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AP Chemistry Unit 5 practice questions

Try 10 multiple-choice questions from Unit 5 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

    For the gas-phase reaction 2N2O5(g)→4NO2(g)+O2(g)2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g), the rate of the reaction can be expressed in terms of the disappearance of reactants or the appearance of products. Which of the following correctly describes the relationship between the rate of appearance of NO2NO_2 and the rate of disappearance of N2O5N_2O_5?

    It is twice the rate of disappearance of N2O5N_2O_5

    It is equal to the rate of disappearance of N2O5N_2O_5

    It is half the rate of disappearance of N2O5N_2O_5

    It is four times the rate of disappearance of N2O5N_2O_5

    Show answer

    Answer: (A)

    Connecting stoichiometry to reaction rates requires using the coefficients from the balanced equation. Since 4 moles of NO2NO_2 form for every 2 moles of N2O5N_2O_5 consumed, the rate of NO2NO_2 appearance is 42\frac{4}{2} or 2 times the rate of N2O5N_2O_5 disappearance.

  2. Question 2

    The decomposition reaction 2A(g)→B(g)2A(g) \rightarrow B(g) is determined to be second order with respect to reactant AA. If the initial concentration of AA is tripled while holding temperature constant, which of the following describes the change in the initial reaction rate?

    The rate increases by a factor of 9

    The rate increases by a factor of 3

    The rate increases by a factor of 6

    The rate increases by a factor of 27

    Show answer

    Answer: (A)

    Analyzing the power relationship in the rate law Rate=k[A]2Rate = k[A]^2 reveals the magnitude of the change. Tripling the concentration (3[A]3[A]) results in a rate proportional to (3)2(3)^2, causing the rate to increase by a factor of 9.

  3. Question 3

    For the elementary gas-phase reaction A(g)+B(g)→C(g)A(g) + B(g) \rightarrow C(g), the rate law is Rate=k[A][B]Rate = k[A][B]. If the volume of the reaction container is halved at constant temperature, how does the reaction rate change?

    The rate quadruples because the concentration of each gas doubles

    The rate doubles because the partial pressure of each gas doubles

    The rate remains constant because the rate constant kk is unchanged

    The rate halves because the volume reduction limits particle motion

    Show answer

    Answer: (A)

    Relating volume to concentration requires recognizing that C=n/VC = n/V, so halving volume doubles the concentration of both gases. Substituting these doubled concentrations into the rate law (Rate=k(2[A])(2[B])Rate = k(2[A])(2[B]) results in a fourfold increase (2×2=42 \times 2 = 4).

  4. Question 4

    The combustion of ammonia proceeds according to the balanced equation 4NH3(g)+5O2(g)→4NO(g)+6H2O(g)4NH_3(g) + 5O_2(g) \rightarrow 4NO(g) + 6H_2O(g). Which of the following correctly describes the relationship between the rate of consumption of oxygen and the rate of production of water?

    The rate of O2O_2 consumption is 5/65/6 times the rate of H2OH_2O production

    The rate of O2O_2 consumption is 6/56/5 times the rate of H2OH_2O production

    The rate of O2O_2 consumption is 4/54/5 times the rate of H2OH_2O production

    The rate of O2O_2 consumption is 5/45/4 times the rate of H2OH_2O production

    Show answer

    Answer: (A)

    Using the stoichiometry of the balanced equation, 55 moles of O2O_2 are consumed for every 66 moles of H2OH_2O produced. Therefore, the rate of O2O_2 change is scaled by the ratio of their coefficients (5/65/6) relative to the rate of H2OH_2O change.

  5. Question 5

    A decomposition reaction of compound XYXY is found to be zero order with respect to XYXY. If the initial concentration of XYXY is doubled, which of the following best describes the effect on the initial reaction rate?

    The initial rate remains the same because it is independent of [XY][XY]

    The initial rate doubles because it is directly proportional to [XY][XY]

    The initial rate quadruples because it depends on the square of [XY][XY]

    The initial rate halves because the reaction time increases with [XY][XY]

    Show answer

    Answer: (A)

    Applying the definition of a zero-order rate law (Rate=k[XY]0Rate = k[XY]^0) reveals that the rate depends only on the rate constant kk. Therefore, changing the concentration of the reactant has no mathematical effect on the reaction rate.

