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AP Chemistry Unit 9 practice questions

Try 10 multiple-choice questions from Unit 9 here. Pick an answer to see why it's right or wrong, then take the timed test above for a full AP-style set.

  1. Question 1

    A researcher plots the standard Gibbs free energy change (ΔG°) on the y-axis versus temperature (T) on the x-axis for a chemical reaction. The resulting line has a negative slope and a positive y-intercept. Based on the linear relationship ΔG° = ΔH° − TΔS°, which statement correctly identifies the signs of the enthalpy and entropy changes indicated by the graph?

    The enthalpy change is positive and the entropy change is positive.

    The enthalpy change is positive and the entropy change is negative.

    The enthalpy change is negative and the entropy change is positive.

    The enthalpy change is negative and the entropy change is negative.

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    Answer: (A)

    Connecting the linear equation y = mx + b to the thermodynamic function shows that the y-intercept corresponds to ΔH° and the slope corresponds to −ΔS°. A positive intercept indicates ΔH° > 0, and a negative slope indicates −ΔS° < 0, meaning ΔS° must be positive.

  2. Question 2

    A heating curve plots temperature versus heat added for a pure substance. The horizontal segment representing vaporization is significantly longer than the horizontal segment representing melting. Which conclusion about the standard entropy of vaporization (ΔS°vap) compared to the standard entropy of fusion (ΔS°fus) is best supported by this graphical difference?

    ΔS°vap > ΔS°fus because the gas phase has greater matter dispersal than the liquid.

    ΔS°vap < ΔS°fus because the higher temperature reduces the calculated entropy change.

    ΔS°vap = ΔS°fus because the mass of the substance remains constant during heating.

    ΔS°vap ≈ 0 because the temperature remains constant throughout the phase transition.

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    Answer: (A)

    Interpreting the graph shows that vaporization requires more energy (longer segment) than fusion. Since ΔS = q/T, the large heat input combined with the massive increase in disorder (dispersal of matter) when forming a gas results in ΔS°vap being significantly larger than ΔS°fus.

  3. Question 3

    A graph of ln K versus 1/T for the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) yields a straight line with a negative y-intercept. Based on the relationship ln K = (−ΔH°/R)(1/T) + (ΔS°/R), what does the y-intercept indicate about the entropy change of the system?

    The entropy change is negative, consistent with the decrease in moles of gas.

    The entropy change is negative, consistent with the increase in moles of gas.

    The entropy change is positive, consistent with the decrease in moles of gas.

    The entropy change is positive, consistent with the increase in moles of gas.

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    Answer: (A)

    Identifying the y-intercept as ΔS°/R allows for the determination of the sign of ΔS°. A negative intercept means ΔS° is negative, which aligns with the reaction stoichiometry showing a reduction from 3 moles of gas reactants to 2 moles of gas products.

  4. Question 4

    Consider the phase changes of water: fusion (H2O(s)→H2O(l)H_2O(s) \rightarrow H_2O(l)) and vaporization (H2O(l)→H2O(g)H_2O(l) \rightarrow H_2O(g)). Based on the dispersal of matter, which statement correctly compares the magnitude of the standard entropy change (ΔS∘\Delta S^\circ) for these processes?

    ΔSvap∘\Delta S^\circ_{vap} is larger than ΔSfus∘\Delta S^\circ_{fus} because the volume increase from liquid to gas is much greater.

    ΔSfus∘\Delta S^\circ_{fus} is larger than ΔSvap∘\Delta S^\circ_{vap} because the structural disruption from solid to liquid is greater.

    ΔSvap∘\Delta S^\circ_{vap} is smaller than ΔSfus∘\Delta S^\circ_{fus} because the intermolecular forces in the liquid state are stronger.

    ΔSfus∘\Delta S^\circ_{fus} is equal to ΔSvap∘\Delta S^\circ_{vap} because both processes involve breaking intermolecular forces.

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    Answer: (A)

    Comparing the relative dispersal of matter in phase changes reveals that the transition from liquid to gas involves a massive increase in volume and particle freedom compared to solid to liquid.

  5. Question 5

    The decomposition of solid calcium carbonate is represented by the equation CaCO3(s)→CaO(s)+CO2(g)CaCO_3(s) \rightarrow CaO(s) + CO_2(g). Based on the states of matter involved, which statement best predicts the sign of the standard entropy change (ΔS∘\Delta S^\circ) for this reaction?

    Positive, because the production of a gas from a solid significantly increases matter dispersal.

    Negative, because the formation of a solid oxide creates a highly ordered crystal lattice structure.

    Positive, because the breaking of ionic bonds in the reactant requires a large input of energy.

    Negative, because the reactant is a complex solid structure that decomposes into simpler substances.

