Chapter 12, Problem 41QAP

### Chemistry: Principles and Reactions

8th Edition
William L. Masterton + 1 other
ISBN: 9781305079373

Chapter
Section

### Chemistry: Principles and Reactions

8th Edition
William L. Masterton + 1 other
ISBN: 9781305079373
Textbook Problem

# At a certain temperature, the equilibrium constant for the following reaction is 0.0472. NO ( g ) + SO 3 ( g ) ⇌ SO 2 ( g ) + NO 2 ( g ) All gases are at an initial pressure of 0.862 atm.(a) Calculate the partial pressure of each gas at equilibrium.(b) Compare the initial total pressure with the total pressure of the gases at equilibrium. Would that relation be true of all gaseous systems?

(a)

Interpretation Introduction

Interpretation: The equilibrium partial pressure of each gas needs to be determined.

Concept Introduction: The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

For a general equilibrium reaction as follows:

A(g)+B(g)C(g)+D(g)

The expression for the equilibrium constant is represented as follows:

K=(PC)(PD)(PA)(PB)

Here, to calculate the equilibrium constant, the values of partial pressure of all the species in reactant and product side are required.

Explanation

The equilibrium reaction is as follows:

â€ƒâ€ƒNO(g)+SO3(g)â‡ŒSO2(g)+NO2(g)

The initial partial pressure of all the gases is 0.862 atm. The value of equilibrium constant is 0.0472 thus, equilibrium partial pressure of all gases can be calculated as follows:

â€ƒâ€ƒNO(g)+SO3(g)â‡ŒSO2(g)+NO2(g)0.862Â Â Â Â Â Â Â Â 0.862Â Â Â Â Â Â Â Â Â Â 0.862Â Â Â Â Â Â Â Â 0.8620.862-xÂ Â Â Â Â Â 0.862-xÂ Â Â Â Â Â Â 0.862+xÂ Â Â Â 0.862+x

The equilibrium constant for the reaction can be calculated as follows:

â€ƒâ€ƒK=(P SO 2 )(P NO 2 )(P NO)(P SO 3 )

Putting the values,

â€ƒâ€ƒ0

(b)

Interpretation Introduction

Interpretation:The initial partial pressure needs to be compared with the total pressure at equilibrium.

Concept Introduction: The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started