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AP Chemistry Help: Pre Equilibrium Approximation

Review real example questions for Pre Equilibrium Approximation in AP Chemistry.

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A reaction is proposed to proceed by the following mechanism. The first step is fast and reversible and is described as establishing a pre-equilibrium prior to the slow step.

Step 1 (fast, reversible; pre-equilibrium): Br2+Fe2+⇌FeBr22+\mathrm{Br_2 + Fe^{2+} \rightleftharpoons FeBr_2^{2+}} Step 2 (slow): FeBr22++Fe2+→2Fe3++2Br−\mathrm{FeBr_2^{2+} + Fe^{2+} \rightarrow 2Fe^{3+} + 2Br^-}

Under these pre-equilibrium conditions, which qualitative rate law is most consistent with the mechanism?

All questions

Question 1

A reaction is proposed to proceed by the following mechanism. The first step is fast and reversible and is described as establishing a pre-equilibrium prior to the slow step.

Step 1 (fast, reversible; pre-equilibrium): Br2+Fe2+⇌FeBr22+\mathrm{Br_2 + Fe^{2+} \rightleftharpoons FeBr_2^{2+}} Step 2 (slow): FeBr22++Fe2+→2Fe3++2Br−\mathrm{FeBr_2^{2+} + Fe^{2+} \rightarrow 2Fe^{3+} + 2Br^-}

Under these pre-equilibrium conditions, which qualitative rate law is most consistent with the mechanism?

  1. Rate ∝[Br2][Fe2+]2\propto [\mathrm{Br_2}][\mathrm{Fe^{2+}}]^2 (correct answer)
  2. Rate ∝[Br2][Fe2+]\propto [\mathrm{Br_2}][\mathrm{Fe^{2+}}]
  3. Rate ∝[FeBr22+][Fe2+]\propto [\mathrm{FeBr_2^{2+}}][\mathrm{Fe^{2+}}]
  4. Rate ∝[Fe3+]2\propto [\mathrm{Fe^{3+}}]^2
  5. Rate ∝[Br−]2\propto [\mathrm{Br^-}]^2

Explanation: The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between Br2\mathrm{Br_2}, Fe2+\mathrm{Fe^{2+}}, and the intermediate FeBr22+\mathrm{FeBr_2^{2+}}, with the equilibrium constant providing a relationship [FeBr22+]=K[Br2][Fe2+][\mathrm{FeBr_2^{2+}}] = K [\mathrm{Br_2}][\mathrm{Fe^{2+}}]. This relationship allows us to express the concentration of the intermediate in terms of the reactants. The slow step then controls the overall rate, giving rate = k [FeBr22+][Fe2+][\mathrm{FeBr_2^{2+}}][\mathrm{Fe^{2+}}] = k K [Br2][Fe2+]2[\mathrm{Br_2}][\mathrm{Fe^{2+}}]^2. A tempting distractor is choice C, which treated the fast step as rate-determining. When an early step is fast and reversible, use it to relate concentrations before the slow step.

Question 2

A mechanism is proposed in which an early step is fast and reversible and is stated to establish a pre-equilibrium before the slow step.

Step 1 (fast, reversible; pre-equilibrium): Fe3++SCN−⇌FeSCN2+\mathrm{Fe^{3+} + SCN^- \rightleftharpoons FeSCN^{2+}} Step 2 (slow): FeSCN2++H2O→Fe2++HSCN+OH−\mathrm{FeSCN^{2+} + H_2O \rightarrow Fe^{2+} + HSCN + OH^-}

Which qualitative rate law is most consistent with the mechanism under pre-equilibrium conditions?

  1. Rate ∝[FeSCN2+][H2O]\propto [\mathrm{FeSCN^{2+}}][\mathrm{H_2O}]
  2. Rate ∝[Fe3+][SCN−][H2O]\propto [\mathrm{Fe^{3+}}][\mathrm{SCN^-}][\mathrm{H_2O}] (correct answer)
  3. Rate ∝[Fe3+][SCN−]\propto [\mathrm{Fe^{3+}}][\mathrm{SCN^-}]
  4. Rate ∝[Fe2+]\propto [\mathrm{Fe^{2+}}]
  5. Rate ∝[SCN−]2\propto [\mathrm{SCN^-}]^2

