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Chapter 1: Problem 7

The following mechanism is proposed for a reaction: \(\begin{array}{ll}{2 \mathrm{A} \rightarrow \mathrm{B}} & {\text { (fast equilibrium) }} \\ {\mathrm{C}+\mathrm{B} \rightarrow \mathrm{D}} & {\text { (slow) }} \\ {\mathrm{D}+\mathrm{A} \rightarrow \mathrm{E}} & {\text { (fast) }}\end{array}\) Which of the following is the correct rate law for compete reaction? (A) Rate \(=k[\mathrm{C}]^{2}[\mathrm{B}]\) (B) Rate \(=k\left[\mathrm{Cl}[\mathrm{A}]^{2}\right.\) (C) Rate \(=k[\mathrm{C}][\mathrm{A}]^{3}\) (D) Rate \(=k[\mathrm{D}][\mathrm{A}]\)

Short Answer

Expert verified
The correct rate law for this reaction mechanism is \(Rate = k[C][A]^2\). This is not included among the given answer options. Hence, none of the given answer choices is correct.

Step by step solution

01

Identify the slow step in the reaction mechanism

According to the given mechanism, the slowest reaction is the second step, \(C + B \rightarrow D\). This is the rate-determining step in the mechanism.
02

Write the rate law expression for the slow step

The slow step is \(C+B \rightarrow D\). From this, it can be seen that the rate depends on the concentration of reactants C and B. So the rate law for this step would be \(Rate = k[C][B]\).
03

Evaluating the rate law for B

Recall that B is formed and used up quickly in the first reaction which is in equilibrium. The concentration of B doesn't build up, as any B produced quickly reacts with C. The rate law for B is determined by the first reaction. The equation for the first reaction is \(2A \rightarrow B\). In assumptions of chemical equilibrium for fast reactions, the rate of forward reaction equals the rate of reverse reaction. This yields \([B]=k1[A]^2\).
04

Substitute the rate law for B into the rate law of the overall reaction

By substituting the concentration of B from the first reaction into the rate law expression we got from the slow step, we get \(Rate = k[C][A]^2\).

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Most popular questions from this chapter

Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: \(\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}\) The concentrations of the solutions in each half-cell are 1.0 M. Which of the following best describes the activity in the salt bridge as the reaction progresses? (A) Electrons flow through the salt bridge from the \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half-cell to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) half-cell. (B) \(\quad \mathrm{Pb}^{2+}\) flows to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) flows to the \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half-cell. (C) \(\mathrm{Na}^{+}\) flows to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) half-cell, and \(\mathrm{Cl}^{-}\) flows to the \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half- cell. (D) \(\quad \mathrm{Na}^{+}\) flows to the \(\mathrm{Pb} / \mathrm{Pb}^{2}\) half- cell, and \(\mathrm{Cl}^{-}\) flows to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) half- cell.

A strong acid/strong base titration is completed using an indicator which changes color at the exact equivalence point of the titration. The protonated form of the indicator is HIn, and the deprotonated form is In. At the equivalence point of the reaction: (A) \([\mathrm{HIn}]=\left[\mathrm{In}^{-}\right]\) (B) \([\mathrm{HIn}]=1 /\left[\mathrm{In}^{-}\right]\) (C) \([\mathrm{HIn}]=2\left[\mathrm{In}^{-}\right]\) (D) \([\mathrm{HIn}]=\left[\mathrm{In}^{-}\right]^{2}\)

A student titrates 20.0 \(\mathrm{mL}\) of 1.0 \(M \mathrm{NaOH}\) with 2.0 \(\mathrm{M}\), \(\mathrm{HCO}_{2} \mathrm{H}\left(K_{\mathrm{a}}=1.8 \times 10^{-4}\right) .\) Formic acid is a monoprotic acid. Which of the following would create a good buffer when dissolved in formic acid? (A) \(\mathrm{NaCO}_{2} \mathrm{H}\) (B) \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) (C) \(\mathrm{NH}_{3}\) (D) \(\mathrm{H}_{2} \mathrm{O}\)

150 \(\mathrm{mL}\) of saturated \(\mathrm{SrF}_{2}\) solution is present in a 250 \(\mathrm{mL}\) beaker at room temperature. The molar solubility of \(\mathrm{SrF}_{2}\) at 298 \(\mathrm{K}\) is \(1.0 \times 10^{-3} \mathrm{M}\) . How could the concentration of \(\mathrm{Sr}^{2+}\) ions in solution be decreased? (A) Adding some \(\operatorname{NaF}(s)\) to the beaker (B) Adding some \(\operatorname{Sr}\left(\mathrm{NO}_{3}\right)_{2}(s)\) to the beaker (C) By heating the solution in the beaker (D) By adding a small amount of water to the beaker, but not dissolving all the solid

A 0.1-molar solution of which of the following acids will be the best conductor of electricity? (A) \(\mathrm{H}_{2} \mathrm{CO}_{3}\) (B) \(\mathrm{H}_{2} \mathrm{S}\) (C) \(\mathrm{HF}\) (D) \(\mathrm{HNO}_{3}\)
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