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Chapter 12: Problem 115

In the gas phase, the production of phosgene from chlorine and carbon monoxide is assumed to proceed by the following mechanism: $$ \mathrm{Cl}_{2} \stackrel{k_{1}}{\rightleftharpoons_{k_{1}}} 2 \mathrm{Cl} $$ $$ \mathrm{Cl}+\mathrm{CO} \stackrel{k_{2}}{\leftrightharpoons_{k-2}} \mathrm{COCl} $$ $$ \mathrm{COCl}+\mathrm{Cl}_{2} \stackrel{k_{3}}{\longrightarrow} \mathrm{COCl}_{2}+\mathrm{Cl} $$ $$ 2 \mathrm{Cl} \stackrel{k}{\longrightarrow} \mathrm{Cl}_{2} $$ Overall reaction: $\mathrm{CO}+\mathrm{Cl}_{2} \longrightarrow \mathrm{COCl}_{2}$ a. Write the rate law for this reaction. b. Which species are intermediates?

Short Answer

Expert verified
a. The rate law for the overall reaction is: Rate = \(k_3 \frac{k_1[\mathrm{Cl}_2]^2 + k_{-2}[\mathrm{COCl}][\mathrm{Cl}_2] - k_2[\mathrm{CO}][\mathrm{Cl}_2]^2}{k_{-1}[\mathrm{Cl}_2]+2k[\mathrm{Cl}_2]}[\mathrm{COCl}]\). b. The intermediate species in this mechanism are Cl and COCl.

Step by step solution

01

Analyze the elementary steps

We have the following elementary steps: 1. \(\mathrm{Cl}_{2} \stackrel{k_{1}}{\rightleftharpoons_{k_{1}}} 2 \mathrm{Cl}\) 2. $\mathrm{Cl}+\mathrm{CO} \stackrel{k_{2}}{\leftrightharpoons_{k-2}} \mathrm{COCl}$ 3. $\mathrm{COCl}+\mathrm{Cl}_{2} \stackrel{k_{3}}{\longrightarrow} \mathrm{COCl}_{2}+\mathrm{Cl}$ 4. \(2 \mathrm{Cl} \stackrel{k}{\longrightarrow} \mathrm{Cl}_{2}\)
02

Write the rate law for each elementary step

Using the rate constants and concentrations of the reactants we can write the rate law for each elementary step: 1. Rate\(_1 = k_1[\mathrm{Cl}_2] - k_{-1}[\mathrm{Cl}]^2\) 2. Rate\(_2 = k_2[\mathrm{Cl}][\mathrm{CO}] - k_{-2}[\mathrm{COCl}]\) 3. Rate\(_3 = k_3[\mathrm{COCl}][\mathrm{Cl}_2]\) 4. Rate\(_4 = k[\mathrm{Cl}]^2\)
03

Determine the rate-determining step (RDS)

The rate-determining step is the slowest step in the mechanism, which determines the overall reaction rate. To find the RDS, we will assume that it is the first step and analyze if it is consistent with the given information: Rate = Rate\(_1\) = \(k_1[\mathrm{Cl}_2] - k_{-1}[\mathrm{Cl}]^2\)
04

Steady-state approximation for intermediates

Using the steady-state approximation, we can assume that the concentration of the intermediates remains essentially constant since their formation and consumption rates are equal. We can write the balance equations for the intermediates: For Cl: \(d[\mathrm{Cl}]/dt = 2 \times Rate_1 + Rate_3 - Rate_2 - 2 \times Rate_4 = 0\)
05

Calculate [Cl]

From the balance equation of Cl, we can calculate the concentration of Cl using the rate expressions of the elementary steps: \(2(k_1[\mathrm{Cl}_2] - k_{-1}[\mathrm{Cl}]^2) + k_3[\mathrm{COCl}][\mathrm{Cl}_2] - (k_2[\mathrm{Cl}][\mathrm{CO}] - k_{-2}[\mathrm{COCl}]) - 2k[\mathrm{Cl}]^2 = 0\) This implies: \([\mathrm{Cl}] = \frac{k_1[\mathrm{Cl}_2] + k_{-2}[\mathrm{COCl}] - k_2[\mathrm{CO}][\mathrm{Cl}_2]}{k_{-1}[\mathrm{Cl}_2]+2k[\mathrm{Cl}_2]}\)
06