Explanation: The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between Fe3+\mathrm{Fe^{3+}}, SCN−\mathrm{SCN^-}, and the intermediate FeSCN2+\mathrm{FeSCN^{2+}}, with the equilibrium constant providing a relationship [FeSCN2+]=K[Fe3+][SCN−[\mathrm{FeSCN^{2+}}] = K [\mathrm{Fe^{3+}}][\mathrm{SCN^-}. This relationship allows us to express the concentration of the intermediate in terms of the reactants. The slow step then controls the overall rate, giving rate = k [FeSCN2+][H2O][\mathrm{FeSCN^{2+}}][\mathrm{H_2O}] = k K [Fe3+][SCN−][H2O[\mathrm{Fe^{3+}}][\mathrm{SCN^-}][\mathrm{H_2O}. A tempting distractor is choice C, which treated the fast step as rate-determining. When an early step is fast and reversible, use it to relate concentrations before the slow step.

Question 3

A two-step mechanism is proposed. The first step is fast and reversible and is stated to establish a pre-equilibrium before the slow step occurs.

Step 1 (fast, reversible; pre-equilibrium): O3+NO⇌NO2+O2\mathrm{O_3 + NO \rightleftharpoons NO_2 + O_2} Step 2 (slow): NO2+O3→NO3+O2\mathrm{NO_2 + O_3 \rightarrow NO_3 + O_2}

Which qualitative rate law is most consistent with this mechanism under the pre-equilibrium assumption?

  1. Rate ∝[NO2][O3]\propto [\mathrm{NO_2}][\mathrm{O_3}]
  2. Rate ∝[NO][O3]\propto [\mathrm{NO}][\mathrm{O_3}]
  3. Rate ∝[NO][O3]2/[O2]\propto [\mathrm{NO}][\mathrm{O_3}]^2/[\mathrm{O_2}] (correct answer)
  4. Rate ∝[NO3]\propto [\mathrm{NO_3}]
  5. Rate ∝[O3]2\propto [\mathrm{O_3}]^2

Explanation: The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between O3, NO, NO2, and O2, with the equilibrium constant providing a relationship [NO2] = K [NO][O3] / [O2]. This relationship allows us to express the concentration of the intermediate in terms of the reactants. The slow step then controls the overall rate, giving rate = k [NO2][O3] = k K [NO][O3]^2 / [O2]. A tempting distractor is choice A, which treated the fast step as rate-determining. When an early step is fast and reversible, use it to relate concentrations before the slow step.

Question 4

A mechanism is proposed in which the first step is fast and reversible and is stated to reach a pre-equilibrium before the slow step.

Step 1 (fast, reversible; pre-equilibrium): CO+Cl2⇌COCl2\mathrm{CO + Cl_2 \rightleftharpoons COCl_2} Step 2 (slow): COCl2+H2O→CO2+2HCl\mathrm{COCl_2 + H_2O \rightarrow CO_2 + 2HCl}

Which qualitative rate law is most consistent with this mechanism under pre-equilibrium conditions?

  1. Rate ∝[COCl2][H2O]\propto [\mathrm{COCl_2}][\mathrm{H_2O}]
  2. Rate ∝[CO][Cl2][H2O]\propto [\mathrm{CO}][\mathrm{Cl_2}][\mathrm{H_2O}] (correct answer)
  3. Rate ∝[CO][Cl2]\propto [\mathrm{CO}][\mathrm{Cl_2}]
  4. Rate ∝[Cl2]2[H2O]\propto [\mathrm{Cl_2}]^2[\mathrm{H_2O}]
  5. Rate ∝[HCl]2\propto [\mathrm{HCl}]^2

Explanation: The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between CO, Cl2, and the intermediate COCl2, with the equilibrium constant providing a relationship [COCl2] = K [CO][Cl2]. This relationship allows us to express the concentration of the intermediate in terms of the reactants. The slow step then controls the overall rate, giving rate = k [COCl2][H2O] = k K [CO][Cl2][H2O]. A tempting distractor is choice C, which treated the fast step as rate-determining. When an early step is fast and reversible, use it to relate concentrations before the slow step.

Question 5

A reaction is proposed to proceed by a mechanism in which the first step is fast and reversible and is described as reaching a pre-equilibrium before the slow step.

