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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.

    Show answer

    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

    A student constructs a Nernst plot (EcellE_{cell} versus log⁡Q\log Q) for a reaction between metal M and metal ion Zn+Z^{n+}. The slope of the line is determined to be −0.010-0.010 V at 298 K. Given that the theoretical slope is −0.0592n\frac{-0.0592}{n} V, what is the balanced net ionic equation represented by this data?

    2M(s)+3Z2+(aq)→2M3+(aq)+3Z(s)2M(s) + 3Z^{2+}(aq) \rightarrow 2M^{3+}(aq) + 3Z(s)

    M(s)+Z3+(aq)→M3+(aq)+Z(s)M(s) + Z^{3+}(aq) \rightarrow M^{3+}(aq) + Z(s)

    M(s)+Z2+(aq)→M2+(aq)+Z(s)M(s) + Z^{2+}(aq) \rightarrow M^{2+}(aq) + Z(s)

    M(s)+Z+(aq)→M+(aq)+Z(s)M(s) + Z^{+}(aq) \rightarrow M^{+}(aq) + Z(s)

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    Answer: (A)

    Solving for nn using the slope relationship (−0.0592/n=−0.010-0.0592/n = -0.010) yields n≈6n \approx 6, indicating a total transfer of 6 electrons, which corresponds to the balanced equation involving two trivalent and three divalent ions.

  5. Question 5

    A graph of current (A) versus time (s) for the electrolysis of a molten chloride salt XClnXCl_n is a rectangle with a height of 10.0 A and width of 965 s. If 0.90 g of metal X is deposited, what is the balanced reduction half-reaction? (F = 96,500 C/mol)

    Al3++3e−→Al(s)Al^{3+} + 3e^- \rightarrow Al(s)

    Mg2++2e−→Mg(s)Mg^{2+} + 2e^- \rightarrow Mg(s)

    Na++e−→Na(s)Na^{+} + e^- \rightarrow Na(s)

    Ca2++2e−→Ca(s)Ca^{2+} + 2e^- \rightarrow Ca(s)

    Show answer

    Answer: (A)

    Calculating total charge (9650 C) and moles of electrons (0.10 mol) allows determination of the equivalent mass (9 g/mol e-), which identifies Aluminum (27 g/mol) as the metal requiring 3 electrons per atom.

  6. Question 6

    In a galvanic cell, a graph of the mass of the anode (Metal A) lost versus the mass of the cathode (Metal B) gained is linear with a slope of 0.30. The atomic mass of A is 65.4 g/mol and B is 107.9 g/mol. What is the balanced cell reaction?

    A(s)+2B+(aq)→A2+(aq)+2B(s)A(s) + 2B^+(aq) \rightarrow A^{2+}(aq) + 2B(s)

    2A(s)+B2+(aq)→2A+(aq)+B(s)2A(s) + B^{2+}(aq) \rightarrow 2A^+(aq) + B(s)

    A(s)+B2+(aq)→A2+(aq)+B(s)A(s) + B^{2+}(aq) \rightarrow A^{2+}(aq) + B(s)

    3A(s)+2B3+(aq)→3A2+(aq)+2B(s)3A(s) + 2B^{3+}(aq) \rightarrow 3A^{2+}(aq) + 2B(s)

    Show answer

    Answer: (A)

    Using the slope (mass ratio) and atomic masses reveals a mole ratio of 1:2 (1×65.42×107.9≈0.30\frac{1 \times 65.4}{2 \times 107.9} \approx 0.30), which corresponds to the stoichiometry of a divalent metal displacing a monovalent metal.

  7. Question 7

    A model of a galvanic cell anode shows cadmium atoms (CdCd) becoming cadmium ions (Cd2+Cd^{2+}). Which combination correctly describes the half-reaction components and the quantitative flow of electrons for every 10 atoms oxidized?

    20 electrons flow from the anode to the cathode

    10 electrons flow from the anode to the cathode

    20 electrons flow from the cathode to the anode

    5 electrons flow from the cathode to the anode

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    Answer: (A)

    Analyzing the oxidation state change shows that each Cd atom loses 2 electrons (Cd→Cd2++2e−Cd \rightarrow Cd^{2+} + 2e^-). Therefore, oxidizing 10 atoms releases 10×2=2010 \times 2 = 20 electrons, which flow from the anode.

  8. Question 8

    A reaction vessel initially contains a mixture of I2I_2 molecules and F−F^- ions. Based on standard reduction potentials (E∘E^\circ for F2F_2 is +2.87 V, E∘E^\circ for I2I_2 is +0.54 V), which statement describes the correct final particulate representation?

    The initial I2I_2 and F−F^- particles remain unchanged

    The initial F⁻ ions are oxidized to F₂ molecules while I₂ remains unchanged

    The initial I₂ molecules are reduced to I⁻ ions while F⁻ remains unchanged

    The initial I₂ and F⁻ particles react to form IF₅ molecules and electrons

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    Answer: (A)

    Comparing reduction potentials indicates that F2F_2 is a stronger oxidant than I2I_2, meaning F−F^- is a weaker reductant than I−I^-. Therefore, I2I_2 cannot oxidize F−F^- spontaneously, and no reaction occurs.

  9. Question 9

    A galvanic cell operates using the reaction Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s). The atomic mass of Zn is 65.38 g/mol and Cu is 63.55 g/mol. Which statement best explains the change in electrode mass based on particulate electron transfer?

    For every mole of reaction, 2 moles of electrons transfer, causing the anode to lose slightly more mass than the cathode gains.

    For every mole of reaction, 1 mole of electrons transfers, causing the anode to gain mass while the cathode loses mass.

    The transfer of electrons from Cu to Zn causes the zinc electrode to increase in mass due to electron accumulation.

    The stoichiometry is 1:1, so the mass lost by the zinc electrode must be exactly equal to the mass gained by the copper electrode.

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    Answer: (A)

    Connecting the 1:1 molar ratio (derived from the 2-electron transfer in both half-reactions) to atomic masses explains the mass difference. Since Zn (anode) is heavier than Cu (cathode), the mass loss exceeds the mass gain.

  10. Question 10

    Electrolysis of a molten metal chloride MClnMCl_n using a current of 5.00 A for 386 seconds deposits 0.010 mol of metal M at the cathode. (Faraday's constant = 96,500 C/mol e−e^-). Which half-reaction represents the reduction of M consistent with this data?

    M2++2e−→MM^{2+} + 2e^- \rightarrow M

    M++e−→MM^+ + e^- \rightarrow M

    M3++3e−→MM^{3+} + 3e^- \rightarrow M

    M4++4e−→MM^{4+} + 4e^- \rightarrow M

    Show answer

    Answer: (A)

    Calculating charge (Q=It=1930Q = It = 1930 C) and converting to moles of electrons (1930/96500=0.021930/96500 = 0.02 mol e−e^-) reveals a 2:1 ratio of electrons to metal moles (0.020.02 mol e−e^- / 0.010.01 mol M), indicating a +2 charge.

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 3268 multiple-choice questions and 100 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