Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 77

For the reaction between hydrogen and iodine, $$\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g})$$ the experimental rate expression is rate \(=k\left[\mathrm{H}_{2}\right] \times\left[\mathrm{I}_{2}\right] .\) Show that this expression is consistent with the mechanism $$\begin{array}{cc}\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) & \text { (fast) } \\ \mathrm{H}_{2}(g)+\mathrm{I}(g)+\mathrm{I}(g) \longrightarrow 2 \mathrm{HI}(g) & \text { (slow) }\end{array}$$

Short Answer

Expert verified
In this exercise, we were given a reaction between hydrogen and iodine, and an experimental rate expression. We were asked to verify whether the rate expression is consistent with a given two-step reaction mechanism. To do this, we determined the rate law for each step in the mechanism, applied the steady-state approximation method to find the overall rate law, and compared it with the given rate expression. We found that the rate expression is consistent with the mechanism, assuming that the rate constant has the observed relationship.

Step by step solution

01

Write the rate law for each step of the mechanism

For each step in the given mechanism, we can write the rate law based on the reactants involved in each step. For the first step (fast equilibrium): $$\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)$$ The forward rate: \(rate_1 = k_1[\mathrm{I}_{2}]\) The reverse rate: \(rate_{-1} = k_{-1}[\mathrm{I}]^2\) For the second step (slow): $$\mathrm{H}_{2}(g)+\mathrm{I}(g)+\mathrm{I}(g) \longrightarrow 2 \mathrm{HI}(g)$$ The rate: \(rate_2 = k_2[\mathrm{H}_{2}][\mathrm{I}]^2\)
02

Apply the steady-state approximation to find the overall rate law

The steady-state approximation states that the rate of the change in concentration of any intermediate species (in this case, I atoms) is zero. Therefore, for I atoms, we have: $$\frac{d[\mathrm{I}]}{dt} = rate_1 - rate_{-1} - rate_2 = 0$$ Substitute the rate laws from Step 1: $$k_1[\mathrm{I}_{2}] - k_{-1}[\mathrm{I}]^2 - k_2[\mathrm{H}_{2}][\mathrm{I}]^2 = 0$$ Now solve for the concentration of I atoms: $$[\mathrm{I}]^2 = \frac{k_1[\mathrm{I}_{2}]}{k_{-1} + k_2[\mathrm{H}_{2}]}$$
03

Express the rate of the overall reaction in terms of the reactants

The overall rate of the reaction is the rate of the slow step, which is given by \(rate_2 = k_2[\mathrm{H}_{2}][\mathrm{I}]^2\). Substitute the expression for \([\mathrm{I}]^2\) obtained in Step 2: $$rate = k_2[\mathrm{H}_{2}]\frac{k_1[\mathrm{I}_{2}]}{k_{-1} + k_2[\mathrm{H}_{2}]}$$
04

Compare the derived rate law with the given rate expression

Now we compare this derived rate law with the given experimental rate expression, which is: $$rate = k[\mathrm{H}_{2}][\mathrm{I}_{2}]$$ To make these rate expressions consistent, the rate constant \(k\) must be equal to: $$k = k_2\frac{k_1}{k_{-1} + k_2[\mathrm{H}_{2}]}$$ Thus, we have shown that the rate expression given in the exercise is consistent with the provided mechanism, assuming that the rate constant \(k\) has the relationship shown above.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A reaction has two reactants \(\mathrm{X}\) and \(\mathrm{Y}\). What is the order with respect to each reactant and the overall order of the reaction described by the following rate expressions? (a) rate \(=k_{1}[\mathrm{X}]^{2} \times[\mathrm{Y}]\) (b) rate \(=k_{2}[\mathrm{X}]\) (c) rate \(=k_{3}[\mathrm{X}]^{2} \times[\mathrm{Y}]^{2}\) (d) rate \(=k_{4}\)

The decomposition of sulfuryl chloride, \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\), to sulfur dioxide and chlorine gases is a first-order reaction. $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ At a certain temperature, the half-life of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is \(7.5 \times 10^{2} \mathrm{~min}\). Consider a sealed flask with \(122.0 \mathrm{~g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (a) How long will it take to reduce the amount of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in the sealed flask to \(45.0 \mathrm{~g}\) ? (b) If the decomposition is stopped after \(29.0 \mathrm{~h}\), what volume of \(\mathrm{Cl}_{2}\) at \(27^{\circ} \mathrm{C}\) and \(1.00\) atm is produced?

Derive the integrated rate law, \([\mathrm{A}]=[\mathrm{A}]_{0}-k t\), for a zero-order reaction. (Hint: Start with the relation \(-\Delta[\mathrm{A}]=k \Delta \mathrm{t}\).)

Argon- 41 is used to measure the rate of gas flow. It has a decay constant of \(6.3 \times 10^{-3} \mathrm{~min}^{-1}\). (a) What is its half-life? (b) How long will it take before only \(1.00 \%\) of the original amount of Ar- 41 is left?

For the reaction $$5 \mathrm{Br}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q)+6 \mathrm{H}^{+}(a q) \longrightarrow 3 \mathrm{Br}_{2}(a q)+3 \mathrm{H}_{2} \mathrm{O}$$ it was found that at a particular instant bromine was being formed at the rate of \(0.039 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\). At that instant, at what rate is (a) water being formed? (b) bromide ion being oxidized? (c) \(\mathrm{H}^{+}\) being consumed?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free