Equilibrium — AP Chemistry Chem Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Equilibrium constants Kc/Kp, reaction quotient Q, Le Chatelier's principle, acid-base equilibria, pH, Ka, buffers, and solubility product Ksp.

You should already know: High-school chemistry, Algebra 2.

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 papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Equilibrium?

Chemical equilibrium is a dynamic state for reversible reactions (denoted with the ⇌ symbol) where the rate of the forward reaction equals the rate of the reverse reaction, so the concentrations of reactants and products remain constant over time, with no net change to macroscopic properties like color, pressure, or pH. This is not a static state: reactants are still forming products, and products are still breaking down into reactants, at equal rates. Equilibrium is the core topic of AP Chemistry Unit 7, and applies to gas-phase reactions, acid-base systems, and solubility of ionic solids, making up 7-9% of your total AP exam score.

2. Equilibrium constants $K_c$ and $K_p$

The equilibrium constant quantifies the ratio of products to reactants at equilibrium for a given reaction and temperature. For a general reversible reaction: $$aA+bB\rightleftharpoons cC+dD$$ $K_c$ uses molar equilibrium concentrations of aqueous or gaseous species, and is written as: $$K_c=\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}$$ Important rule: Solids (s) and pure liquids (l) are excluded from K expressions, as their chemical activity is defined as 1, so they do not affect the ratio.

For gas-phase reactions, you may use $K_p$, which uses partial pressures of gaseous species in place of concentrations: $$K_p=\frac{(P_C{)}^c(P_D{)}^d}{(P_A{)}^a(P_B{)}^b}$$ You can convert between $K_p$ and $K_c$ using the formula: $$K_p=K_c(RT{)}^{\Delta n}$$ Where $\Delta n=\text{moles of gaseous products}-\text{moles of gaseous reactants}$, $R=0.0821\text{ Lcdotpatm/(molcdotpK)}$, and $T$ is temperature in Kelvin.

Worked Example

For the Haber reaction $N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g)$ at 472°C, equilibrium concentrations are $[N_2]=0.12\text{ M}$, $[H_2]=0.08\text{ M}$, $[N{H}_3]=0.02\text{ M}$. Calculate $K_c$ and $K_p$.

1. Calculate $K_c$: $$K_c=\frac{[N{H}_3{]}^2}{[N_2][H_2{]}^3}=\frac{(0.02{)}^2}{(0.12)(0.08{)}^3}=\frac{0.0004}{0.00006144}\approx 6.5$$
2. $\Delta n=2-(1+3)=-2$, $T=472+273=745\text{ K}$: $$K_p=6.5\times (0.0821\times 745{)}^{-2}\approx 6.5\times (61.16{)}^{-2}\approx 1.7\times 10^{-3}$$ Exam tip: Examiners often test the rule excluding solids/liquids, so cross out all non-gas, non-aqueous species before writing your K expression.

3. Reaction quotient $Q$

The reaction quotient $Q$ uses the same formula as $K_c$ or $K_p$, but uses current, non-equilibrium concentrations or partial pressures instead of equilibrium values. Comparing $Q$ to $K$ tells you which direction the reaction will shift to reach equilibrium:

• $Q<K$: The ratio of products to reactants is lower than at equilibrium, so the forward reaction proceeds to make more products.
• $Q=K$: The system is at equilibrium, no net change occurs.
• $Q>K$: The ratio of products to reactants is higher than at equilibrium, so the reverse reaction proceeds to make more reactants.

Worked Example

For the same Haber reaction with $K_c=6.5$ at 472°C, a reaction vessel contains $[N_2]=0.2\text{ M}$, $[H_2]=0.1\text{ M}$, $[N{H}_3]=0.05\text{ M}$. Which direction will the reaction shift?

1. Calculate $Q_c$: $$Q_c=\frac{(0.05{)}^2}{(0.2)(0.1{)}^3}=\frac{0.0025}{0.0002}=12.5$$
2. $Q_c=12.5>K_c=6.5$, so the reaction shifts left, consuming $N{H}_3$ and producing more $N_2$ and $H_2$ until equilibrium is reached.

