   Chapter 12, Problem 65QAP 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

Isopropyl alcohol is the main ingredient in rubbing alcohol. It can decompose into acetone (the main ingredient in nail polish remover) and hydrogen gas according to the following reaction: C 3 H 7 OH ( g ) ⇌ C 2 H 6 CO ( g ) + H 2 ( g ) At 180°C, the equilibrium constant for the decomposition is 0.45. If 20.0 mL ( d = 0.785   g / mL ) of isopropyl alcohol is placed in a 5.00-L vessel and heated to 180°C, what percent- age remains undissociated at equilibrium?

Interpretation Introduction

Interpretation:

The percentage of isopropyl alcohol that remains undissociated in the system at equilibrium needs to be calculated.

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 is as follows:

C3H7OH(g)C2H6CO(g)+H2(g)

The density of the isopropyl alcohol and its volume is given 0.785 g/mL and 20.0 mL respectively.

The mass can be calculated as follows:

m=d×V

Putting the value,

m=(0.785 g/mL)(20 mL)=15.7 g

The molar mass of isopropyl alcohol is 60.1 g/mol thus, number of moles can be calculated as follows:

n=mM

Putting the values,

n=15.7 g60.1 g/mol=0.26123 mol

From the number of moles, pressure can be calculated as follows:

P=nRTV

The volume of the vessel is given 5.0 L at 180C.

The temperature can be converted into K as follows:

0 C=273.15 K

Or,

180 C=180+273.15 K=453.15 K

Putting the values,

P=(0.26123 mol)(0.082 L atm/K mol)(453.15 K)(5.0 L)=1.9413 atm

The equilibrium constant can be calculated as follows:

K=(PH2)(PC2H6CO)(PC3H7OH)

From the equilibrium reaction:

C3H7OH(g)C2H6CO(g)+H2(g)    1

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