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Chapter 14: Problem 71

The following mechanism has been proposed for the gasphase reaction of \(\mathrm{H}_{2}\) with ICl: $$ \begin{array}{l} \mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g) \\ \mathrm{HI}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g) \end{array} $$ (a) Write the balanced equation for the overall reaction. (b) Identify any intermediates in the mechanism. (c) If 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
(a) \(\mathrm{H}_{2}(g) + 2\mathrm{ICl}(g) \rightarrow \mathrm{I}_{2}(g) + 2\mathrm{HCl}(g)\). (b) Intermediate: \(\mathrm{HI}(g)\). (c) Rate law: \(\text{Rate} = k[\mathrm{H}_{2}][\mathrm{ICl}]\).

Step by step solution

01

Write the Balanced Equation for the Overall Reaction

First, sum the two given elementary steps to find the overall reaction. The intermediate species produced in the first step and consumed in the second are canceled out. Thus, the intermediate is HI. Sum the two reactions:\[\begin{align*} \text{Step 1:} & \ \mathrm{H}_{2}(g) + \mathrm{ICl}(g) \rightarrow \mathrm{HI}(g) + \mathrm{HCl}(g) \ \text{Step 2:} & \ \mathrm{HI}(g) + \mathrm{ICl}(g) \rightarrow \mathrm{I}_{2}(g) + \mathrm{HCl}(g) \end{align*}\]Cancellation of the intermediate HI results in the overall reaction:\[\mathrm{H}_{2}(g) + 2\mathrm{ICl}(g) \rightarrow \mathrm{I}_{2}(g) + 2\mathrm{HCl}(g)\]This is the balanced equation for the overall reaction.
02

Identify Intermediate

An intermediate is a species that is produced in one step of the mechanism and consumed in another. From the given mechanism, \(\mathrm{HI}(g)\) is formed in the first step and consumed in the second, so \(\mathrm{HI}(g)\) is the intermediate species in the reaction mechanism.
03

Determine the Rate Law for the Overall Reaction

The overall rate law depends on the rate-determining step, which is the slowest step in the mechanism. Given that the first step is slow, this will determine the overall rate. The rate law for the slow step is based on its reactants. For step 1:\[\mathrm{H}_{2}(g) + \mathrm{ICl}(g) \rightarrow \mathrm{HI}(g) + \mathrm{HCl}(g)\]The rate law is:\[\text{Rate} = k[\mathrm{H}_{2}][\mathrm{ICl}]\]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.

Intermediates in Reactions
In chemical reaction mechanisms, intermediates play a crucial role. These are species that form in one step of a mechanism and are consumed in another. They are neither reactants nor final products; instead, they exist temporarily as the reaction occurs. Intermediates can sometimes be tricky to identify because they do not appear in the overall balanced chemical equation. For example, in the reaction mechanism given,
  • The intermediate identified is \(\mathrm{HI}(g)\), which is produced in the first reaction step and consumed in the second.
  • This substance is key to understanding the progress of the reaction at the molecular level, but it does not appear in the final overall reaction equation.
When looking at mechanisms, identifying intermediates helps chemists understand how a reaction proceeds and can provide insights into controlling the reaction speed and path.To identify intermediates, look for species that are produced and then consumed. They should cancel out when writing the overall balanced equation from the mechanism's elementary steps. Therefore, they never appear in the net equation because they are used up during the process.
Rate Law Determination
Understanding rate law determination is a key aspect of analyzing reaction mechanisms. The rate law gives insight into which molecules are involved in the slowest, or rate-determining step, of the mechanism.For the reaction \(\mathrm{H}_{2}(g) + \mathrm{ICl}(g) \rightarrow \mathrm{HI}(g) + \mathrm{HCl}(g)\),
  • The overall rate is determined by the slowest step, which in many complex reactions is the first step.
  • In the given mechanism, since the first step is slower, it dictates the reaction鈥檚 rate.
  • Consequently, the rate law can be written using the reactants from this first step.
The expected rate law is \(\text{Rate} = k[\mathrm{H}_{2}][\mathrm{ICl}]\). This rate law indicates that the reaction rate depends linearly on the concentration of both \(\mathrm{H}_2\) and \(\mathrm{ICl}\). This implies that both species are involved in the slowest part of the reaction. Understanding which step controls the rate allows chemists to manipulate reactant concentrations to optimize the reaction speed.
Elementary Steps in Reactions
Elementary steps in a reaction provide a detailed view of the mechanism by dissecting larger complex reactions into simpler, individual stages. Each step represents a single molecular event, such as the collision between reactant molecules, that must occur for the reaction to proceed.
  • In the given problem, there are two elementary steps:
  • The first step: \(\mathrm{H}_{2}(g) + \mathrm{ICl}(g) \rightarrow \mathrm{HI}(g) + \mathrm{HCl}(g)\), which is the slow step.
  • The second step: \(\mathrm{HI}(g) + \mathrm{ICl}(g) \rightarrow \mathrm{I}_{2}(g) + \mathrm{HCl}(g)\), which is faster.
These elementary steps, when summed, yield the reaction's overall balanced equation.Being able to discern elementary steps allows for a better understanding of reaction dynamics. It involves identifying intermediates and understanding which steps contribute to the rate law. Each step also has its own rate and is affected by the concentration of the reacting species involved. This structured approach in breaking down reactions helps in predicting reaction behavior and in planning reaction conditions effectively.

