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Chapter 14: Problem 71

The following mechanism has been proposed for the gasphase reaction of \(\mathrm{H}_{2}\) with ICl: $$\begin{array}{c}{\mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g)} \\ {\mathrm{HI}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g)}\end{array}$$ \(\begin{array}{l}{\text { (a) Write the balanced equation for the overall reaction. }} \\ {\text { (b) Identify any intermediates in the mechanism. (c) If }}\end{array}\) the first step is slow and the second one is fast, which rate law do you expect to be observed for the overall reaction?

Short Answer

Expert verified
The balanced equation for the overall reaction is \(H_2(g) + 2ICl(g) \longrightarrow I_2(g) + 2HCl(g)\). The intermediate in this mechanism is HI. The expected rate law for the overall reaction, given that the first step is slow and the second step is fast, is \(Rate = k[H_2][ICl]\), where k is the rate constant and [H_2] and [ICl] are the concentrations of H_2 and ICl, respectively.

Step by step solution

01

Write Balanced Equation for Overall Reaction

Add the two given steps of the reaction mechanism and then cancel any species that appear on both sides of the equation. This will provide us with the balanced equation for the overall reaction. Given steps: \(1. \quad H_2(g) + ICl(g) \longrightarrow HI(g) + HCl(g)\) \(2. \quad HI(g) + ICl(g) \longrightarrow I_2(g) + HCl(g)\) Add the two steps: \(H_2(g) + 2ICl(g) + HI(g) \longrightarrow HI(g) + HCl(g) + I_2(g) + HCl(g)\) Now, cancel species that appear on both sides: \(HI\) appears on both sides and can be cancelled. This leaves us with the balanced equation for the overall reaction: \(H_2(g) + 2ICl(g) \longrightarrow I_2(g) + 2HCl(g)\)
02

Identify Intermediates

Intermediates are species that are produced in one step of a mechanism and consumed in another step. Analyze the given mechanism steps and identify any intermediates. From the two given steps, we can see that HI is produced in the first step and consumed in the second step. Therefore, HI is an intermediate in this mechanism.
03

Determine Rate Law

We are given that the first step is slow and the second step is fast. The slow step determines the rate law for the overall reaction because it acts as the bottleneck in the reaction progress. Our rate law will depend only on the reactants in this slow step: Given slow step: \(H_2(g) + ICl(g) \longrightarrow HI(g) + HCl(g)\) The rate law can be written in the form: \(Rate = k[H_2][ICl]\) where k is the rate constant, and [H_2] and [ICl] are the concentrations of H_2 and ICl, respectively. This is the expected rate law for the overall reaction.

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Most popular questions from this chapter

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 \([\mathrm{A}]_{0}\) is tripled. (b) The rate increases by a factor of 9 when \([\mathrm{A}]_{0}\) is tripled. (c) When \([\mathrm{A}]_{0}\) is doubled, the rate increases by a factor of \(8 .\)

Consider the following reaction: $$\mathrm{CH}_{3} \mathrm{Br}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Br}^{-}(a q)$$ The rate law for this reaction is first order in \(\mathrm{CH}_{3} \mathrm{Br}\) and first order in \(\mathrm{OH}^{-} .\) When \(\left[\mathrm{CH}_{3} \mathrm{Br}\right]\) is \(5.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.050 \mathrm{M},\) the reaction rate at 298 \(\mathrm{K}\) is 0.0432 \(\mathrm{M} / \mathrm{s}\) . (a) What is the value of the rate constant? (\mathbf{b} )What are the units of the rate constant? (c) What would happen to the rate if the concentration of OH \(^{-}\) were tripled? (d) What would happen to the rate if the concentration of both reactants were tripled?

The enzyme urease catalyzes the reaction of urea, \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right),\) with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \mathrm{s}^{-1}\) at \(100^{\circ} \mathrm{C} .\) In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(3.4 \times 10^{4} \mathrm{s}^{-1}\) at \(21^{\circ} \mathrm{C}\) . (a) Write out the balanced equation for the reaction catalyzed by urease. (b) If the rate of the catalyzed reaction were the same at \(100^{\circ} \mathrm{C}\) as it is at \(21^{\circ} \mathrm{C},\) what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions? (c) In actuality, what would you expect for the rate of the catalyzed reaction at \(100^{\circ} \mathrm{Cas} \mathrm{com}-\) pared to that at \(21^{\circ} \mathrm{C} ?(\mathbf{d})\) On the basis of parts \((\mathrm{c})\) and \((\mathrm{d}),\) what can you conclude about the difference in activation energies for the catalyzed and uncatalyzed reactions?

Consider two reactions. Reaction \((1)\) has a constant half-life, whereas reaction \((2)\) has a half life that gets longer as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?

The reaction between ethyl bromide \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right)\) and hydroxide ion in ethyl alcohol at 330 \(\mathrm{K}\) , \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{Br}^{-}(a l c)\) is first order each in ethyl bromide and hydroxide ion. When \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right]\) is 0.0477 \(\mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.100 \mathrm{M},\) the rate of disappearance of ethyl bromide is \(1.7 \times 10^{-7} \mathrm{M} / \mathrm{s}\) (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) How would the rate of disappearance of ethyl bromide change if the solution were diluted by adding an equal volume of pure ethyl alcohol to the solution?
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