4. Le Chatelier's principle

Le Chatelier's principle states that if a stress is applied to a system at equilibrium, the system will shift to counteract the stress and re-establish equilibrium. The three common stresses are:

1. Concentration change: Adding a reactant shifts equilibrium right to consume the added reactant; adding a product shifts equilibrium left. Removing a species shifts equilibrium toward the side of the removed species.
2. Pressure/volume change: Only affects gas-phase reactions with unequal moles of gas on each side. Increasing pressure (decreasing volume) shifts equilibrium to the side with fewer moles of gas; decreasing pressure shifts to the side with more moles of gas. If $\Delta n=0$, pressure changes have no effect.
3. Temperature change: The only stress that changes the value of $K$. For endothermic reactions ($\Delta H>0$, heat is a reactant), increasing temperature shifts equilibrium right, increasing $K$. For exothermic reactions ($\Delta H<0$, heat is a product), increasing temperature shifts equilibrium left, decreasing $K$. Note: Catalysts speed up the rate of both forward and reverse reactions equally, so they do not affect equilibrium position or the value of $K$, they only reduce the time taken to reach equilibrium.

Worked Example

The Haber reaction is exothermic, $\Delta H=-92\text{ kJ/mol}$. What is the effect of each stress?

• Add $H_2$: Shifts right
• Decrease volume: $\Delta n=-2$, fewer moles on product side, shifts right
• Increase temperature: Heat is a product, shifts left, $K$ decreases
• Add iron catalyst: No change to equilibrium position or $K$

5. Acid-base equilibria — pH, $K_a$, buffers

Acid-base equilibria apply the general equilibrium rules to weak acid and base dissociation in aqueous solution. Start with the autoionization of water: $$H_{2}O(l)\rightleftharpoons H^{+}(aq)+O{H}^{-}(aq)$$ The ion product of water $K_w=[H^{+}][O{H}^{-}]=1.0\times 10^{-14}$ at 25°C. pH is defined as $pH=-\log [H^{+}]$, $pOH=-\log [O{H}^{-}]$, so $pH+pOH=14$ at 25°C.

For a weak acid $HA$ dissociating as $HA(aq)\rightleftharpoons H^{+}(aq)+A^{-}(aq)$, the acid dissociation constant is: $$K_a=\frac{[H^{+}][A^{-}]}{[HA]}$$ $p{K}_a=-\log K_a$, so smaller $p{K}_a$ values correspond to stronger weak acids.

Buffers are solutions that resist pH change when small amounts of strong acid or base are added, made of a weak acid and its conjugate base (or weak base and its conjugate acid) in comparable concentrations. The Henderson-Hasselbalch equation calculates buffer pH: $$pH=p{K}_a+\log \left(\frac{[A^{-}]}{[HA]}\right)$$ Buffer capacity is highest when $[A^{-}]=[HA]$, so $pH=p{K}_a$, and when buffer component concentrations are high.

Worked Example

Calculate the pH of a buffer made of 0.2 M acetic acid ($K_a=1.8\times 10^{-5}$, $p{K}_a=4.74$) and 0.3 M sodium acetate. $$pH=4.74+\log \left(\frac{0.3}{0.2}\right)=4.74+0.18=4.92$$ If 0.01 mol HCl is added to 1 L of this buffer, $H^{+}$ reacts with $A^{-}$ to form $HA$, so new $[A^{-}]=0.29$ M, $[HA]=0.21$ M: $$pH=4.74+\log \left(\frac{0.29}{0.21}\right)\approx 4.88$$ Only a 0.04 pH drop occurs, compared to a drop to pH 2 if the same HCl was added to pure water.

