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ChemistryChemistry: Principles and Reactions8th Edition

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Chapter 12, Problem 81QAP
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Chemistry: Principles and Reactions

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

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Problem 64QAP
Problem 65QAP
Problem 66QAP
Problem 67QAP
Problem 68QAP
Problem 69QAP
Problem 70QAP
Problem 71QAP
Problem 72QAP
Problem 73QAP
Problem 74QAP
Problem 75QAP
Problem 76QAP
Problem 77QAP
Problem 78QAP
Problem 79QAP
Problem 80QAP
Problem 81QAP
Problem 82QAP
Problem 83QAP
Problem 84QAP
Problem 85QAP
Problem 86QAP
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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)(

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