/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 100 In a hydrocarbon solution, the g... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 100

In a hydrocarbon solution, the gold compound \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\) decomposes into ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) and a dif- ferent gold compound, \(\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3}\). The following mechanism has been proposed for the decomposition of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}:\) Step 1: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} \underset{k_{-1}}{\stackrel{k_{1}}{\rightleftharpoons}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}+\mathrm{PH}_{3} \quad \text { (fast) }}\) Step 2: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au} \stackrel{\mathrm{k}_{2}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}+\left(\mathrm{CH}_{3}\right) \mathrm{Au} \quad\) (slow) Step 3: \(\left(\mathrm{CH}_{3}\right) \mathrm{Au}+\mathrm{PH}_{3} \stackrel{k_{2}}{\longrightarrow}\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3} \quad\) (fast) (a) What is the overall reaction? (b) What are the intermediates in the mechanism? (c) What is the molecularity of each of the elementary steps? (d) What is the ratedetermining step? (e) What is the rate law predicted by this mechanism? (f) What would be the effect on the reaction rate of adding \(\mathrm{PH}_{3}\) to the solution of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} ?\)

Short Answer

Expert verified
The overall reaction is \( (CH_3)_3AuPH_3 \rightarrow C_2H_6 + (CH_3)AuPH_3 \). The intermediates are \( (CH_3)_3Au \) and \( (CH_3)Au \). The molecularity of each step is: Step 1 - Bimolecular, Step 2 - Unimolecular, and Step 3 - Bimolecular. The rate-determining step is Step 2. The rate law predicted by this mechanism is Rate = \( k_2 \frac{K_1[(CH_3)_3AuPH_3]}{[PH_3]} \). Adding PH鈧� to the solution of \( (CH_3)_3AuPH_3 \) will decrease the reaction rate.

Step by step solution

01

(a) Overall Reaction

To find the overall reaction, we need to sum up all the elementary steps (Step 1, Step 2, and Step3) and cancel out any intermediate species that appear on both sides of the reaction. We are given the following steps: 1. \( (CH_3)_3AuPH_3 \rightleftharpoons (CH_3)_3Au + PH_3 \) 2. \( (CH_3)_3Au \rightarrow C_2H_6 + (CH_3)Au \) 3. \( (CH_3)Au + PH_3 \rightarrow (CH_3)AuPH_3 \) Summing up all the steps: \( (CH_3)_3AuPH_3 \rightarrow C_2H_6 + (CH_3)AuPH_3 \) So, the overall reaction is: \( (CH_3)_3AuPH_3 \rightarrow C_2H_6 + (CH_3)AuPH_3 \)
02

(b) Intermediates

Intermediates are species that are formed and consumed during the course of the reaction. They are present in the reaction mechanism but not in the overall reaction. From the given reaction steps, the intermediates are: 1. \( (CH_3)_3Au \) 2. \( (CH_3)Au \)
03

(c) Molecularity of each step

Molecularity is the number of reacting species (molecules, atoms, or ions) involved in an elementary step. We determine the molecularity of each step by counting the number of reactants in each step. Step 1: Bimolecular (as there are 2 reactants - \( (CH_3)_3AuPH_3 \) and \( PH_3 \)) Step 2: Unimolecular (as there is 1 reactant - \( (CH_3)_3Au \)) Step 3: Bimolecular (as there are 2 reactants - \( (CH_3)Au \) and \( PH_3 \))
04

(d) Rate-determining step

In a reaction mechanism, the rate-determining step is the slowest step. In this case, the slow step is Step 2, involving the decomposition of \( (CH_3)_3Au \) to form ethane and a different gold compound.
05

(e) Rate law

To find the rate law, we must consider the rate-determining step of the reaction mechanism (Step 2). Step 2 involves the decomposition of \( (CH_3)_3Au \), and its rate can be represented as: Rate = \( k_2[(CH_3)_3Au] \) However, we need the rate in terms of the initial reactant, \( (CH_3)_3AuPH_3 \). To do this, we will use the equilibrium expression from Step 1: \( K_1 = \frac{[(CH_3)_3Au][PH_3]}{[(CH_3)_3AuPH_3]} \) Solving for \( (CH_3)_3Au \), we get: \( (CH_3)_3Au = \frac{K_1[(CH_3)_3AuPH_3]}{[PH_3]} \) Substituting this into the rate law, we obtain: Rate = \( k_2 \frac{K_1[(CH_3)_3AuPH_3]}{[PH_3]} \)
06

(f) Effect of adding PH鈧�

According to the derived rate law, the rate of reaction depends on the concentration of \( PH_3 \) in the denominator. Increasing the concentration of \( PH_3 \) would decrease the reaction rate, as the ratio of \( \frac{(CH_3)_3AuPH_3}{PH_3} \) would decrease. Thus, adding \( PH_3 \) to the solution of \( (CH_3)_3AuPH_3 \) will decrease the reaction rate.

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Key Concepts

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

Elementary Steps
In chemical reactions, an elementary step is a single reaction in a series that makes up a reaction mechanism. It involves the direct conversion of reactants to products in a single catalyst-free step. Each elementary step is characterized by its molecularity, which indicates the number of reactant molecules involved.
  • Step 1: In the given exercise, the first step is bimolecular because it involves two reactant molecules, \((CH_3)_3 AuPH_3\) and \(PH_3\).

