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AP Chemistry covers atomic structure, bonding, thermodynamics, kinetics, equilibrium, electrochemistry, and organic chemistry fundamentals. The exam's free-response section rewards students who can explain their reasoning quantitatively and qualitatively. Consistent practice with calculation-heavy problems and written justifications is the key to a 4 or 5.

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Unlike generic question banks, every Crameleon exam reflects the specific framing, terminology, and difficulty level of your course. You'll practise answering questions that look and feel like the real thing — which is exactly what the research on retrieval practice says you should be doing. Students who test themselves with practice exams consistently outperform students who re-read notes, regardless of study time. The generation takes under 2 minutes and your first exam is completely free.

Topics typically tested

  • Atomic structure & periodic trends
  • Chemical bonding & molecular geometry
  • Stoichiometry & limiting reagents
  • Thermodynamics (ΔH, ΔS, ΔG)
  • Chemical kinetics & rate laws
  • Equilibrium (Ka, Kb, Ksp) & Le Chatelier's Principle

Sample AP Chemistry exam questions

Three original questions in the style of a real AP Chemistry exam, each with a worked solution. Your own practice exam can be built from your course's notes and syllabus.

Question 1Multiple choiceChemical kinetics & rate laws

The initial rate of the reaction A+B→C\mathrm{A} + \mathrm{B} \rightarrow \mathrm{C} was measured at 25 °C for three sets of starting concentrations:

Experiment[A][\mathrm{A}] (M)[B][\mathrm{B}] (M)Initial rate (M/s)
10.100.102.0×10−42.0\times10^{-4}
20.200.108.0×10−48.0\times10^{-4}
30.200.302.4×10−32.4\times10^{-3}

What is the rate law for the reaction, and what is the value of the rate constant kk?

  1. A.

    rate=k[A]2[B]\text{rate} = k[\mathrm{A}]^2[\mathrm{B}], with k=0.20 M−2 s−1k = 0.20\,\mathrm{M^{-2}\,s^{-1}}

  2. B.

    rate=k[A][B]\text{rate} = k[\mathrm{A}][\mathrm{B}], with k=0.020 M−1 s−1k = 0.020\,\mathrm{M^{-1}\,s^{-1}}

  3. C.

    rate=k[A]2[B]2\text{rate} = k[\mathrm{A}]^2[\mathrm{B}]^2, with k=2.0 M−3 s−1k = 2.0\,\mathrm{M^{-3}\,s^{-1}}

  4. D.

    rate=k[A][B]2\text{rate} = k[\mathrm{A}][\mathrm{B}]^2, with k=0.20 M−2 s−1k = 0.20\,\mathrm{M^{-2}\,s^{-1}}

Worked solution
Answer:A

Find each order by comparing two experiments in which only one concentration changes.

Order in A: experiments 1 and 2 keep [B][\mathrm{B}] at 0.10 M while [A][\mathrm{A}] doubles. The rate rises from 2.0×10−42.0\times10^{-4} to 8.0×10−48.0\times10^{-4} M/s, a factor of 4. Since 2m=42^{m} = 4, m=2m = 2: the reaction is second order in A.

Order in B: experiments 2 and 3 keep [A][\mathrm{A}] at 0.20 M while [B][\mathrm{B}] triples. The rate rises from 8.0×10−48.0\times10^{-4} to 2.4×10−32.4\times10^{-3} M/s, a factor of 3. Since 3n=33^{n} = 3, n=1n = 1: first order in B.

So the rate law is rate=k[A]2[B]\text{rate} = k[\mathrm{A}]^2[\mathrm{B}] and the overall order is 3. Solve for kk using experiment 1:

k=rate[A]2[B]=2.0×10−4 M/s(0.10 M)2(0.10 M)=2.0×10−41.0×10−3=0.20 M−2 s−1k = \frac{\text{rate}}{[\mathrm{A}]^2[\mathrm{B}]} = \frac{2.0\times10^{-4}\ \text{M/s}}{(0.10\ \text{M})^2(0.10\ \text{M})} = \frac{2.0\times10^{-4}}{1.0\times10^{-3}} = 0.20\ \mathrm{M^{-2}\,s^{-1}}

The units follow from an overall third-order reaction: M/s÷M3=M−2 s−1\text{M/s} \div \text{M}^3 = \mathrm{M^{-2}\,s^{-1}}. Confirm with experiment 3: 0.20×(0.20)2×0.30=2.4×10−30.20\times(0.20)^2\times0.30 = 2.4\times10^{-3} M/s, exactly as measured.

