   Chapter 12, Problem 81QAP Chemistry: Principles and Reactions

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

Solutions

Chapter
Section Chemistry: Principles and Reactions

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

For the system SO 3 ( g ) ⇌ SO 2 ( g ) + 1 2 O 2 ( g ) at 1000 K, K = 0.45 . Sulfur trioxide, originally at 1.00 atm pressure, partially dissociates to SO2 and O2 at 1000 K. What is its partial pressure at equilibrium?

Interpretation Introduction

Interpretation:

The equilibrium partial pressure of sulfur trioxide needs to be determined.

Concept introduction:

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.

The number of moles of gas can be calculated from pressure as follows:

n=PVRT

Here, P is pressure, V is volume, R is Universal gas constant and T is temperature.

The mass of a substance can be calculated from number of moles and mass as follows:

m=n×M

Here, n is number of moles and M is molar mass.

Explanation

The equilibrium reaction for the system is shown as follows:

SO3(g)SO2(g)+12O2(g)

The initial pressure of sulfur trioxide is 1 atm thus, the partial pressure of each gases at equilibrium can be calculated as follows:

SO3(g)SO2(g)+12O2(g)1                   -                 -1-p                  p             12p

The expression for the equilibrium constant will be:

K=(pSO2)(

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 