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Most popular questions from this chapter

Heterogeneous catalysts that perform hydrogenation reactions, as illustrated in Figure \(14.23,\) are subject to "poisoning," which shuts down their catalytic ability. Compounds of sulfur are often poisons. Suggest a mechanism by which such compounds might act as poisons.

A colored dye compound decomposes to give a colorless product. The original dye absorbs at \(608 \mathrm{nm}\) and has an extinction coefficient of \(4.7 \times 10^{4} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at that wavelength. You perform the decomposition reaction in a \(1-\mathrm{cm}\) cuvette in a spectrometer and obtain the following data: \begin{tabular}{cc} \hline Time (min) & Absorbance at \(608 \mathrm{nm}\) \\ \hline 0 & 1.254 \\ 30 & 0.941 \\ 60 & 0.752 \\ 90 & 0.672 \\ 120 & 0.545 \\ \hline \end{tabular} From these data, determine the rate law for the reaction "dye \(\longrightarrow\) product" and determine the rate constant.

At \(25^{\circ} \mathrm{C}\), raw milk sours in \(6.0 \mathrm{~h}\) but takes \(60 \mathrm{~h}\) to sour in a refrigerator at \(5{ }^{\circ} \mathrm{C}\). Estimate the activation energy in \(\mathrm{kJ} / \mathrm{mol}\) for the reaction that leads to the souring of milk.

Consider the following reaction: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) The rate law for this reaction is first order in \(\mathrm{H}_{2}\) and second order in NO. Write the rate law, \((\mathbf{b})\) If the rate constant for this reaction at \(1000 \mathrm{~K}\) is \(6.0 \times 10^{4} \mathrm{M}^{-2} \mathrm{~s}^{-1}\) what is the reaction rate when \([\mathrm{NO}]=0.035 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=0.015 \mathrm{M} ?(\mathbf{c})\) What is the reaction rate at \(1000 \mathrm{~K}\) when the concentration of NO is increased to \(0.10 \mathrm{M}\) while the concentration of \(\mathrm{H}_{2}\) is \(0.010 \mathrm{M} ?\) (d) What is the reaction rate at \(1000 \mathrm{~K}\) if \([\mathrm{NO}]\) is decreased to \(0.010 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]\) is increased to \(0.030 \mathrm{M} ?\)

You perform a series of experiments for the reaction \(\mathrm{A} \rightarrow 2 \mathrm{~B}\) 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) The rate increases by a factor of \(6.25,\) when \([A]_{0}\) is increased by a factor of \(2.5 .\) (b) There is no rate change when \([A]_{0}\) is increased by a factor of \(4 .(\mathbf{c})\) The rate decreases by a factor of \(1 / 2\), when \([A]\) is cut in half.
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