Rate =k[A]2[B]= k[\mathrm{A}]^2[\mathrm{B}] with k=0.20 M−2 s−1k = 0.20\,\mathrm{M^{-2}\,s^{-1}} (option A).

Option B is the trap of reading the orders off the coefficients in the balanced equation; orders must come from experimental data. Option C misreads the factor of 3 in experiment 3 as a squared effect (that would need a factor of 9), and option D swaps the two orders, which fits neither pair of experiments.

Question 2Short answerThermodynamics (ΔH, ΔS, ΔG)

For the reaction

2 SO2(g)+O2(g)→2 SO3(g)2\,\mathrm{SO_2}(g) + \mathrm{O_2}(g) \rightarrow 2\,\mathrm{SO_3}(g)

ΔH∘=−197.8 kJ mol−1\Delta H^\circ = -197.8\,\mathrm{kJ\,mol^{-1}} and ΔS∘=−187.9 J mol−1 K−1\Delta S^\circ = -187.9\,\mathrm{J\,mol^{-1}\,K^{-1}}, both per mole of reaction as written. (a) Calculate ΔG∘\Delta G^\circ at 298 K and state whether the reaction is thermodynamically favourable under standard conditions. (b) Assuming ΔH∘\Delta H^\circ and ΔS∘\Delta S^\circ do not change with temperature, estimate the temperature above which the reaction becomes unfavourable.

Worked solution
Answer:

(a) ΔG° = −141.8 kJ per mole of reaction; favourable (negative). (b) About 1053 K (roughly 780 °C)

(a) Use the Gibbs relation ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ. The two given quantities are in different units, so first convert the entropy change to kJ:

ΔS∘=−187.9 J mol−1 K−1=−0.1879 kJ mol−1 K−1\Delta S^\circ = -187.9\ \mathrm{J\,mol^{-1}\,K^{-1}} = -0.1879\ \mathrm{kJ\,mol^{-1}\,K^{-1}}

Substitute at 298 K, keeping the signs explicit because the double negative is where errors happen:

ΔG∘=−197.8 kJ/mol−(298 K)(−0.1879 kJ mol−1 K−1)=−197.8+56.0=−141.8 kJ/mol\Delta G^\circ = -197.8\ \text{kJ/mol} - (298\ \text{K})(-0.1879\ \mathrm{kJ\,mol^{-1}\,K^{-1}}) = -197.8 + 56.0 = -141.8\ \text{kJ/mol}

ΔG∘\Delta G^\circ is negative, so the reaction is thermodynamically favourable at 298 K under standard conditions. This makes sense: the reaction is strongly exothermic, and at room temperature the unfavourable entropy term TΔS∘T\Delta S^\circ is only about a quarter the size of ΔH∘\Delta H^\circ. (Favourable does not mean fast; without a catalyst this reaction is very slow at 298 K.)

(b) Because ΔH∘\Delta H^\circ and ΔS∘\Delta S^\circ are both negative, the reaction is enthalpy-driven and becomes unfavourable at high temperature, where TΔS∘T\Delta S^\circ outweighs ΔH∘\Delta H^\circ. The crossover is where ΔG∘=0\Delta G^\circ = 0:

T=ΔH∘ΔS∘=−197.8 kJ/mol−0.1879 kJ mol−1 K−1=1053 KT = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{-197.8\ \text{kJ/mol}}{-0.1879\ \mathrm{kJ\,mol^{-1}\,K^{-1}}} = 1053\ \text{K}

Above about 1053 K (roughly 780 °C) ΔG∘\Delta G^\circ becomes positive and the reverse reaction, decomposition of SO3\mathrm{SO_3}, is favoured.

