Pre-Equilibrium Approximation
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AP Chemistry › Pre-Equilibrium Approximation
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?
Rate $\propto[\mathrm{N_2O_4}][\mathrm{CO}]$
Rate $\propto[\mathrm{NO_2}]^2[\mathrm{CO}]$
Rate $\propto[\mathrm{CO}]^2$
Rate $\propto[\mathrm{NO}]$
Rate $\propto[\mathrm{NO_2}][\mathrm{CO}]$
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.
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?
Rate $\propto[\mathrm{FeBr_2^{2+}}][\mathrm{Fe^{2+}}]$
Rate $\propto[\mathrm{Fe^{3+}}]^2$
Rate $\propto[\mathrm{Br_2}][\mathrm{Fe^{2+}}]$
Rate $\propto[\mathrm{Br^-}]^2$
Rate $\propto[\mathrm{Br_2}][\mathrm{Fe^{2+}}]^2$
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.
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): $\mathrm{FeSCN^{2+} + H_2O \rightarrow Fe^{2+} + HSCN + OH^-}$
Which qualitative rate law is most consistent with the mechanism under pre-equilibrium conditions?
Rate $\propto[\mathrm{SCN^-}]^2$
Rate $\propto[\mathrm{Fe^{3+}}][\mathrm{SCN^-}]$
Rate $\propto[\mathrm{Fe^{2+}}]$
Rate $\propto[\mathrm{Fe^{3+}}][\mathrm{SCN^-}][\mathrm{H_2O}]$
Rate $\propto[\mathrm{FeSCN^{2+}}][\mathrm{H_2O}]$
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.
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?
Rate $\propto[\mathrm{Cl_2}]^2$
Rate $\propto[\mathrm{NO}][\mathrm{Cl_2}]$
Rate $\propto[\mathrm{NOCl}]^2$
Rate $\propto[\mathrm{NO}]^2[\mathrm{Cl_2}]$
Rate $\propto[\mathrm{NOCl_2}][\mathrm{NO}]$
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.
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?
Rate $\propto[\mathrm{O_3}]^2$
Rate $\propto[\mathrm{NO}][\mathrm{O_3}]^2/[\mathrm{O_2}]$
Rate $\propto[\mathrm{NO}][\mathrm{O_3}]$
Rate $\propto[\mathrm{NO_2}][\mathrm{O_3}]$
Rate $\propto[\mathrm{NO_3}]$
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.
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?
Rate $\propto[\mathrm{CO}][\mathrm{Cl_2}]$
Rate $\propto[\mathrm{HCl}]^2$
Rate $\propto[\mathrm{Cl_2}]^2[\mathrm{H_2O}]$
Rate $\propto[\mathrm{COCl_2}][\mathrm{H_2O}]$
Rate $\propto[\mathrm{CO}][\mathrm{Cl_2}][\mathrm{H_2O}]$
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.
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?
Rate $\propto[\mathrm{CD}][\mathrm{D}]$
Rate $\propto[\mathrm{CD_2}]$
Rate $\propto[\mathrm{D}]$
Rate $\propto[\mathrm{C}][\mathrm{D}]^2$
Rate $\propto[\mathrm{C}][\mathrm{D}]$
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.
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?
Rate $\propto[\mathrm{CH_3OH\cdots Br^-}]$
Rate $\propto[\mathrm{CH_3Br}][\mathrm{OH^-}]$
Rate $\propto[\mathrm{OH^-}]^2$
Rate $\propto[\mathrm{CH_3Br}]$
Rate $\propto[\mathrm{Br^-}]$
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.
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?
Rate $\propto[\mathrm{NO}]^2[\mathrm{O_2}]$
Rate $\propto[\mathrm{N_2O_2}][\mathrm{O_2}]$
Rate $\propto[\mathrm{O_2}]^2$
Rate $\propto[\mathrm{NO}][\mathrm{O_2}]$
Rate $\propto[\mathrm{NO_2}]^2$
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.
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?
Rate $\propto[\mathrm{HOCl}][\mathrm{I^-}]$
Rate $\propto[\mathrm{ClO^-}][\mathrm{I^-}]$
Rate $\propto[\mathrm{Cl^-}]$
Rate $\propto[\mathrm{H^+}]^2[\mathrm{I^-}]$
Rate $\propto[\mathrm{ClO^-}][\mathrm{H^+}][\mathrm{I^-}]$
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.