Pre-Equilibrium Approximation - AP Chemistry

1. 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): $\mathrm{NH_3 + H^+ \rightleftharpoons NH_4^+}$ Step 2 (slow): $\mathrm{NH_4^+ + NO_2^- \rightarrow N_2 + 2H_2O}$ Which qualitative rate law is most consistent with this mechanism under pre-equilibrium conditions?
   Answer: Rate $\propto [\mathrm{NH_3}][\mathrm{H^+}][\mathrm{NO_2^-}]$ The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between NH3, $H^+$, and the intermediate $NH4^+$, with the equilibrium constant providing a relationship [$NH4^+$] = K [NH3][$H^+$]. 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 [$NH4^+$][$NO2^-$] = k K [NH3][$H^+$][$NO2^-$]. 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.
2. 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): $\mathrm{C + D \rightleftharpoons CD}$ Step 2 (slow): $\mathrm{CD + D \rightarrow CD_2}$ Which qualitative rate law is most consistent with this mechanism under pre-equilibrium conditions?
   Answer: Rate $\propto [\mathrm{C}][\mathrm{D}]^2$ 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.
3. A reaction is proposed to proceed via the mechanism below. The first step is fast and reversible and is described as establishing a pre-equilibrium before the slow step. Step 1 (fast, reversible; pre-equilibrium): $\mathrm{2ClO_2 \rightleftharpoons Cl_2O_4}$ Step 2 (slow): $\mathrm{Cl_2O_4 + OH^- \rightarrow ClO_3^- + ClO_2^- + H^+}$ Which qualitative rate law is most consistent with the mechanism under pre-equilibrium conditions?
   Answer: Rate $\propto [\mathrm{ClO_2}]^2[\mathrm{OH^-}]$ The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between ClO2 and the intermediate Cl2O4, with the equilibrium constant providing a relationship [Cl2O4] = K [ClO2]^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 [Cl2O4][$OH^-$] = k K [ClO2]^2 [$OH^-$]. 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.
4. A proposed mechanism includes a fast, reversible first step that is stated to establish a pre-equilibrium before the slow step. Step 1 (fast, reversible; pre-equilibrium): $\mathrm{2HCl \rightleftharpoons H_2Cl^+ + Cl^-}$ Step 2 (slow): $\mathrm{H_2Cl^+ + Zn \rightarrow ZnCl^+ + H_2}$ Which qualitative rate law is most consistent with this mechanism under pre-equilibrium conditions?
   Answer: Rate $\propto [\mathrm{HCl}]^2[\mathrm{Zn}]/[\mathrm{Cl^-}]$ This question tests the pre-equilibrium approximation. In the pre-equilibrium approximation, the fast reversible first step reaches equilibrium quickly, establishing a constant ratio of concentrations defined by the equilibrium constant K = [H₂Cl⁺][Cl⁻] / [HCl]². The slow second step then determines the overall rate, so rate = k₂ [H₂Cl⁺][Zn]. To express this in terms of measurable species, we solve for the intermediate [H₂Cl⁺] from the equilibrium expression, giving [H₂Cl⁺] = K [HCl]² / [Cl⁻], leading to rate ∝ [HCl]²[Zn]/[Cl⁻]. A tempting distractor is choice A, Rate ∝ [H₂Cl⁺][Zn], which is incorrect because it treats the intermediate as if its concentration were independent and measurable, failing to apply the pre-equilibrium substitution. When an early step is fast and reversible, use the equilibrium constant to relate the intermediate's concentration to the reactants before substituting into the rate-determining step's rate law.
5. The following mechanism is proposed. 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): $\mathrm{H_2O_2 + I^- \rightleftharpoons HOI + OH^-}$ Step 2 (slow): $\mathrm{HOI + I^- + H^+ \rightarrow I_2 + H_2O}$ Which qualitative rate law is most consistent with this mechanism under pre-equilibrium conditions?
   Answer: Rate $\propto [\mathrm{H_2O_2}][\mathrm{I^-}]^2[\mathrm{H^+}]/[\mathrm{OH^-}]$ The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between H2O2, $I^-$, HOI, and $OH^-$, with the equilibrium constant providing a relationship [HOI] = K [H2O2][$I^-$] / [$OH^-$]. 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 [HOI][$I^-$][$H^+$] = k K [H2O2][$I^-$]^2 [$H^+$] / [$OH^-$]. 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.
6. 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): $\mathrm{H^+ + A^- \rightleftharpoons HA}$ Step 2 (slow): $\mathrm{HA + B^- \rightarrow AB^- + H^+}$ Which qualitative rate law is most consistent with this mechanism under pre-equilibrium conditions?