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    Answer: (A)

    Analyzing the states of matter shows that generating a gaseous product from a solid reactant drastically increases the number of accessible microstates and disorder.

  6. Question 6

    When solid ammonium nitrate (NH4NO3NH_4NO_3) dissolves in water, the particles dissociate into ions. Which explanation best describes the primary driving force for the increase in entropy during this process?

    Entropy increases because the crystal lattice breaks down, dispersing ions throughout the solvent.

    Entropy decreases because water molecules form ordered hydration shells around the dissolved ions.

    Entropy increases because the endothermic dissolution process absorbs heat from the surroundings.

    Entropy decreases because the ions lose their vibrational energy upon entering the liquid phase.

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    Answer: (A)

    Connecting the dissolution process to particle dispersal reveals that breaking the ordered solid lattice to allow ions to move freely in the solvent volume drives the entropy increase.

  7. Question 7

    A rigid container is divided into two equal bulbs by a valve. One bulb contains 1.0 mol of He(g)He(g) at 298 K, and the other is evacuated. When the valve is opened, the gas expands to fill both bulbs. Which statement best explains the change in entropy?

    Entropy increases because the gas occupies a larger volume, increasing the number of accessible microstates.

    Entropy remains constant because the temperature of the gas is unchanged during the expansion into a vacuum.

    Entropy decreases because the pressure of the gas drops significantly as it fills the total volume.

    Entropy increases because the gas molecules gain kinetic energy while moving into the empty bulb.

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    Answer: (A)

    Applying statistical probability principles indicates that expanding into a larger volume increases the number of possible positions (microstates) for the particles.

  8. Question 8

    Aqueous solutions of lead(II) nitrate and potassium iodide are mixed, forming a precipitate of lead(II) iodide according to the equation Pb2+(aq)+2I−(aq)→PbI2(s)Pb^{2+}(aq) + 2I^-(aq) \rightarrow PbI_2(s). Which statement correctly predicts the sign of ΔS∘\Delta S^\circ for this reaction?

    Negative, because mobile ions in solution combine to form a rigid, highly ordered solid structure.

    Positive, because mixing two aqueous solutions increases the total volume available to the ions.

    Negative, because the molar mass of the solid product is greater than that of the individual ions.

    Positive, because the reaction is exothermic and releases energy that increases solvent disorder.

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    Answer: (A)

    Predicting the entropy change involves recognizing that transitioning from free-moving aqueous ions to a fixed solid lattice significantly reduces particle freedom.

  9. Question 9

    The synthesis of methanol is represented by the reaction CO(g)+2H2(g)→CH3OH(g)CO(g) + 2H_2(g) \rightarrow CH_3OH(g). Which of the following best explains the sign of ΔS∘\Delta S^\circ for this reaction?

    Negative, because three moles of gaseous reactants combine to form one mole of gaseous product.

    Positive, because the product molecule is more complex and has more internal vibrational modes.

    Negative, because the formation of stronger covalent bonds releases energy to the surroundings.

    Positive, because the product has a higher molar mass than either of the individual reactants.

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    Answer: (A)

    Counting gas moles reveals a decrease from 3 to 1, which corresponds to a significant reduction in the number of particles and thus a negative entropy change.

  10. Question 10

    Which of the following gas-phase reactions results in the largest increase in the standard entropy (ΔS∘\Delta S^\circ) of the system?

    The decomposition of 2H2O(g)2H_2O(g) into 2H2(g)2H_2(g) and O2(g)O_2(g)

    The synthesis of 2NH3(g)2NH_3(g) from N2(g)N_2(g) and 3H2(g)3H_2(g)

    The oxidation of 2NO(g)2NO(g) by O2(g)O_2(g) to form 2NO2(g)2NO_2(g)

    The combination of H2(g)H_2(g) and Cl2(g)Cl_2(g) to form 2HCl(g)2HCl(g)

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    Answer: (A)

    Analyzing the stoichiometry shows this reaction produces a net increase of +1 mole of gas (2 mol →\to 3 mol), resulting in the most positive entropy change.

What's on the AP Chemistry Unit 9 practice test

The standard version takes about 60 minutes and follows the timing and directions of the AP exam, sized to one unit:

  • 20 Multiple Choice Questions (30 minutes)
  • 1 FRQ – FRQ 1 – Long Answer (15 minutes)
  • 1 FRQ – FRQ 4 – Short Answer (15 minutes)

Questions are drawn from 4101 multiple-choice questions and 114 FRQs written for Unit 9, Thermodynamics and Electrochemistry.

Practice AP Chemistry Unit 9 FRQs

Answer the unit's free-response questions one at a time and get a score on every part.

Review Unit 9 before you start

Other AP Chemistry unit tests