/*! 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 91 The reaction $$\mathrm{H}_{2}(g)... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 91

The reaction $$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g)$$ may occur by the following mechanism: $$\begin{aligned}& \mathrm{I}_{2} \frac{k_{1}}{k_{-1}} & 2 \mathrm{I} \\ \mathrm{I}+\mathrm{I}+\mathrm{H}_{2} & \stackrel{k_{2}}{\longrightarrow} & 2 \mathrm{HI} \end{aligned}$$ What rate law is predicted by the mechanism?

Short Answer

Expert verified
The rate law is \( \text{Rate} = k' [\mathrm{I}_{2}] [\mathrm{H}_{2}] \).

Step by step solution

01

Identify the Rate-Determining Step

The rate-determining step is often the slowest step in a reaction mechanism. In this case, the second step, where \( \mathrm{I} + \mathrm{I} + \mathrm{H}_{2} \rightarrow 2 \mathrm{HI} \), is typically slower because it involves a termolecular reaction, which is less likely to occur frequently compared to bimolecular reactions.
02

Write the Rate Law for the Rate-Determining Step

For the rate-determining step (\( \mathrm{I} + \mathrm{I} + \mathrm{H}_{2} \rightarrow 2 \mathrm{HI} \)), the rate law can be expressed as: \( \text{Rate} = k_{2} [\mathrm{I}]^2 [\mathrm{H}_{2}] \), where \( k_{2} \) is the rate constant for the step.
03

Express Intermediate Concentrations

The intermediate \( \mathrm{I} \) is formed in the fast pre-equilibrium step: \( \mathrm{I}_{2} \leftrightharpoons 2\mathrm{I} \). The equilibrium expression for this step is \( K = \frac{[\mathrm{I}]^2}{[\mathrm{I}_{2}]} \). Rearranging gives \( [\mathrm{I}]^2 = K [\mathrm{I}_{2}] \), where \( K = \frac{k_{1}}{k_{-1}} \).
04

Substitute Intermediate Concentrations Into the Rate Law

Substitute \( [\mathrm{I}]^2 \) from Step 3 into the rate law from Step 2: \( \text{Rate} = k_{2} [\mathrm{I}]^2 [\mathrm{H}_{2}] = k_{2} (K [\mathrm{I}_{2}]) [\mathrm{H}_{2}] \). This simplifies to \( \text{Rate} = k_{2} K [\mathrm{I}_{2}] [\mathrm{H}_{2}] \).
05

Write the Final Rate Law

Combine all constants into a single constant: \( k' = k_{2} K \). Thus, the rate law simplifies to \( \text{Rate} = k' [\mathrm{I}_{2}] [\mathrm{H}_{2}] \). This is the predicted rate law for the reaction according to the mechanism.