  6. Question 6

    For a reaction with the rate law Rate=k[X][Y]2Rate = k[X][Y]^2, an experiment is performed where the concentration of XX is doubled and the concentration of YY is halved simultaneously. How does the new initial reaction rate compare to the original rate?

    The rate becomes half the original rate

    The rate becomes double the original rate

    The rate remains equal to the original rate

    The rate becomes four times the original rate

    Show answer

    Answer: (A)

    Substituting the changes into the rate law reveals the net effect: Ratenew=k(2[X])(0.5[Y])2Rate_{new} = k(2[X])(0.5[Y])^2. The factors become 2×(0.5)2=2×0.25=0.52 \times (0.5)^2 = 2 \times 0.25 = 0.5. Thus, the new rate is 0.50.5 times the original rate.

  7. Question 7

    A solid cube of reactant is cut into eight smaller, identical cubes. Assuming the reaction rate is directly proportional to the surface area of the solid reactant, which of the following describes the change in the initial reaction rate?

    The rate doubles because the total surface area doubles

    The rate quadruples because the total surface area quadruples

    The rate remains constant because the total mass is constant

    The rate increases by a factor of 8 due to particle number

    Show answer

    Answer: (A)

    Analyzing the geometry shows that cutting a cube of side LL into 8 cubes of side L/2L/2 increases the total surface area from 6L26L^2 to 8×6(L/2)2=12L28 \times 6(L/2)^2 = 12L^2. Since rate is proportional to surface area, the rate doubles.

  8. Question 8

    A graph of concentration versus time for the decomposition of N2O5N_2O_5 shows a curve. A tangent line drawn at t=0t=0 passes through the points (0 s,0.50 M)(0 \text{ s}, 0.50 \text{ M}) and (50 s,0.30 M)(50 \text{ s}, 0.30 \text{ M}). Based on this line, what is the initial rate of disappearance of N2O5N_2O_5?

    0.0040 M/s0.0040 \text{ M/s}

    0.0060 M/s0.0060 \text{ M/s}

    0.0100 M/s0.0100 \text{ M/s}

    0.2000 M/s0.2000 \text{ M/s}

    Show answer

    Answer: (A)

    Calculating the slope of the tangent line provides the instantaneous rate. The slope is 0.30−0.5050−0=−0.0040 M/s\frac{0.30 - 0.50}{50 - 0} = -0.0040 \text{ M/s}, and the rate of disappearance is the magnitude of this slope.

  9. Question 9

    For a first-order reaction, a plot of ln⁡[A]\ln[\text{A}] versus time is a straight line. The line passes through the points (0 s,−0.50)(0 \text{ s}, -0.50) and (100 s,−2.50)(100 \text{ s}, -2.50). What is the value of the rate constant kk for this reaction?

    0.020 s−10.020 \text{ s}^{-1}

    0.025 s−10.025 \text{ s}^{-1}

    0.050 s−10.050 \text{ s}^{-1}

    2.000 s−12.000 \text{ s}^{-1}

    Show answer

    Answer: (A)

    Identifying the slope of the linearized plot yields the rate constant. For a first-order reaction, the slope of ln⁡[A]\ln[\text{A}] vs. time equals −k-k. The slope is −2.50−(−0.50)100=−0.020\frac{-2.50 - (-0.50)}{100} = -0.020, so k=0.020 s−1k = 0.020 \text{ s}^{-1}.

  10. Question 10

    Molecular kinetic energy distributions were measured at two temperatures, T1T_1 and T2T_2. The distribution at T2T_2 is broader, and its maximum occurs at a higher kinetic energy than the maximum at T1T_1. Which conclusion is supported by these observations?

    T2>T1T_2 > T_1, causing a higher reaction rate

    T2<T1T_2 < T_1, causing a lower reaction rate

    T2>T1T_2 > T_1, causing a lower reaction rate

    T2<T1T_2 < T_1, causing a higher reaction rate

    Show answer

    Answer: (A)

    Interpreting the Maxwell-Boltzmann distribution shape reveals the temperature relationship. A flatter curve shifted to higher energies indicates a higher temperature (T2>T1T_2 > T_1), which increases the fraction of molecules with energy exceeding EaE_a, thus increasing the rate.

What's on the AP Chemistry Unit 5 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 3072 multiple-choice questions and 56 FRQs written for Unit 5, Kinetics.

Practice AP Chemistry Unit 5 FRQs

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

Review Unit 5 before you start

Other AP Chemistry unit tests