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An 8.00-g sample of SO 3 was placed in an evacuated container, where it decomposed at 600°C according to the following reaction: SO 3 ( g ) ⇌ SO 2 ( g ) + 1 2 O 2 ( g ) At equilibrium the total pressure and the density of the gaseous mixtures were 1.80 atm and 1.60 g/L, respectively. Calculate K p for this reaction.

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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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Section
BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 12, Problem 109CP
Textbook Problem
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An 8.00-g sample of SO3 was placed in an evacuated container, where it decomposed at 600°C according to the following reaction:

SO 3 ( g ) SO 2 ( g ) + 1 2 O 2 ( g )

At equilibrium the total pressure and the density of the gaseous mixtures were 1.80 atm and 1.60 g/L, respectively. Calculate Kp for this reaction.

Interpretation Introduction

Interpretation: The decomposition reaction of a given mass of SO3 is given. The total pressure and the density of the gaseous mixtures have been stated. The value of the equilibrium constant (Kp) for the reaction is to be calculated.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction.

When the equilibrium constant is expressed in terms of pressure, it is represented as (Kp)

To determine: The value of the equilibrium constant (Kp) for the reaction.

Explanation of Solution

Explanation

Given

The stated reaction is,

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

The given mass of SO3 is 8.0g.

The molar mass of SO3 =S+3O=((32)+3(16))g/mol=80g/mol

The number of moles of a substance is calculated by the formula,

Numberofmoles=GivenmassMolarmass

Substitute the value of the given mass and the molar mass of SO3 in the above expression.

Numberofmoles=8.00g80g/mol=0.0999mol_

The volume of SO3 is 5.0L_.

The density of a substance is calculated by the formula,

Density=MassVolume

Rearrange the above expression,

Volume=MassDensity

Substitute the given values of mass and density of SO3 in the above expression.

Volume=8.0g1.60g/L=5.0L_

The initial pressure of SO3 is 1.432atm_.

The given temperature is 600°C(873K)

The initial pressure of a SO3 is calculated by the ideal gas equation.

The ideal gas equation is,

PV=nRT

Where,

  • P is the partial pressure.
  • V is the volume of the container.
  • n is the number of moles.
  • R is the universal gas constant (0.0821LatmK1mol1)
  • T is the temperature.

Rearrange the ideal gas equation.

P=nRTV

Substitute the values of V,R,n and T for P4(g) in the above expression.

P=(0.0999mol)(0.0821LatmK1mol1)(873K)(5.0L)=0.1432atm_

The equilibrium constant expression for the given reaction is, Kp=(x)(12x)12(1.432x)

The initial pressure of SO3 is 1.432atm.

The change in the partial pressure of SO3 is assumed to be x.

The equilibrium pressures are represented as,

SO3(g)SO2(g)+12O2(g)Initial(atm)1

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Chapter 12 Solutions

Chemistry: An Atoms First Approach
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