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Key Concepts

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

Rate-Determining Step
In a chemical reaction mechanism, the rate-determining step is crucial because it acts like a bottleneck for the whole process. It is the slowest step, and thus determines the overall rate of the reaction. In our example, the second step of the reaction: \( \mathrm{I} + \mathrm{I} + \mathrm{H}_{2} \rightarrow 2 \mathrm{HI} \) is identified as the rate-determining step.
There are a few reasons why this particular step is slower:
  • It involves three molecules colliding at once, which is much rarer than two molecules interacting. This is known as a termolecular reaction.
  • The formation of such complex associations takes more time, making the reaction slower.
Understanding which step is rate-determining helps us focus on what affects the entire speed of the reaction. Adjustments or conditions that can influence this step will likely have the greatest impact on how fast the reaction proceeds.
Rate Law
The rate law is a mathematical expression that relates the rate of a chemical reaction to the concentration of its reactants. For the rate-determining step in our reaction, the rate law derives from its chemical equation: \( \text{Rate} = k_{2} [\mathrm{I}]^2 [\mathrm{H}_{2}] \), where \( k_{2} \) is the constant specific to this step.
Key points to remember about rate laws include:
  • The exponents reflect the number of molecules involved in the rate-determining step: \([\mathrm{I}]^2\) indicates that two iodine radicals are necessary.
  • The concentrations of intermediate species, like \(\mathrm{I}\), affect the reaction rate, but these are often replaced by more stable reactant concentrations using pre-equilibrium steps.
Understanding the rate law is essential for predicting how changes in reactant concentrations can influence the speed of a reaction.
Chemical Kinetics
Chemical kinetics is the branch of chemistry that examines the rates of chemical reactions and the factors affecting these rates. In our reaction mechanism, kinetics provides insight into how each step progresses, especially those like the rate-determining step.
By studying chemical kinetics, we can:
  • Create models that predict reaction speeds under various conditions.
  • Identify ways to control and optimize industrial chemical processes.
  • Understand how to increase or decrease reaction speeds to align with desired outcomes.
Knowledge of kinetics is crucial in fields from pharmaceuticals to environmental science, where reaction rates directly impact product development and policy decisions.
Equilibrium Expression
Equilibrium expressions are used to relate the concentrations of reactants and products at equilibrium. In our reaction's context, the pre-equilibrium step: \( \mathrm{I}_{2} \leftrightharpoons 2\mathrm{I} \) reveals how iodine is temporarily balanced between its molecular and radical forms.
The equilibrium expression here is given by \( K = \frac{[\mathrm{I}]^2}{[\mathrm{I}_{2}]} \), simplifying the intermediate concentrations:
  • This provides a bridge between transient species and more stable, measurable quantities.
  • \( K = \frac{k_{1}}{k_{-1}} \) relates the forward and reverse rate constants, embodying the balance at equilibrium.
These expressions help walk through the complexities of reactions, allowing deeper comprehension of how instantaneous states lead to the stable conditions observable experimentally.

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Most popular questions from this chapter

A reaction of the form \(a \mathrm{~A} \longrightarrow\) Products is second order with a half-life of \(425 \mathrm{~s}\). What is the rate constant of the reaction if the initial concentration of \(\mathrm{A}\) is \(5.99 \times 10^{-3} \mathrm{~mol} / \mathrm{L}\) ?

Hydrogen sulfide is oxidized by chlorine in aqueous solution. $$\mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{S}(s)+2 \mathrm{HCl}(a q)$$ The experimental rate law is $$\text { Rate }=k\left[\mathrm{H}_{2} \mathrm{~S}\right]\left[\mathrm{Cl}_{2}\right]$$ What is the reaction order with respect to \(\mathrm{H}_{2} \mathrm{~S}\) ? with respect to \(\mathrm{Cl}_{2}\) ? What is the overall order?

To obtain the rate of the reaction $$3 \mathrm{I}^{-}(a q)+\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+2 \mathrm{H}^{+}(a q) \longrightarrow$$ you might follow the \(\mathrm{I}^{-}\) concentration or the \(\mathrm{I}_{3}^{-}\) concentration. How are the rates in terms of these species related?

In a series of experiments on the decomposition of dinitrogen pentoxide, \(\mathrm{N}_{2} \mathrm{O}_{5}\), rate constants were determined at two different temperatures. At \(35^{\circ} \mathrm{C}\), the rate constant was \(1.4 \times\) \(10^{-4} / \mathrm{s}\); at \(45^{\circ} \mathrm{C}\), the rate constant was \(5.0 \times 10^{-4} / \mathrm{s}\). What is the activation energy for this reaction? What is the value of the rate constant at \(55^{\circ} \mathrm{C} ?\)

Nitrogen dioxide decomposes when heated. $$2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ During an experiment, the concentration of \(\mathrm{NO}_{2}\) varied with time in the following way: \(\begin{array}{ll}\text { Time } & {\left[\mathrm{NO}_{2}\right]} \\ 0.0 \mathrm{~min} & 0.1103 M \\ 1.0 \mathrm{~min} & 0.1076 M \\ 2.0 \mathrm{~min} & 0.1050 M \\ 3.0 \mathrm{~min} & 0.1026 M\end{array}\) Obtain the average rate of decomposition of \(\mathrm{NO}_{2}\) in units of \(M / \mathrm{s}\) for each time interval.
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