)}}{2(8)}

x = \frac{-1 \pm \sqrt{33}}{16}

Choose the positive root:

x \approx 0.2965

So:

[N_2O_4]_{eq} \approx 0.7035

[NO_2]_{eq} \approx 0.593

This shows why quadratics often appear when a “small x” approximation is not justified.

Example 2: When the “small x” approximation is reasonable

Approximations are common when K is very small (reactants favored) or very large (products favored) and the change is small compared with a starting concentration. For example:

H_2(g) + I_2(g) \rightleftharpoons 2HI(g)

If K_c is very large and you start with large amounts of reactants and no product, the reaction may proceed almost to completion and leave only a small amount of reactant. In that situation, you might approximate:

[H_2]_{eq} = [H_2]_{initial} - x \approx [H_2]_{initial}

A common rule of thumb is that the approximation should change the initial value by less than about 5%. If not, solve exactly.

Working with K_p using partial pressures

ICE-table logic is the same with partial pressures. For:

PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)

The equilibrium expression is:

K_p = \frac{P_{PCl_3}P_{Cl_2}}{P_{PCl_5}}

You can run the entire ICE process in atm (or another consistent unit).

Common “hidden” skills in equilibrium calculations

Successful solutions depend on matching stoichiometry to variable changes (coefficients control the change ratios), using Q to decide shift direction (and therefore the signs in the ICE row), and enforcing physical constraints like nonnegative equilibrium values.

Exam Focus

Typical question patterns include using an ICE table with a given K_c to find equilibrium concentrations, computing K_c from initial/equilibrium data, and solving gas equilibrium problems using K_p and partial pressures. Common mistakes include choosing the wrong sign for changes (especially when the reaction shifts left), forgetting to apply coefficients as exponents in the expression, and using an approximation without checking its validity.

Le Châtelier’s Principle: How Systems Respond to Stress (and How Q Explains It)

Le Châtelier’s principle says that if a system at equilibrium is disturbed, it shifts in the direction that counteracts the disturbance and helps re-establish equilibrium. A dependable way to apply this is to think in two stages: the stress immediately changes concentrations/pressures (or temperature), which immediately changes Q, and then the system shifts in whichever direction restores the condition Q = K.

Concentration changes

If you add a reactant, the denominator of Q increases and Q decreases; if that makes Q < K, the reaction proceeds forward to form more products. If you add a product, the numerator increases and Q increases; if Q > K, the system shifts left.

In the Haber process,

N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)

adding N_2 or H_2 shifts right to create more NH_3, while adding NH_3 shifts left to create more N_2 and H_2. If N_2 or H_2 is removed, the system shifts left to make up for the loss by forming more reactants.

Adding or removing a pure solid or pure liquid does not change Q because those substances are not in the equilibrium expression.

A concrete gas-phase example:

CO(g) + Cl_2(g) \rightleftharpoons COCl_2(g)

Q_p = \frac{P_{COCl_2}}{P_{CO}P_{Cl_2}}

If you increase P_{Cl_2} suddenly, the denominator increases so Q_p decreases; if Q_p < K_p, the system shifts forward to make more COCl_2.

Pressure and volume changes (gases)

Pressure/volume stresses matter most when gases are involved and the total moles of gas differ between sides. Decreasing volume (increasing pressure) shifts toward fewer moles of gas; increasing volume (decreasing pressure) shifts toward more moles of gas. If both sides have the same moles of gas, pressure/volume changes do not shift equilibrium (even though the pressures change).

For the Haber process, there are 4 moles of gas on the left and 2 on the right. Increasing pressure shifts right (toward fewer moles), while decreasing pressure shifts left.

Inert (noble) gas addition

This is a frequent exam trap. If you add an inert gas at constant volume, the partial pressures of reacting gases do not change (since their n_i, T, and V are unchanged), so there is no shift. If you add an inert gas at constant pressure, the volume increases, partial pressures of reacting gases decrease, and the system may shift toward the side with more moles of gas.

Temperature changes (the only stress that changes K)

Temperature is special because changing temperature changes the value of K, not just Q. Treat heat like a reactant or product. For the exothermic Haber process,

N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) + energy

increasing temperature shifts left (toward reactants), and decreasing temperature shifts right (toward products).

