   # A gaseous material XY( g ) dissociates to some extent to produce X( g ) and Y( g ): XY ( g ) ⇌ X ( g ) + Y ( g ) A 2.00-g sample of XY (molar mass = 165 g/mol) is placed in a container with a movable piston at 25°C. The pressure is held constant at 0.967 atm. As XY begins to dissociate, the piston moves until 35.0 mole percent of the original XY has dissociated and then remains at a constant position. Assuming ideal behavior, calculate the density of the gas in the container after the piston has stopped moving, and determine the value of K for this reaction of 25°C. ### Chemistry: An Atoms First Approach

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

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 12, Problem 119MP
Textbook Problem
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## A gaseous material XY(g) dissociates to some extent to produce X(g) and Y(g): XY ( g ) ⇌ X ( g ) + Y ( g ) A 2.00-g sample of XY (molar mass = 165 g/mol) is placed in a container with a movable piston at 25°C. The pressure is held constant at 0.967 atm. As XY begins to dissociate, the piston moves until 35.0 mole percent of the original XY has dissociated and then remains at a constant position. Assuming ideal behavior, calculate the density of the gas in the container after the piston has stopped moving, and determine the value of K for this reaction of 25°C.

Interpretation Introduction

Interpretation: The density of the gas in the container is to be calculated, after the piston has been stopped and behavior is to be assumed ideal. At the temperature 25°C value of the K for the dissociation reaction is to be calculated.

Concept introduction: The equilibrium constant K describes the ratio of the reactant to the product on the equilibrium conditions in terms of molar concentration.

The equilibrium constant depends upon temperature.

Law of mass action is applicable on the equilibrium reactions.

The Le Chatelier’s principle states that the addition of the reactants shifts the equilibrium to the right while the addition of product shifts the equilibrium to the left at constant temperature.

To determine: The density of the gas in the container after the piston has stopped and the value of equilibrium constant for the given reaction at the temperature 25°C .

### