AP Chemistry Quiz: Pre Equilibrium Approximation

What this quiz covers

This quiz focuses on Pre Equilibrium Approximation, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Rate $\propto [\mathrm{Br_2}][\mathrm{Fe^{2+}}]^2$ (correct answer)
2. Rate $\propto [\mathrm{Br_2}][\mathrm{Fe^{2+}}]$
3. Rate $\propto [\mathrm{FeBr_2^{2+}}][\mathrm{Fe^{2+}}]$
4. Rate $\propto [\mathrm{Fe^{3+}}]^2$
5. Rate $\propto [\mathrm{Br^-}]^2$

Explanation: The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between $\mathrm{Br_2}$, $\mathrm{Fe^{2+}}$, 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.

Question 2

1. Rate $\propto [\mathrm{FeSCN^{2+}}][\mathrm{H_2O}]$
2. Rate $\propto [\mathrm{Fe^{3+}}][\mathrm{SCN^-}][\mathrm{H_2O}]$ (correct answer)
3. Rate $\propto [\mathrm{Fe^{3+}}][\mathrm{SCN^-}]$
4. Rate $\propto [\mathrm{Fe^{2+}}]$
5. Rate $\propto [\mathrm{SCN^-}]^2$

Explanation: The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between $\mathrm{Fe^{3+}}$, $\mathrm{SCN^-}$, and the intermediate $\mathrm{FeSCN^{2+}}$, with the equilibrium constant providing a relationship $[\mathrm{FeSCN^{2+}}] = K [\mathrm{Fe^{3+}}][\mathrm{SCN^-}$. 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 $[\mathrm{FeSCN^{2+}}][\mathrm{H_2O}]$ = k K $[\mathrm{Fe^{3+}}][\mathrm{SCN^-}][\mathrm{H_2O}$. 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

Question 11

1. Rate $\propto [\mathrm{NO}]^2[\mathrm{O_2}]$ (correct answer)
2. Rate $\propto [\mathrm{N_2O_2}][\mathrm{O_2}]$
3. Rate $\propto [\mathrm{NO}][\mathrm{O_2}]$
4. Rate $\propto [\mathrm{NO_2}]^2$
5. Rate $\propto [\mathrm{O_2}]^2$

Explanation: 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 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 12

1. Rate $\propto [\mathrm{H_2Cl^+}][\mathrm{Zn}]$
2. Rate $\propto [\mathrm{HCl}]^2[\mathrm{Zn}]/[\mathrm{Cl^-}]$ (correct answer)
3. Rate $\propto [\mathrm{HCl}][\mathrm{Zn}]$
4. Rate $\propto [\mathrm{H_2}]$
5. Rate $\propto [\mathrm{Cl^-}][\mathrm{Zn}]$

Explanation: 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.

Question 13

1. Rate $\propto [\mathrm{Cl_2O_4}][\mathrm{OH^-}]$
2. Rate $\propto [\mathrm{ClO_2}]^2[\mathrm{OH^-}]$ (correct answer)
3. Rate $\propto [\mathrm{ClO_2}][\mathrm{OH^-}]$
4. Rate $\propto [\mathrm{ClO_3^-}]$
5. Rate $\propto [\mathrm{OH^-}]^2$

Explanation: The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between 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.

Question 14

1. Rate $\propto [\mathrm{PQ}][\mathrm{R}]$
2. Rate $\propto [\mathrm{P}][\mathrm{Q}][\mathrm{R}]$ (correct answer)
3. Rate $\propto [\mathrm{P}][\mathrm{R}]$
4. Rate $\propto [\mathrm{Q}][\mathrm{R}]$
5. Rate $\propto [\mathrm{PR}]$

Explanation: 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.

Question 15

1. Rate $\propto [\mathrm{SO_2}][\mathrm{O_2}]$
2. Rate $\propto [\mathrm{SO_4}][\mathrm{SO_2}]$
3. Rate $\propto [\mathrm{SO_2}]^2[\mathrm{O_2}]$ (correct answer)
4. Rate $\propto [\mathrm{SO_3}]^2$
5. Rate $\propto [\mathrm{O_2}]^2$

Explanation: 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.

Question 16

1. Rate $\propto [\mathrm{HOI}][\mathrm{I^-}][\mathrm{H^+}]$
2. Rate $\propto [\mathrm{H_2O_2}][\mathrm{I^-}]^2[\mathrm{H^+}]/[\mathrm{OH^-}]$ (correct answer)
3. Rate $\propto [\mathrm{H_2O_2}][\mathrm{I^-}]$
4. Rate $\propto [\mathrm{I_2}]$
5. Rate $\propto [\mathrm{H_2O_2}][\mathrm{H^+}]$

Explanation: 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.

Question 17

1. Rate $\propto [\mathrm{HA}][\mathrm{B^-}]$
2. Rate $\propto [\mathrm{A^-}][\mathrm{B^-}]$
3. Rate $\propto [\mathrm{H^+}][\mathrm{A^-}][\mathrm{B^-}]$ (correct answer)
4. Rate $\propto [\mathrm{AB^-}]$
5. Rate $\propto [\mathrm{H^+}]^2[\mathrm{B^-}]$

Explanation: 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.

Question 18

1. Rate $\propto [\mathrm{NO_2}]^2[\mathrm{F_2}]/[\mathrm{NO_2F}]$ (correct answer)
2. Rate $\propto [\mathrm{F}][\mathrm{NO_2}]$
3. Rate $\propto [\mathrm{NO_2F}]$
4. Rate $\propto [\mathrm{NO_2}][\mathrm{F_2}]$
5. Rate $\propto [\mathrm{F_2}]^2$

Explanation: The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between NO2, F2, NO2F, and F, with the equilibrium constant providing a relationship [F] = K [NO2][F2] / [NO2F]. 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 [F][NO2] = k K [NO2]^2 [F2] / [NO2F]. 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 19

1. Rate $\propto [\mathrm{X_2Y}]$
2. Rate $\propto [\mathrm{X}][\mathrm{Y}]$
3. Rate $\propto [\mathrm{X}]^2[\mathrm{Y}]$ (correct answer)
4. Rate $\propto [\mathrm{XY}][\mathrm{X}]$
5. Rate $\propto [\mathrm{Y}]^2$

Explanation: 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.

Question 20

1. Rate $\propto [\mathrm{A}][\mathrm{C}]$
2. Rate $\propto [\mathrm{A}][\mathrm{B}][\mathrm{C}]$ (correct answer)
3. Rate $\propto [\mathrm{AB}][\mathrm{C}]$
4. Rate $\propto [\mathrm{B}][\mathrm{C}]$
5. Rate $\propto [\mathrm{AC}]$

Explanation: The skill being tested is the pre-equilibrium approximation. In this mechanism, the fast reversible step establishes an equilibrium between A, B, and the intermediate AB, with the equilibrium constant providing a relationship [AB] = K [A][B]. 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 [AB][C] = k K [A][B][C]. 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.