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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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Key Concepts

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

Gas-Phase Reaction
A gas-phase reaction occurs when reactants in a gaseous state interact to form products. These reactions are influenced by the properties of gases, such as temperature and pressure. In the given problem, the reaction between hydrogen gas \(H_2(g)\) and iodine chloride \(ICl(g)\) is a typical example.
  • The process involves different transition states, where molecules collide, leading to transformations.
  • It's crucial to understand that the outcome depends on these collisions' energy and orientation.
Gas-phase reactions often have unique dynamics compared to those occurring in liquid or solid phases. The speed and efficiency of these reactions are also contingent on the reactants' concentrations and the inherent kinetics of the involved molecules.
Intermediates
Intermediates are crucial to understanding reaction mechanisms. They are species formed in one step and consumed in another, thus not appearing in the overall balanced equation. In our problem:
  • \(HI\) is generated during the first step \(H_2(g) + ICl(g) \to HI(g) + HCl(g)\).
  • Then, it is consumed in the second step \(HI(g) + ICl(g) \to I_2(g) + HCl(g)\).
As an intermediate, \(HI\) doesn't appear in the final balanced equation \(H_2(g) + 2ICl(g) \to I_2(g) + 2HCl(g)\).
Understanding intermediates allows chemists to deeply know how each step contributes to the overall reaction, aiding in predicting and controlling reaction behaviors.
Rate Law
The rate law expresses how the rate of a chemical reaction relates to the concentration of its reactants. In our scenario, the reaction's rate is determined by the slowest or 'rate-determining' step. For our gas-phase reaction:
  • The slow step is \(H_2(g) + ICl(g) \to HI(g) + HCl(g)\).
  • This determines the rate law as \(Rate = k[H_2][ICl]\).
Here, \(k\) is the rate constant, and [H_2] and [ICl] are the reactants' concentrations.

The rate law helps predict the reaction speed and analyze how changes in concentration affect reaction dynamics. Understanding the rate law is crucial for controlling industrial chemical processes and optimizing lab reactions.

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Most popular questions from this chapter

Based on their activation energies and energy changes and assuming that all collision factors are the same, rank the following reactions from slowest to fastest.\( \begin{aligned} \text { (a) } E_{a} &=45 \mathrm{kJ} / \mathrm{mol} ; \Delta E=-25 \mathrm{kJ} / \mathrm{mol} \\ \text { (b) } E_{a} &=35 \mathrm{kJ} / \mathrm{mol} ; \Delta E=-10 \mathrm{kJ} / \mathrm{mol} \\ \text { (c) } E_{a} &=55 \mathrm{kJ} / \mathrm{mol} ; \Delta E=10 \mathrm{kJ} / \mathrm{mol} \end{aligned}\)

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 oxidation of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\) is accelerated by \(\mathrm{NO}_{2} .\) The reaction proceeds according to: $$ \begin{array}{l}{\mathrm{NO}_{2}(g)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{SO}_{3}(g)} \\ {2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)}\end{array}$$ (a) Show that, with appropriate coefficients, the two reactions can be summed to give the overall oxidation of \(S O_{2}\) by \(\mathrm{O}_{2}\) to give \(S O_{3} .(\mathbf{b})\) Do we consider \(N O_{2}\) a catalyst or an intermediate in this reaction? (c) Would you classify NO as a catalyst or as an intermediate? { ( d ) } Is this an example of homogeneous catalysis or heterogeneous catalysis?

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 a hypothetical reaction between \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) that is first order in \(\mathrm{A},\) zero order in \(\mathrm{B},\) and second order in C. (a) Write the rate law for the reaction. (b) How does the rate change when [A] is doubled and the other reactant concentrations are held constant? (c) How does the rate change when [B] is tripled and the other reactant concentrations are held constant? (d) How does the rate change when \([C]\) is tripled and the other reactant concentrations are held constant? (e) By what factor does the rate change when the concentrations of all three reactants are tripled? (f) By what factor does the rate change when the concentrations of all three reactants are cut in half?
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