Step 1 (fast, reversible; pre-equilibrium): CH3Br+OH−⇌CH3OH⋯Br−\mathrm{CH_3Br + OH^- \rightleftharpoons CH_3OH\cdots Br^-} Step 2 (slow): CH3OH⋯Br−→CH3OH+Br−\mathrm{CH_3OH\cdots Br^- \rightarrow CH_3OH + Br^-}

Which qualitative rate law is most consistent with the mechanism under the pre-equilibrium assumption?

  1. Rate ∝[CH3Br][OH−]\propto [\mathrm{CH_3Br}][\mathrm{OH^-}] (correct answer)
  2. Rate ∝[CH3OH⋯Br−]\propto [\mathrm{CH_3OH\cdots Br^-}]
  3. Rate ∝[Br−]\propto [\mathrm{Br^-}]
  4. Rate ∝[CH3Br]\propto [\mathrm{CH_3Br}]
  5. Rate ∝[OH−]2\propto [\mathrm{OH^-}]^2

Explanation: The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between CH3Br, OH^-, and the intermediate CH3OH···Br^-, with the equilibrium constant providing a relationship [CH3OH···Br^-] = K [CH3Br][OH^-]. This relationship allows us to express the concentration of the intermediate in terms of the reactants. The slow step then controls the overall rate, giving rate = k [CH3OH···Br^-] = k K [CH3Br][OH^-]. A tempting distractor is choice D, which treated the fast step as rate-determining. When an early step is fast and reversible, use it to relate concentrations before the slow step.

Question 6

A reaction is proposed to occur by the mechanism below. The first step is fast and reversible and is stated to establish a pre-equilibrium before the slow step.

Step 1 (fast, reversible; pre-equilibrium): ClO−+H+⇌HOCl\mathrm{ClO^- + H^+ \rightleftharpoons HOCl} Step 2 (slow): HOCl+I−→HOI+Cl−\mathrm{HOCl + I^- \rightarrow HOI + Cl^-}

Which qualitative rate law is most consistent with this mechanism under the pre-equilibrium assumption?

  1. Rate ∝[HOCl][I−]\propto [\mathrm{HOCl}][\mathrm{I^-}]
  2. Rate ∝[ClO−][H+][I−]\propto [\mathrm{ClO^-}][\mathrm{H^+}][\mathrm{I^-}] (correct answer)
  3. Rate ∝[ClO−][I−]\propto [\mathrm{ClO^-}][\mathrm{I^-}]
  4. Rate ∝[H+]2[I−]\propto [\mathrm{H^+}]^2[\mathrm{I^-}]
  5. Rate ∝[Cl−]\propto [\mathrm{Cl^-}]

Explanation: The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between ClO^-, H^+, and the intermediate HOCl, with the equilibrium constant providing a relationship [HOCl] = K [ClO^-][H^+]. This relationship allows us to express the concentration of the intermediate in terms of the reactants. The slow step then controls the overall rate, giving rate = k [HOCl][I^-] = k K [ClO^-][H^+][I^-]. A tempting distractor is choice C, which treated the fast step as rate-determining. When an early step is fast and reversible, use it to relate concentrations before the slow step.

Question 7

A mechanism is proposed where the first step is fast and reversible and is stated to reach a pre-equilibrium before the slow step.

Step 1 (fast, reversible; pre-equilibrium): 2A⇌A2\mathrm{2A \rightleftharpoons A_2} Step 2 (slow): A2+B→AB+A\mathrm{A_2 + B \rightarrow AB + A}

Which qualitative rate law is most consistent with this mechanism under pre-equilibrium conditions?

  1. Rate ∝[A2][B]\propto [\mathrm{A_2}][\mathrm{B}]
  2. Rate ∝[A]2[B]\propto [\mathrm{A}]^2[\mathrm{B}] (correct answer)
  3. Rate ∝[AB]\propto [\mathrm{AB}]
  4. Rate ∝[A][B]\propto [\mathrm{A}][\mathrm{B}]
  5. Rate ∝[B]2\propto [\mathrm{B}]^2

Explanation: The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between A and the intermediate A2, with the equilibrium constant providing a relationship [A2] = K [A]^2. This relationship allows us to express the concentration of the intermediate in terms of the reactants. The slow step then controls the overall rate, giving rate = k [A2][B] = k K [A]^2 [B]. A tempting distractor is choice C, which treated the fast step as rate-determining. When an early step is fast and reversible, use it to relate concentrations before the slow step.

Question 8

A reaction is proposed to occur by the following mechanism. The first step is fast and reversible and is stated to establish a pre-equilibrium before the slow step.

