Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 15

When nitrogen monoxide reacts with oxygen to produce nitrogen dioxide, an equilibrium is established. a. Write the balanced equation. b. Write the equilibrium constant expression.

Short Answer

Expert verified
Balanced equation: 2NO + O2 -> 2NO2. Equilibrium constant expression: Keq = [NO2]² / ([NO]² [O2])

Step by step solution

01

Balancing the chemical equation

The chemical reaction is between Nitrogen monoxide (NO) and Oxygen (O2) to form Nitrogen Dioxide (NO2). Unbalanced, it looks like this: NO + O2 -> NO2. To balance this equation, you see that there are two Oxygen atoms on the left-hand side, and there are only two on the right-hand side, so you should have two NO molecules reacting with one O2 molecule to form two NO2 molecules. Hence, the balanced chemical equation is: 2NO + O2 -> 2NO2.
02

Writing the equilibrium constant expression

The equilibrium constant expression Keq is given by the molar concentrations of the products divided by the molar concentrations of the reactants, each raised to the power equal to their stoichiometric coefficients in the balanced chemical equation. The equilibrium expression for this reaction is, therefore, Keq = [NO2]² / ([NO]² [O2])

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

Why might an industrial process be operated at high temperature even though the reaction is more favorable at lower temperatures?

Vinegar-a solution of acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH},\) in water \(-\) is used in varying concentrations for different household tasks. The following equilibrum exists in vinegar. $$\begin{array}{c}{\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows} \\ {\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)}\end{array}$$ If the concentration of the acetic acid solution at equilibrium is 3.00 \(\mathrm{M}\) and the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]=7.22 \times 10^{-3},\) what is the \(K_{e q}\) value for acetic acid?

Define complex ion and give two examples.

Does the reverse reaction rate ever equal zero? Why or why not?

When does a pressure change affect a chemical equilibrium?
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemistry Branches

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