Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 13

The following rates of reaction were obtained in three experiments with the reaction \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NOCl}(\mathrm{g}).\) $$\begin{array}{llll} \hline & \text { Initial } & \text { Initial } & \text { Initial Rate of } \\ \text { Expt } & \text { [NO], M } & \text { [Cl }_{2} \text { ], M } & \text { Reaction, } \mathrm{M} \mathrm{s}^{-1} \\ \hline 1 & 0.0125 & 0.0255 & 2.27 \times 10^{-5} \\ 2 & 0.0125 & 0.0510 & 4.55 \times 10^{-5} \\ 3 & 0.0250 & 0.0255 & 9.08 \times 10^{-5} \\ \hline \end{array}$$ What is the rate law for this reaction?

Short Answer

Expert verified
The rate law for this reaction is \[ \text{rate} = k[\text{NO}]^2[\text{Cl}_2] \].

Step by step solution

01

Determine The Order Of Reaction With Respect to NO

Comparing experiments 1 and 3 will help to determine the order of reaction with respect to NO. The concentration of Cl₂ is the same in both experiments, but the concentration of NO is doubled in experiment 3 compared to experiment 1. As a result, we can see how the rate of reaction is affected when the concentration of NO changes. The rate is quadrupled when the concentration of NO is doubled which implies that the reaction is second order with respect to NO.
02

Determine The Order Of Reaction With Respect to Cl₂

Comparing experiments 1 and 2 will help to determine the order of reaction with respect to Cl₂. The concentration of NO is the same in both experiments, but the concentration of Cl₂ is doubled in experiment 2 compared to experiment 1. As a result, we can see how the rate of reaction is affected when the concentration of Cl₂ changes. The rate is doubled when the concentration of Cl₂ is doubled which implies that the reaction is first order with respect to Cl₂.
03

Formulate The Rate Law

The order of reaction with respect to each reactant has been found. The reaction is second order with respect to NO and first order with respect to Cl₂. Therefore, the rate law for this reaction is \[ \text{rate} = k[\text{NO}]^2[\text{Cl}_2] \]. Note that the brackets denote concentration and 'k' represents the rate constant.

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

We have used the terms order of a reaction and molecularity of an elementary process (that is, unimolecular, bimolecular). What is the relationship, if any, between these two terms?

The first-order reaction \(A \longrightarrow\) products has \(t_{1 / 2}=180 \mathrm{s}\) (a) What percent of a sample of A remains unreacted \(900 \mathrm{s}\) after a reaction has been started? (b) What is the rate of reaction when \([\mathrm{A}]=0.50 \mathrm{M} ?\)

The decomposition of ethylene oxide at \(690 \mathrm{K}\) is monitored by measuring the total gas pressure as a function of time. The data obtained are \(t=10 \mathrm{min}, P_{\text {tot }}= 139.14 \mathrm{mmHg} ; 20 \mathrm{min}, 151.67 \mathrm{mmHg} ; 40 \mathrm{min}, 172.65 \mathrm{mmHg} ; 60 \mathrm{min}, 189.15 \mathrm{mmHg} ;\) \(100 \mathrm{min}, 212.34\) \(\mathrm{mmHg} ; 200 \mathrm{min}, 238.66 \mathrm{mmHg} ; \infty, 249.88 \mathrm{mmHg}\) What is the order of the reaction \(\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) ?\)

The rate constant for the reaction \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \longrightarrow\) \(2 \mathrm{HI}(\mathrm{g})\) has been determined at the following temperatures: \(599 \mathrm{K}, k=5.4 \times 10^{-4} \mathrm{M}^{-1} \mathrm{s}^{-1} ; 683 \mathrm{K}, k=2.8 \times 10^{-2} \mathrm{M}^{-1} \mathrm{s}^{-1} .\) Calculate the activation energy for the reaction.

For the first-order reaction $$\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ \(t_{1 / 2}=22.5 \mathrm{h}\) at \(20^{\circ} \mathrm{C}\) and \(1.5 \mathrm{h}\) at \(40^{\circ} \mathrm{C}.\) (a) Calculate the activation energy of this reaction. (b) If the Arrhenius constant \(A=2.05 \times 10^{13} \mathrm{s}^{-1}\) determine the value of \(k\) at \(30^{\circ} \mathrm{C}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Kinetics

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