(a) ΔG∘=−141.8\Delta G^\circ = -141.8 kJ per mole of reaction, favourable; (b) unfavourable above about 1053 K.

Question 3Worked problemEquilibrium (Ka, Kb, Ksp) & Le Chatelier's Principle

The solubility product of calcium fluoride, CaF2\mathrm{CaF_2}, is Ksp=3.9×10−11K_{sp} = 3.9\times10^{-11} at 25 °C.

(a) Write the solubility equilibrium and the KspK_{sp} expression, and calculate the molar solubility of CaF2\mathrm{CaF_2} in pure water.

(b) Calculate the molar solubility of CaF2\mathrm{CaF_2} in 0.10 M sodium fluoride solution. State any approximation you make and justify it.

(c) Use Le Chatelier's principle to explain the difference between (a) and (b).

Worked solution
Answer:

(a) s = 2.1 × 10⁻⁴ M; (b) s = 3.9 × 10⁻⁹ M; (c) the added F⁻ (common ion) shifts the dissolution equilibrium toward solid CaF₂, lowering the solubility about 55,000-fold

(a) Dissolution releases one calcium ion and two fluoride ions per formula unit:

CaF2(s)⇌Ca2+(aq)+2 F−(aq),Ksp=[Ca2+][F−]2\mathrm{CaF_2}(s) \rightleftharpoons \mathrm{Ca^{2+}}(aq) + 2\,\mathrm{F^-}(aq), \qquad K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{F^-}]^2

Let ss be the molar solubility. Then [Ca2+]=s[\mathrm{Ca^{2+}}] = s and [F−]=2s[\mathrm{F^-}] = 2s, so the fluoride concentration is both doubled and squared:

Ksp=(s)(2s)2=4s3⇒s=3.9×10−1143=9.75×10−123=2.1×10−4 MK_{sp} = (s)(2s)^2 = 4s^3 \quad\Rightarrow\quad s = \sqrt[3]{\frac{3.9\times10^{-11}}{4}} = \sqrt[3]{9.75\times10^{-12}} = 2.1\times10^{-4}\ \text{M}

(b) In 0.10 M NaF the fluoride ion is already present before any CaF2\mathrm{CaF_2} dissolves. An ICE table gives [Ca2+]=s′[\mathrm{Ca^{2+}}] = s' and [F−]=0.10+2s′[\mathrm{F^-}] = 0.10 + 2s'. Because the solubility in pure water was only 2×10−42\times10^{-4} M and the common ion will suppress it further, 2s′2s' is negligible next to 0.10, so [F−]≈0.10[\mathrm{F^-}] \approx 0.10 M:

Ksp=s′(0.10)2⇒s′=3.9×10−111.0×10−2=3.9×10−9 MK_{sp} = s'(0.10)^2 \quad\Rightarrow\quad s' = \frac{3.9\times10^{-11}}{1.0\times10^{-2}} = 3.9\times10^{-9}\ \text{M}

The approximation is justified: 2s′=7.8×10−92s' = 7.8\times10^{-9} M, which is less than one ten-millionth of 0.10 M.

(c) Dissolution is an equilibrium, and adding fluoride (the common ion) raises the concentration of a product. By Le Chatelier's principle the system responds by consuming the added ion, shifting toward solid CaF2\mathrm{CaF_2}. Less salt can dissolve before the product [Ca2+][F−]2[\mathrm{Ca^{2+}}][\mathrm{F^-}]^2 reaches KspK_{sp}. Because fluoride enters the expression squared, the effect is dramatic: the solubility drops by a factor of 2.1×10−4/3.9×10−9≈5.5×1042.1\times10^{-4} / 3.9\times10^{-9} \approx 5.5\times10^{4}.

(a) s=2.1×10−4s = 2.1\times10^{-4} M; (b) s=3.9×10−9s = 3.9\times10^{-9} M in 0.10 M NaF; (c) the common-ion effect shifts the equilibrium toward undissolved CaF2\mathrm{CaF_2}.

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