Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 45

A \(1.00-\mathrm{L}\) flask was filled with 2.00 moles of gaseous \(\mathrm{SO}_{2}\) and 2.00 moles of gaseous \(\mathrm{NO}_{2}\) and heated. After equilibrium was reached, it was found that 1.30 moles of gaseous NO was present. Assume that the reaction $$\mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g)$$, occurs under these conditions. Calculate the value of the equilibrium constant, \(K\), for this reaction.

Short Answer

Expert verified
The equilibrium constant, \(K\), for the given reaction can be calculated using the formula: \( K = \frac{[\mathrm{SO_3}][\mathrm{NO}]}{[\mathrm{SO_2}][\mathrm{NO_2}]} \). By determining the change in concentrations and finding the equilibrium concentrations, we calculate the value of \(K \approx 3.45\).

Step by step solution

01

Write the initial concentrations for each reactant and product

We know the initial amounts of SO2 and NO2 are 2.00 moles and the volume of the flask is 1.00 L. The initial concentrations can be calculated as follows: \[ [\mathrm{SO_2}]_{initial} = \frac{2.00 \ \text{moles}}{1.00 \ \text{L}} = 2.00 \ \mathrm{M} \] \[ [\mathrm{NO_2}]_{initial} = \frac{2.00 \ \text{moles}}{1.00 \ \text{L}} = 2.00 \ \mathrm{M} \] Initially, there is no SO3 and NO in the flask, hence: \[ [\mathrm{SO_3}]_{initial} = [\mathrm{NO}]_{initial} = 0 \ \mathrm{M} \]
02

Calculate the change in concentrations

Given that 1.30 moles of NO are present at equilibrium, we can calculate x, the change in concentration. For every mole of NO formed, one mole of SO2 and NO2 each are consumed. \[ x = \frac{1.30 \ \text{moles}}{1.00 \ \text{L}} = 1.30 \ \mathrm{M} \]
03

Determine the equilibrium concentrations

Using the calculated value of x, we can find the concentrations at equilibrium: \[ [\mathrm{SO_2}]_{eq} = [\mathrm{SO_2}]_{initial} - x = 2.00 \ \mathrm{M} - 1.30 \ \mathrm{M} = 0.70 \ \mathrm{M} \] \[ [\mathrm{NO_2}]_{eq} = [\mathrm{NO_2}]_{initial} - x = 2.00 \ \mathrm{M} - 1.30 \ \mathrm{M} = 0.70 \ \mathrm{M} \] \[ [\mathrm{SO_3}]_{eq} = [\mathrm{SO_3}]_{initial} + x = 0 \ \mathrm{M} + 1.30 \ \mathrm{M} = 1.30 \ \mathrm{M} \] \[ [\mathrm{NO}]_{eq} = [\mathrm{NO}]_{initial} + x = 0 \ \mathrm{M} + 1.30 \ \mathrm{M} = 1.30 \ \mathrm{M} \]
04

Calculate the equilibrium constant, K

Now that we have the equilibrium concentrations, we can calculate K: \[ K = \frac{[\mathrm{SO_3}]_{eq}[\mathrm{NO}]_{eq}}{[\mathrm{SO_2}]_{eq}[\mathrm{NO_2}]_{eq}} = \frac{(1.30 \ \mathrm{M})(1.30 \ \mathrm{M})}{(0.70 \ \mathrm{M})(0.70 \ \mathrm{M})} = \frac{1.69}{0.49} \] Thus, \[ K \approx 3.45 \]

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

Consider the following reactions:\(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g) \quad\) and \(\quad \mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \longrightarrow 2 \mathrm{HI}(g)\).List two property differences between these two reactions that relate to equilibrium.

At \(2200^{\circ} \mathrm{C}, K_{\mathrm{p}}=0.050\) for the reaction $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$.What is the partial pressure of NO in equilibrium with \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) that were placed in a flask at initial pressures of 0.80 and 0.20 atm, respectively?

For the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g),\) consider two possibilities: (a) you mix 0.5 mole of each reactant, allow the system to come to equilibrium, and then add another mole of \(\mathrm{H}_{2}\) and allow the system to reach equilibrium again, or (b) you mix 1.5 moles of \(\mathrm{H}_{2}\) and 0.5 mole of \(\mathrm{I}_{2}\) and allow the system to reach equilibrium. Will the final equilibrium mixture be different for the two procedures? Explain.

The reaction $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ has \(K_{\mathrm{p}}=109\) at \(25^{\circ} \mathrm{C}\). If the equilibrium partial pressure of \(\mathrm{Br}_{2}\) is 0.0159 atm and the equilibrium partial pressure of NOBr is 0.0768 atm, calculate the partial pressure of NO at equilibrium.

A 1.604 -g sample of methane \(\left(\mathrm{CH}_{4}\right)\) gas and 6.400 g oxygen gas are sealed into a 2.50 -L vessel at \(411^{\circ} \mathrm{C}\) and are allowed to reach equilibrium. Methane can react with oxygen to form gaseous carbon dioxide and water vapor, or methane can react with oxygen to form gaseous carbon monoxide and water vapor. At equilibrium, the pressure of oxygen is 0.326 atm, and the pressure of water vapor is 4.45 atm. Calculate the pressures of carbon monoxide and carbon dioxide present at equilibrium.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic 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