Step 1 (fast, reversible; pre-equilibrium): 2NO2⇌N2O4\mathrm{2NO_2 \rightleftharpoons N_2O_4} Step 2 (slow): N2O4+CO→NO+NO3+CO\mathrm{N_2O_4 + CO \rightarrow NO + NO_3 + CO}

Which qualitative rate law is most consistent with this mechanism under pre-equilibrium conditions?

  1. Rate ∝[N2O4][CO]\propto [\mathrm{N_2O_4}][\mathrm{CO}]
  2. Rate ∝[NO2]2[CO]\propto [\mathrm{NO_2}]^2[\mathrm{CO}] (correct answer)
  3. Rate ∝[NO2][CO]\propto [\mathrm{NO_2}][\mathrm{CO}]
  4. Rate ∝[NO]\propto [\mathrm{NO}]
  5. Rate ∝[CO]2\propto [\mathrm{CO}]^2

Explanation: The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between NO2 and the intermediate N2O4, with the equilibrium constant providing a relationship [N2O4] = K [NO2]^2. This relationship allows us to express the concentration of the intermediate in terms of the reactants. The slow step then controls the overall rate, giving rate = k [N2O4][CO] = k K [NO2]^2 [CO]. A tempting distractor is choice C, which treated the fast step as rate-determining. When an early step is fast and reversible, use it to relate concentrations before the slow step.

Question 9

A proposed mechanism for the reaction overall is shown below. The first step is explicitly stated to be fast and reversible, and it establishes a pre-equilibrium before the slow step occurs.

Step 1 (fast, reversible; pre-equilibrium): NO+Cl2⇌NOCl2\mathrm{NO + Cl_2 \rightleftharpoons NOCl_2} Step 2 (slow): NOCl2+NO→2NOCl\mathrm{NOCl_2 + NO \rightarrow 2NOCl}

Which qualitative rate law form is most consistent with this mechanism under the pre-equilibrium assumption?

  1. Rate ∝[NO]2[Cl2]\propto [\mathrm{NO}]^2[\mathrm{Cl_2}] (correct answer)
  2. Rate ∝[NO][Cl2]\propto [\mathrm{NO}][\mathrm{Cl_2}]
  3. Rate ∝[NOCl2][NO]\propto [\mathrm{NOCl_2}][\mathrm{NO}]
  4. Rate ∝[NOCl]2\propto [\mathrm{NOCl}]^2
  5. Rate ∝[Cl2]2\propto [\mathrm{Cl_2}]^2

Explanation: The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between NO, Cl2, and the intermediate NOCl2, with the equilibrium constant providing a relationship [NOCl2] = K [NO][Cl2]. This relationship allows us to express the concentration of the intermediate in terms of the reactants. The slow step then controls the overall rate, giving rate = k [NOCl2][NO] = k K [NO]^2 [Cl2]. A tempting distractor is choice B, which treated the fast step as rate-determining. When an early step is fast and reversible, use it to relate concentrations before the slow step.

Question 10

A reaction is proposed to proceed via the mechanism below. The first step is fast and reversible and is stated to establish a pre-equilibrium before the slow step.

Step 1 (fast, reversible; pre-equilibrium): C+D⇌CD\mathrm{C + D \rightleftharpoons CD} Step 2 (slow): CD+D→CD2\mathrm{CD + D \rightarrow CD_2}

Which qualitative rate law is most consistent with this mechanism under pre-equilibrium conditions?

  1. Rate ∝[CD][D]\propto [\mathrm{CD}][\mathrm{D}]
  2. Rate ∝[C][D]2\propto [\mathrm{C}][\mathrm{D}]^2 (correct answer)
  3. Rate ∝[C][D]\propto [\mathrm{C}][\mathrm{D}]
  4. Rate ∝[CD2]\propto [\mathrm{CD_2}]
  5. Rate ∝[D]\propto [\mathrm{D}]

Explanation: The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between C, D, and the intermediate CD, with the equilibrium constant providing a relationship [CD] = K [C][D]. This relationship allows us to express the concentration of the intermediate in terms of the reactants. The slow step then controls the overall rate, giving rate = k [CD][D] = k K [C][D]^2. A tempting distractor is choice C, which treated the fast step as rate-determining. When an early step is fast and reversible, use it to relate concentrations before the slow step.