Dilution (aqueous equilibria)

Adding water to an aqueous equilibrium changes concentrations and can shift equilibrium. For example:

Fe^{3+}(aq) + SCN^{-}(aq) \rightleftharpoons FeSCN^{2+}(aq)

Diluting by adding water lowers all aqueous concentrations. The system shifts toward the side with more aqueous particles to counteract that dilution, which is the left side (two aqueous ions versus one aqueous complex). If water evaporates, concentrations increase and the system shifts toward the side with fewer aqueous particles, which is the right side.

Catalysts

A catalyst lowers activation energy and speeds up both forward and reverse reactions. Equilibrium is reached faster, but the equilibrium position does not change and K does not change.

Changes in the equilibrium constant

When concentration or pressure changes, the equilibrium constant stays the same and the product-to-reactant ratio will revert to the original equilibrium value after shifting. When temperature changes, the equilibrium constant changes to a new value because temperature affects the equilibrium itself (not just the current ratio).

Seeing Le Châtelier in concentration-time graphs

If a stress is applied at time t, concentration-time graphs often show an instantaneous jump for any species directly added or removed, followed by a gradual change as the system shifts to a new equilibrium (where the curves level off again). Volume changes can create immediate changes to multiple gaseous species at once.

Exam Focus

Typical question patterns include predicting shift direction and which species increase/decrease after concentration, pressure/volume, temperature, or dilution changes; justifying shifts using Q vs. K; and interpreting graphs showing a disturbance and re-establishment of equilibrium. Common mistakes include saying K changes when concentration/pressure changes (only temperature changes K), shifting the wrong way on pressure problems by miscounting (count only gaseous moles), and treating catalysts as changing equilibrium composition rather than just rates.

Representations of Equilibrium: Particles, Graphs, and What Changes (and What Doesn’t)

Equilibrium is tested in multiple representations: macroscopic measurements (concentrations/pressures), particulate diagrams (relative numbers of particles), and symbolic representations (balanced equations, equilibrium expressions, and Q vs. K comparisons). Being able to translate among these is a core AP skill.

In a particle diagram, equilibrium means the numbers of each type of particle remain constant over time in a closed system, even though reactions continue in both directions. If a diagram shows many more product particles than reactant particles at equilibrium, that corresponds to a larger K (products favored). However, particulate diagrams alone rarely determine a numerical K unless volume scaling and proportionality are explicitly stated.

On concentration vs. time graphs, equilibrium is reached when curves become horizontal. Disturbances can often be diagnosed by the “shape” of the response: a sudden vertical jump suggests direct addition/removal of that species, while a sudden change affecting all gases at once often suggests a volume/pressure change. After the disturbance, gradual adjustment indicates shifting until a new equilibrium is reached.

A strong mental model is to separate what happens immediately from what happens during re-equilibration. Adding/removing a dissolved species changes that species immediately. Changing volume changes all partial pressures immediately. Changing temperature changes rate constants immediately and drives the system toward a new equilibrium with a new K.

Example: graph reasoning (qualitative)

For:

A(g) \rightleftharpoons 2B(g)

suppose that at a certain time both [A] and [B] suddenly drop, and then [B] slowly rises while [A] slowly falls further until leveling off. The sudden drop in both suggests an increase in volume (dilution in concentration terms, or a pressure drop). After volume increases, the system shifts toward the side with more moles of gas (the right side), so B increases and A decreases during re-equilibration.

Exam Focus

Typical question patterns include translating between particle diagrams and equilibrium concepts (dynamic nature, relative amounts, favored side), interpreting concentration vs. time graphs with disturbances, and identifying which stress occurred based on immediate and gradual changes. Common mistakes include concluding “equilibrium” from equal concentrations rather than constant concentrations, misreading a volume change as a substance being added, and forgetting the two-stage nature of disturbances (immediate change, then shifting).

Multi-Step Equilibrium Reasoning: Combining Q, K, and Le Châtelier in AP-Style Scenarios

Many AP free-response prompts are explanation problems rather than pure computation. The core skill is moving fluently between the balanced reaction, the equilibrium expression, the effect of a stress on Q, and the shift that restores equilibrium.

A reliable reasoning template

When asked “what happens if…,” a strong structure is:

1. Write Q for the reaction (even if you do not compute a number).
2. Identify which term is changed by the stress.
3. State how Q changes immediately.
4. Compare Q to K (greater than or less than).
5. Conclude the direction that restores Q = K.
6. State what happens to key concentrations/pressures at the new equilibrium.