6. Solubility — $K_{sp}$

The solubility product constant $K_{sp}$ describes the equilibrium between a sparingly soluble ionic solid and its dissolved ions in solution. For a solid $M_x{A}_y(s)\rightleftharpoons x{M}^{y+}(aq)+y{A}^{x-}(aq)$: $$K_{sp}=[M^{y+}{]}^x[A^{x-}{]}^y$$ The solid is excluded from the expression, as it is a pure solid. Molar solubility $s$ is the moles of solid that dissolve per liter of solution, and can be related to $K_{sp}$ using the dissociation stoichiometry. For example:

• $AgCl(s)\rightleftharpoons A{g}^{+}+C{l}^{-}$: $K_{sp}=s^2$, so $s=\sqrt{K_{sp}}$
• $Ca{F}_2(s)\rightleftharpoons C{a}^{2+}+2F^{-}$: $K_{sp}=s(2s{)}^2=4s^3$, so $s=\sqrt[3]{K_{sp}/4}$

The common ion effect states that adding a soluble salt with an ion common to the sparingly soluble solid decreases its solubility, as the added ion increases $Q_{sp}$ above $K_{sp}$, causing precipitation.

Worked Example

$K_{sp}$ of $AgCl$ is $1.8\times 10^{-10}$ at 25°C. Calculate its molar solubility in pure water and in 0.1 M $AgN{O}_3$.

1. Pure water: $s=\sqrt{1.8\times 10^{-10}}\approx 1.3\times 10^{-5}$ M
2. 0.1 M $AgN{O}_3$: $[A{g}^{+}]=0.1$ M, so $K_{sp}=(0.1+s)s\approx 0.1s$ (since $s$ is negligible compared to 0.1). $s=1.8\times 10^{-10}/0.1=1.8\times 10^{-9}$ M, ~7000x lower than in pure water.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Including solids and pure liquids in $K/Q$ calculations. Why you do it: You forget condensed phases have activity = 1. Correct move: Cross out all (s) and (l) species first before writing any equilibrium expression, only keep (aq) and (g) species.
• Wrong move: Assuming all pressure changes affect equilibrium. Why you do it: You memorize pressure shifts without checking gas mole counts. Correct move: Calculate $\Delta n$ for gaseous species first. If $\Delta n=0$, pressure/volume changes have no effect on equilibrium position.
• Wrong move: Thinking all stresses change the value of $K$. Why you do it: You confuse shifts from $Q$ changes vs $K$ changes. Correct move: Only temperature changes alter $K$. Concentration, pressure, and catalyst changes only change $Q$, $K$ stays constant for a given temperature.
• Wrong move: Using the Henderson-Hasselbalch equation for strong acid/base mixtures. Why you do it: You assume any acid-base mixture is a buffer. Correct move: Only use HH for solutions with a weak conjugate acid-base pair present in comparable concentrations. Strong acid + strong base produces a salt and water, no buffer.
• Wrong move: Forgetting to raise ion concentrations to stoichiometric coefficients in $K_{sp}$ calculations. Why you do it: You rush through problems and ignore the solid's formula. Correct move: Write the full dissociation equation first, map each ion concentration to $s$ with its coefficient, then write the $K_{sp}$ expression before plugging in numbers.

8. Practice Questions (AP Chemistry Style)

Question 1

For the reaction $2S{O}_2(g)+O_2(g)\rightleftharpoons 2S{O}_3(g)$, $\Delta H=-196\text{ kJ/mol}$. At 1000 K, $K_p=2.5\times 10^2$. A reaction vessel at 1000 K contains partial pressures $P_{S{O}_2}=0.1\text{ atm}$, $P_{O_2}=0.05\text{ atm}$, $P_{S{O}_3}=0.8\text{ atm}$. a) Calculate $Q_p$ and state the direction the reaction will shift to reach equilibrium. b) State the effect of increasing the reaction temperature on the value of $K_p$, justify your answer.

Worked Solution

a) $$Q_p=\frac{(P_{S{O}_3}{)}^2}{(P_{S{O}_2}{)}^2{P}_{O_2}}=\frac{(0.8{)}^2}{(0.1{)}^2(0.05)}=\frac{0.64}{0.0005}=1280$$ $Q_p=1280>K_p=250$, so the reaction shifts left (reverse direction) to consume $S{O}_3$ and produce more $S{O}_2$ and $O_2$. b) The reaction is exothermic ($\Delta H<0$), so heat acts as a product. Increasing temperature adds stress to the product side, so equilibrium shifts left to consume excess heat. Since temperature is the only stress that changes $K$, shifting left means $K_p$ decreases at higher temperatures.