  • Step 2: This step is unimolecular, involving just one reactant, \((CH_3)_3 Au\). It is the slowest step and hence the rate-determining step.

  • Step 3: The last step is again bimolecular, involving \((CH_3)Au\) and \(PH_3\).
Understanding these steps will help in predicting how a reaction will proceed under different conditions.
Rate Law
The rate law of a reaction provides an equation that relates the rate of reaction to the concentration of reactants. It is often derived from the rate-determining step, which is the slowest elementary step in a mechanism. In our case, Step 2 is the slowest.
The rate law for this reaction can be originally represented as: \[\text{Rate} = k_2[(CH_3)_3Au]\] However, since we need the rate law expressed with the initial reactant, \((CH_3)_3AuPH_3\), we use the equilibrium expression from Step 1. This allows us to substitute into the rate law, giving:
\[\text{Rate} = k_2 \frac{K_1[(CH_3)_3AuPH_3]}{[PH_3]}\] Here, \(K_1\) is the equilibrium constant from Step 1. This expression shows how reactant concentrations influence the reaction rate.
Reaction Intermediates
In a reaction mechanism, intermediates are species that appear in the mechanism's steps but not in the overall reaction. These are neither reactants nor products but form and consume during the progression of the reaction.
  • For the given mechanism, the intermediates are:
  • \((CH_3)_3Au\): This species is formed in Step 1 and consumed in Step 2.

  • \((CH_3)Au\): Formed in Step 2 and consumed in Step 3.
Intermediates are crucial as they often lead to the rate-determining step and thus control the kinetics of the entire reaction.
Rate-Determining Step
A rate-determining step is the slowest step in a reaction mechanism and governs the rate of the overall chemical reaction. For our reaction, this is Step 2, where \((CH_3)_3Au\) decomposes to \(C_2H_6\) and \((CH_3)Au\). This step has the largest activation energy barrier, making it slower than the other steps.
Identifying the rate-determining step is key, as it allows chemists to focus on optimizing or modifying this part of the reaction to influence the overall rate. In practice, knowing this step helps in developing strategies to accelerate reactions for better efficiency.

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Most popular questions from this chapter

(a) What is meant by the term molecularity? (b) Why are termolecular elementary reactions so rare? (c) What is an intermediate in a mechanism?

The reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) is second order in \(\mathrm{NO}\) and first order in \(\mathrm{O}_{2}\). When \([\mathrm{NO}]=0.040 \mathrm{M}\) and \(\left[\mathrm{O}_{2}\right]=0.035 \mathrm{M}\), the observed rate of disappearance of \(\mathrm{NO}\) is \(9.3 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the rate of disappearance of \(\mathrm{O}_{2}\) at this moment? (b) What is the value of the rate constant? (c) What are the units of the rate constant? (d) What would happen to the rate if the concentration of NO were increased by a factor of \(1.8\) ?

The following is a quote from an article in the August 18,1998 , issue of The New York Times about the breakdown of cellulose and starch: "A drop of 18 degrees Fahrenheit [from \(77^{\circ} \mathrm{F}\) to \(\left.59{ }^{\circ} \mathrm{F}\right]\) lowers the reaction rate six times; a 36-degree drop [from \(77^{\circ} \mathrm{F}\) to \(\left.41{ }^{\circ} \mathrm{F}\right]\) produces a fortyfold decrease in the rate." (a) Calculate activation energies for the breakdown process based on the two estimates of the effect of temperature on rate. Are the values consistent? (b) Assuming the value of \(E_{a}\) calculated from the 36 -degree drop and that the rate of breakdown is first order with a half-life at \(25^{\circ} \mathrm{C}\) of \(2.7\) years, calculate the half-life for breakdown at a temperature of \(-15^{\circ} \mathrm{C}\).

You perform a series of experiments for the reaction \(\mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C}\) and find that the rate law has the form rate \(=k[\mathrm{~A}]^{x}\). Determine the value of \(x\) in each of the following cases: (a) There is no rate change when [A] is tripled. (b) The rate increases by a factor of 9 when [A] is tripled. (c) When [A] is doubled, the rate increases by a factor of 8 . [Section 14.3]

Consider the following reaction between mercury(II) chloride and oxalate ion: \(2 \mathrm{HgCl}_{2}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}(a q)\) \(2 \mathrm{Cl}^{-}(a q)+2 \mathrm{CO}_{2}(g)+\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)\) The initial rate of this reaction was determined for several concentrations of \(\mathrm{HgCl}_{2}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\), and the rate constant in units of \(\mathrm{s}^{-1}\). (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to \(0.100\) ? $$ \begin{array}{llll} \text { Experiment } & {\left[\mathrm{HgCl}_{2} \mathrm{~J}(\mathrm{M})\right.} & {\left[\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right](M)} & \text { Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 1 & 0.164 & 0.15 & 3.2 \times 10^{-5} \\ 2 & 0.164 & 0.45 & 2.9 \times 10^{-4} \\ 3 & 0.082 & 0.45 & 1.4 \times 10^{-4} \\ 4 & 0.246 & 0.15 & 4.8 \times 10^{-5} \\ \hline \end{array} $$ (a) What is the rate law for this reaction? (b) What is the value of the rate constant? (c) What is the reaction rate when the concentration of \(\mathrm{HgCl}_{2}\) is \(0.100 \mathrm{M}\) and that of \(\left(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2}\right)\) is \(0.25 \mathrm{M}\), if the temperature is the same as that used to obtain the data shown?
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