   Chapter 12, Problem 70QAP 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 reaction C ( s ) + CO 2 ( g ) ⇌ 2 CO ( g ) K = 168 at 1273 K. If one starts with 0.3 atm of CO2 and 12.0 g of C at 1273 K, will the equilibrium mixture contain(a) mostly CO2?(b) mostly CO?(c) roughly equal amounts of CO2 and CO?(d) only C?

Interpretation Introduction

Interpretation:

The correct option needs to be chosen for the amount of gases in the mixture at equilibrium.

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

Reason for correct option:

The given reaction is as follows:

C(s)+CO2(g)2CO(g)

The equilibrium constant for the reaction is 168 at 1273 K.

The initial pressure of carbon dioxide gas is 0.3 atm and the mass of C is 12.0 g.

Since, the molar mass of carbon is 12 g/mol, the number of moles of C can be calculated as follows:

n=mM

Putting the values,

n=12 g12 g/mol=1 mol

From the equilibrium reaction, the partial pressure of each gases can be calculated as follows:

C(s)+CO2(g)2CO(g)          0.3               --           0.3-p              2p

The expression for equilibrium constant will be:

K=PCO2PCO2

Putting the values,

168=(2p)20.3p

On rearranging,

p2+42p12

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