Write the rate law for the overall reaction

Using the RDS and the relation for [Cl] from the steady-state approximation, we can write the rate law for the overall reaction: Rate = Rate\(_3\) = \(k_3[\mathrm{COCl}][\mathrm{Cl}_2]\) Substituting [Cl] expression, we get: Rate = \(k_3 \frac{k_1[\mathrm{Cl}_2]^2 + k_{-2}[\mathrm{COCl}][\mathrm{Cl}_2] - k_2[\mathrm{CO}][\mathrm{Cl}_2]^2}{k_{-1}[\mathrm{Cl}_2]+2k[\mathrm{Cl}_2]}[\mathrm{COCl}]\) a. This is the rate law for the overall reaction.
07

Identify the intermediate species

From the mechanism, the intermediate species are formed and consumed in the reaction but do not appear in the overall reaction. b. In this mechanism, Cl and COCl are intermediates as they participate in the elementary steps but do not appear in the overall reaction.

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Most popular questions from this chapter

Consider two reaction vessels, one containing \(\mathrm{A}\) and the other containing \(\mathrm{B}\) , with equal concentrations at \(t=0 .\) If both substances decompose by first-order kinetics, where $$ \begin{array}{l}{k_{\mathrm{A}}=4.50 \times 10^{-4} \mathrm{s}^{-1}} \\\ {k_{\mathrm{B}}=3.70 \times 10^{-3} \mathrm{s}^{-1}}\end{array} $$ how much time must pass to reach a condition such that \([\mathrm{A}]=\) 4.00\([\mathrm{B}]\) ?

Write the rate laws for the following elementary reactions. a. \(\mathrm{CH}_{3} \mathrm{NC}(g) \rightarrow \mathrm{CH}_{3} \mathrm{CN}(g)\) b. $\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{NO}_{2}(g)$ c. \(\mathrm{O}_{3}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{O}(g)\) d. \(\mathrm{O}_{3}(g)+\mathrm{O}(g) \rightarrow 2 \mathrm{O}_{2}(g)\)

Two isomers (A and B) of a given compound dimerize as follows: $$ \begin{array}{l}{2 \mathrm{A} \stackrel{k_{1}}{\longrightarrow} A_{2}} \\ {2 \mathrm{B} \stackrel{k_{2}}{\longrightarrow} \mathrm{B}_{2}}\end{array} $$ Both processes are known to be second order in reactant, and \(k_{1}\) is known to be 0.250 \(\mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\) at $25^{\circ} \mathrm{C}\( . In a particular experiment \)\mathrm{A}\( and \)\mathrm{B}$ were placed in separate containers at \(25^{\circ} \mathrm{C},\) where \([\mathrm{A}]_{0}=1.00 \times 10^{-2} M\) and $[\mathrm{B}]_{0}=2.50 \times 10^{-2} M\( It was found that after each reaction had progressed for \)3.00 \mathrm{min},[\mathrm{A}]=3.00[\mathrm{B}]$ . In this case the rate laws are defined as $$ \begin{array}{l}{\text { Rate }=-\frac{\Delta[\mathrm{A}]}{\Delta t}=k_{1}[\mathrm{A}]^{2}} \\ {\text { Rate }=-\frac{\Delta[\mathrm{B}]}{\Delta t}=k_{2}[\mathrm{B}]^{2}}\end{array} $$ a. Calculate the concentration of \(\mathrm{A}_{2}\) after 3.00 \(\mathrm{min}\) . b. Calculate the value of \(k_{2}\) . c. Calculate the half-life for the experiment involving A.

Each of the statements given below is false. Explain why. a. The activation energy of a reaction depends on the overall energy change \((\Delta E)\) for the reaction. b. The rate law for a reaction can be deduced from examination of the overall balanced equation for the reaction. c. Most reactions occur by one-step mechanisms.

A certain reaction has the following general form: $$ \mathrm{aA} \longrightarrow \mathrm{bB} $$ At a particular temperature and \([\mathrm{A}]_{0}=2.00 \times 10^{-2} M,\) con- centration versus time data were collected for this reaction, and a plot of \(\ln [\mathrm{A}]\) versus time resulted in a straight line with a slope value of \(-2.97 \times 10^{-2} \mathrm{min}^{-1}\) . a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. Calculate the half-life for this reaction. c. How much time is required for the concentration of A to decrease to $2.50 \times 10^{-3} M ?$
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