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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate-Determining Step
The rate-determining step in a chemical reaction mechanism is like the bottleneck of the process. Imagine a busy highway that suddenly narrows down to one lane; despite how fast the cars can move elsewhere, the flow of traffic is determined by the slowest point. Similarly, the rate-determining step is the slowest step in a reaction mechanism. It controls the overall pace of the reaction because further steps can't proceed faster than molecules arrive from this stage. In the given exercise, the reaction between C and B to form D is identified as the slow step.
  • This means the rate of the entire reaction depends primarily on this step.
  • Thus, the concentration of reactants involved in this step are pivotal in determining the reaction rate.
Each reaction has its own unique slow step, and spotting it is crucial to understanding how fast the reaction will occur and writing the correct rate law.
Reaction Mechanism
Understanding a reaction mechanism is like putting together the pieces of a puzzle to see the entire picture of a chemical reaction. A reaction mechanism details each step that occurs as reactants transform into products. It's like the play-by-play account of what's happening on a microscopic level when substances react.
  • Reactions often happen through multiple stages or "elementary steps." Each step contributes to the overall transformation.
  • The mechanism proposed in our example involves three steps: a fast equilibrium between A and B, the slow reaction between C and B, and a fast reaction that forms the product E.
A comprehensive understanding of the reaction mechanism helps chemists write accurate rate laws, predict reaction behavior, and even design more efficient reactions by influencing each step strategically.
Chemical Equilibrium
A chemical equilibrium is like a balanced seesaw, where two different reactions occur at equal rates. When the rate of the forward reaction equates to the rate of the reverse reaction, a steady state is reached. At this point, the concentration of reactants and products remain constant, even if the reactions keep happening.
  • In our exercise, the first step where 2A forms B is in fast equilibrium.
  • This means any B formed can revert back to reactants A or be used up in subsequent steps.
Understanding chemical equilibrium in reaction mechanisms is critical because it helps us determine the concentration of intermediates, like B, which are essential in calculating the overall rate law for the reaction. By analyzing these equilibria, we can derive expressions that explain how different substances are interrelated during the process.

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Most popular questions from this chapter

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ How many moles of electrons must be transferred to create 127 g of copper? (A) 1 mole of electrons (B) 2 moles of electrons (C) 3 moles of electrons (D) 4 moles of electrons

\(2 \mathrm{ClF}(g)+\mathrm{O}_{2}(g) \leftrightarrow \mathrm{Cl}_{2} \mathrm{O}(g)+\mathrm{F}_{2} \mathrm{O}(g) \Delta H=167 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) During the reaction above, the product yield can be increased by increasing the temperature of the reaction. Why is this effective? (A) The reaction is endothermic; therefore adding heat will shift it to the right. (B) Increasing the temperature increases the speed of the molecules, meaning there will be more collisions between them. (C) The reactants are less massive than the products, and an increase in temperature will cause their kinetic energy to increase more than that of the products. (D) The increase in temperature allows for a higher percentage of molecular collisions to occur with the proper orientation to create the product.

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ What will happen in the salt bridge as the reaction progresses? (A) The Na' ions will flow to the Cu/Cu' half-cell. (B) The Br' ions will flow to the Cu/Cu' half-cell. (C) Electrons will transfer from the Cu/Cu' half-cell to the Zn/Zn" half cell. (D) Electrons will transfer from the \(\mathrm{Zn} / \mathrm{Zn}^{2+}\) half- cell to the Cu/Cu' half cell.

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. \(\mathrm{CH}_{3} \mathrm{NH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \leftrightarrow \mathrm{OH}^{-}(a q)+\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}(a q)\) The above equation represents the reaction between the base methylamine \(\left(K_{\mathrm{b}}=4.38 \times 10^{-4}\right)\) and water. Which of the following best represents the.concentrations of the various species at equilibrium? (A) \(\left[\mathrm{OH}^{-}\right]>\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]\) (B) \(\left[\mathrm{OH}^{-}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]\) (C) \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]>\left[\mathrm{OH}^{-}\right]>\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]\) (D) \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]>\left[\mathrm{OH}^{-}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{3}+\right]\)

When calcium chloride \(\left(\mathrm{CaCl}_{2}\right)\) dissolves in water, the temperature of the water increases dramatically. During this reaction, heat transfers from (A) the reactants to the products (B) the reactants to the system (C) the system to the surroundings (D) the products to the surroundings
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