Question 2

A buffer is prepared by mixing 500 mL of 0.4 M formic acid ($HCOOH$, $K_a=1.8\times 10^{-4}$) and 500 mL of 0.2 M sodium formate ($HCOONa$). a) Calculate the pH of the buffer. b) Calculate the pH after adding 10 mL of 1.0 M HCl to the buffer, assume volume is additive.

Worked Solution

a) Moles of $HCOOH=0.5\text{ L}\times 0.4\text{ M}=0.2\text{ mol}$. Moles of $HCO{O}^{-}=0.5\text{ L}\times 0.2\text{ M}=0.1\text{ mol}$. $p{K}_a=-\log (1.8\times 10^{-4})\approx 3.74$. $$pH=3.74+\log \left(\frac{0.1}{0.2}\right)=3.74-0.30=3.44$$ b) Moles of $H^{+}$ added = $0.01\text{ L}\times 1.0\text{ M}=0.01\text{ mol}$. $H^{+}$ reacts with $HCO{O}^{-}$ to form $HCOOH$, so new moles of $HCO{O}^{-}=0.1-0.01=0.09\text{ mol}$, new moles of $HCOOH=0.2+0.01=0.21\text{ mol}$. $$pH=3.74+\log \left(\frac{0.09}{0.21}\right)\approx 3.74-0.37=3.37$$

Question 3

The $K_{sp}$ of lead(II) iodide ($Pb{I}_2$) is $9.8\times 10^{-9}$ at 25°C. a) Calculate the molar solubility of $Pb{I}_2$ in pure water. b) Will a precipitate form when 100 mL of $2.0\times 10^{-4}$ M $Pb(N{O}_3{)}_2$ is mixed with 100 mL of $1.0\times 10^{-3}$ M $NaI$? Justify your answer.

Worked Solution

a) Dissociation equation: $Pb{I}_2(s)\rightleftharpoons P{b}^{2+}(aq)+2I^{-}(aq)$. Let $s=$ molar solubility, so $[P{b}^{2+}]=s$, $[I^{-}]=2s$. $$K_{sp}=s(2s{)}^2=4s^3$$ $$s^3=\frac{9.8\times 10^{-9}}{4}=2.45\times 10^{-9}$$ $$s\approx 1.3\times 10^{-3}\text{ M}$$ b) After mixing, total volume = 200 mL, so concentrations are halved: $[P{b}^{2+}]=1.0\times 10^{-4}\text{ M}$, $[I^{-}]=5.0\times 10^{-4}\text{ M}$. $$Q_{sp}=[P{b}^{2+}][I^{-}{]}^2=1.0\times 10^{-4}\times (5.0\times 10^{-4}{)}^2=2.5\times 10^{-11}$$ $Q_{sp}=2.5\times 10^{-11}<K_{sp}=9.8\times 10^{-9}$, so no precipitate forms.

9. Quick Reference Cheatsheet

10. What's Next

Equilibrium is a foundational topic that connects to almost every later unit in the AP Chemistry syllabus. You will apply these core concepts to thermodynamics, where you will relate $K$ to Gibbs free energy using $\Delta G^{\circ }=-RT\ln K$, to electrochemistry, where you will calculate non-standard cell potentials using $Q$, and to advanced acid-base titration curve analysis. Mastering the rules for $Q$, $K$, and Le Chatelier's principle now will eliminate 30-40% of common errors in these higher-weight units, which make up a combined 25-35% of your AP exam score.

If you struggle with any of the calculations, rule applications, or practice questions in this guide, you can ask Ollie, our AI tutor, for personalized step-by-step explanations, extra practice problems, or targeted review sessions aligned to your weak areas. You can also find more AP Chemistry topic guides and full-length practice exams on the homepage to build your test readiness in the months leading up to your exam.