This mirrors how scoring rubrics reward justification.

Example: using the template (no heavy math)

Reaction:

2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)

Expression:

Q_p = \frac{(P_{SO_3})^2}{(P_{SO_2})^2(P_{O_2})}

Stress: increase P_{SO_3} by adding product. The numerator increases, so Q_p increases. With temperature constant, K_p is unchanged, so now Q_p > K_p. The system shifts left to reduce SO_3 and increase reactants until Q_p = K_p again. At the new equilibrium, P_{SO_2} and P_{O_2} are higher than immediately after the addition, and P_{SO_3} is lower than immediately after the addition (though it may still be higher than before the stress).

Quantitative + qualitative combo

If a prompt gives initial pressures and K_p, computing Q_p and stating the direction of shift is often sufficient unless the problem explicitly asks for equilibrium amounts.

Common trap statements

• “Adding reactant increases K.” False; K depends only on temperature.
• “If you compress the container, K_p changes.” False; K_p depends only on temperature.
• “At equilibrium, rates are zero.” False; rates are equal.

Exam Focus

Typical question patterns include written justifications of shifts using Q compared to K, predicting whether different starting mixtures shift forward or reverse, and mixed conceptual tasks distinguishing immediate changes from shift-driven changes. Common mistakes include mixing up the immediate effect of a stress with the later shift, stating a shift without referencing Q or the equilibrium expression when justification is required, and claiming a new equilibrium restores original concentrations (it restores the equilibrium relationship, not the original amounts).

Solubility Equilibria: Solubility, K_{sp}, Molar Solubility, and the Common Ion Effect

Solubility equilibria apply equilibrium ideas to dissolving ionic solids. A practical (and commonly used) rule-of-thumb definition is that a salt is considered soluble if 1 gram of the salt dissolves in 100 milliliters of water. In many AP-style problems, “soluble salts” are assumed to dissolve completely in water, while “insoluble” salts establish an equilibrium between solid and dissolved ions.

Most solids increase in solubility as temperature increases because greater molecular motion can help break apart the solid lattice, but some solids show decreased solubility with increasing temperature; this trend can be hard to predict without data.

Solubility product constant K_{sp}

K_{sp} measures how much a salt dissolves in a solution. The larger the K_{sp} value, the more soluble the salt is (under the specified conditions).

For a generic dissolution such as:

A_aB_b(s) \rightleftharpoons aA(aq) + bB(aq)

the solubility product expression is:

K_{sp} = [A]^a[B]^b

There is typically no denominator because the reactant is a pure solid, and pure solids are omitted from equilibrium expressions.

Molar solubility and ion concentrations

The molar solubility of a salt is the equilibrium amount of solid that dissolves per liter, and it connects directly to ion concentrations through the dissolution stoichiometry. A key idea is that the solubility of a salt corresponds to the concentrations of ions it produces.

For example, if the molar solubility of A_aB_b is 1.3 \times 10^{-4} \text{ M}, then the equilibrium ion concentrations depend on the coefficients: the concentration of the A ion would be the molar solubility multiplied by a, and the concentration of the B ion would be the molar solubility multiplied by b. The salt-to-ion ratio matters.

The common ion effect

The common ion effect occurs when an ion already present in solution suppresses the dissolution of a slightly soluble salt by shifting the dissolution equilibrium back toward the solid.

Example: If AgCl is placed in water, it dissolves only in small amounts. If NaCl is added to the solution, it dissolves completely and does not directly affect Ag^+ ions. The Na^+ ions can generally be ignored in the solubility equilibrium, but all Cl^- ions from both NaCl and AgCl must be considered because Cl^- is the common ion. The added Cl^- shifts the equilibrium to reduce dissolved Ag^+ and therefore decreases the solubility of AgCl.

Exam Focus

Typical question patterns include writing K_{sp} expressions (omitting solids), relating molar solubility to equilibrium ion concentrations using stoichiometry, comparing solubilities using relative K_{sp} values, and predicting solubility changes with a common ion present. Common mistakes include putting solids into the K_{sp} expression, forgetting to multiply molar solubility by stoichiometric coefficients to get ion concentrations, and ignoring a common ion contributed by a different soluble salt (for example, neglecting Cl^- from NaCl when analyzing AgCl dissolution).