   Answer: Rate $\propto [\mathrm{H^+}][\mathrm{A^-}][\mathrm{B^-}]$ The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between $H^+$, $A^-$, and the intermediate HA, with the equilibrium constant providing a relationship [HA] = K [$H^+$][$A^-$]. 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 [HA][$B^-$] = k K [$H^+$][$A^-$][$B^-$]. 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.
7. A 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): $\mathrm{E + F \rightleftharpoons EF}$ Step 2 (slow): $\mathrm{EF + G \rightarrow EG + F}$ Which qualitative rate law is most consistent with this mechanism under pre-equilibrium conditions?
   Answer: Rate $\propto [\mathrm{E}][\mathrm{F}][\mathrm{G}]$ The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between E, F, and the intermediate EF, with the equilibrium constant providing a relationship [EF] = K [E][F]. 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 [EF][G] = k K [E][F][G]. 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.
8. A reaction is proposed to proceed via 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): $\mathrm{X + Y \rightleftharpoons XY}$ Step 2 (slow): $\mathrm{XY + X \rightarrow X_2Y}$ Which qualitative rate law is most consistent with this mechanism under pre-equilibrium conditions?
   Answer: Rate $\propto [\mathrm{X}]^2[\mathrm{Y}]$ This question tests the pre-equilibrium approximation. In the pre-equilibrium approximation, the fast reversible first step reaches equilibrium quickly, establishing a constant ratio of concentrations defined by the equilibrium constant K = [XY] / [X][Y]. The slow second step then determines the overall rate, so rate = k₂ [XY][X]. To express this in terms of measurable species, we solve for the intermediate [XY] from the equilibrium expression, giving [XY] = K [X][Y], leading to rate ∝ [X]²[Y]. A tempting distractor is choice A, Rate ∝ [XY][X], which is incorrect because it treats the intermediate as if its concentration were independent and measurable, failing to apply the pre-equilibrium substitution. When an early step is fast and reversible, use the equilibrium constant to relate the intermediate's concentration to the reactants before substituting into the rate-determining step's rate law.
9. 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): $\mathrm{Br_2 + Fe^{2+} \rightleftharpoons FeBr_2^{2+}}$ Step 2 (slow): $\mathrm{FeBr_2^{2+} + Fe^{2+} \rightarrow 2Fe^{3+} + 2Br^-}$ Under these pre-equilibrium conditions, which qualitative rate law is most consistent with the mechanism?
   Answer: Rate $\propto [\mathrm{Br_2}][\mathrm{Fe^{2+}}]^2$ The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between $\mathrm{Br_2}$, $\mathrm{Fe^{2+}}$, and the intermediate $\mathrm{FeBr_2^{2+}}$, with the equilibrium constant providing a relationship $[\mathrm{FeBr_2^{2+}}] = K [\mathrm{Br_2}][\mathrm{Fe^{2+}}]$. 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 $[\mathrm{FeBr_2^{2+}}][\mathrm{Fe^{2+}}]$ = k K $[\mathrm{Br_2}][\mathrm{Fe^{2+}}]^2$. 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.
10. 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): $\mathrm{2NO_2 \rightleftharpoons N_2O_4}$ Step 2 (slow): $\mathrm{N_2O_4 + CO \rightarrow NO + NO_3 + CO}$ Which qualitative rate law is most consistent with this mechanism under pre-equilibrium conditions?
    Answer: Rate $\propto [\mathrm{NO_2}]^2[\mathrm{CO}]$ 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.
11. A mechanism is proposed in which an initial association step is fast and reversible and is stated to establish a pre-equilibrium before the slow step. Step 1 (fast, reversible; pre-equilibrium): $\mathrm{SO_2 + O_2 \rightleftharpoons SO_4}$ Step 2 (slow): $\mathrm{SO_4 + SO_2 \rightarrow 2SO_3}$ Which qualitative rate law is most consistent with this mechanism under pre-equilibrium conditions?
    Answer: Rate $\propto [\mathrm{SO_2}]^2[\mathrm{O_2}]$ The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between SO2, O2, and the intermediate SO4, with the equilibrium constant providing a relationship [SO4] = K [SO2][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 [SO4][SO2] = k K [SO2]^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.
12. A multistep mechanism is proposed. The first step is fast and reversible and is explicitly stated to establish a pre-equilibrium before the slow step. Step 1 (fast, reversible; pre-equilibrium): $\mathrm{2NO \rightleftharpoons N_2O_2}$ Step 2 (slow): $\mathrm{N_2O_2 + O_2 \rightarrow 2NO_2}$ Which qualitative rate law is most consistent with this mechanism under pre-equilibrium conditions?
    Answer: Rate $\propto [\mathrm{NO}]^2[\mathrm{O_2}]$ The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between NO and the intermediate N2O2, with the equilibrium constant providing a relationship [N2O2] = K [NO]^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 [N2O2][O2] = k K [NO]^2 [O2]. A tempting distractor is choice B, the slow step's rate law, which still contains the intermediate N₂O₂. Choice C misses the squared dependence on NO. When an early step is fast and reversible, use it to relate concentrations before the slow step.
13. 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): $\mathrm{P + Q \rightleftharpoons PQ}$ Step 2 (slow): $\mathrm{PQ + R \rightarrow PR + Q}$ Which qualitative rate law is most consistent with this mechanism under pre-equilibrium conditions?
    Answer: Rate $\propto [\mathrm{P}][\mathrm{Q}][\mathrm{R}]$ This question tests the pre-equilibrium approximation. In the pre-equilibrium approximation, the fast reversible first step reaches equilibrium quickly, establishing a constant ratio of concentrations defined by the equilibrium constant K = [PQ] / [P][Q]. The slow second step then determines the overall rate, so rate = k₂ [PQ][R]. To express this in terms of measurable species, we solve for the intermediate [PQ] from the equilibrium expression, giving [PQ] = K [P][Q], leading to rate ∝ [P][Q][R]. A tempting distractor is choice A, Rate ∝ [PQ][R], which is incorrect because it treats the intermediate as if its concentration were independent and measurable, failing to apply the pre-equilibrium substitution. When an early step is fast and reversible, use the equilibrium constant to relate the intermediate's concentration to the reactants before substituting into the rate-determining step's rate law.
14. 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): $\mathrm{CO + Cl_2 \rightleftharpoons COCl_2}$ Step 2 (slow): $\mathrm{COCl_2 + H_2O \rightarrow CO_2 + 2HCl}$ Which qualitative rate law is most consistent with this mechanism under pre-equilibrium conditions?
    Answer: Rate $\propto [\mathrm{CO}][\mathrm{Cl_2}][\mathrm{H_2O}]$ 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.
15. 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): $\mathrm{ClO^- + H^+ \rightleftharpoons HOCl}$ Step 2 (slow): $\mathrm{HOCl + I^- \rightarrow HOI + Cl^-}$ Which qualitative rate law is most consistent with this mechanism under the pre-equilibrium assumption?
    Answer: Rate $\propto [\mathrm{ClO^-}][\mathrm{H^+}][\mathrm{I^-}]$ 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.
16. 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): $\mathrm{NO + Cl_2 \rightleftharpoons NOCl_2}$ Step 2 (slow): $\mathrm{NOCl_2 + NO \rightarrow 2NOCl}$ Which qualitative rate law form is most consistent with this mechanism under the pre-equilibrium assumption?
    Answer: Rate $\propto [\mathrm{NO}]^2[\mathrm{Cl_2}]$ 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.
17. 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): $\mathrm{2A \rightleftharpoons A_2}$ Step 2 (slow): $\mathrm{A_2 + B \rightarrow AB + A}$ Which qualitative rate law is most consistent with this mechanism under pre-equilibrium conditions?
    Answer: Rate $\propto [\mathrm{A}]^2[\mathrm{B}]$ 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.
18. 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): $\mathrm{O_3 + NO \rightleftharpoons NO_2 + O_2}$ Step 2 (slow): $\mathrm{NO_2 + O_3 \rightarrow NO_3 + O_2}$ Which qualitative rate law is most consistent with this mechanism under the pre-equilibrium assumption?
    Answer: Rate $\propto [\mathrm{NO}][\mathrm{O_3}]^2/[\mathrm{O_2}]$ 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.
19. 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): $\mathrm{CH_3Br + OH^- \rightleftharpoons CH_3OH\cdots Br^-}$ Step 2 (slow): $\mathrm{CH_3OH\cdots Br^- \rightarrow CH_3OH + Br^-}$ Which qualitative rate law is most consistent with the mechanism under the pre-equilibrium assumption?
    Answer: Rate $\propto [\mathrm{CH_3Br}][\mathrm{OH^-}]$ 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.
20. 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): $\mathrm{Fe^{3+} + SCN^- \rightleftharpoons FeSCN^{2+}}$ Step 2 (slow): $\mathrm{FeSCN^{2+} + SCN^- \rightarrow Fe(SCN)_2^{+}}$ Which qualitative rate law is most consistent with the mechanism under pre-equilibrium conditions?
    Answer: Rate $\propto [\mathrm{Fe^{3+}}][\mathrm{SCN^-}]^2$ Step 2 is slow, so rate = k₂[FeSCN²⁺][SCN⁻]. FeSCN²⁺ is an intermediate. The fast pre-equilibrium in Step 1 gives [FeSCN²⁺] = K₁[Fe³⁺][SCN⁻]. Substituting gives rate = k₂K₁[Fe³⁺][SCN⁻]², which is choice B. Choice A still contains the intermediate. Choice C treats Step 1 as rate-determining and leaves out the second SCN⁻ from Step 2.

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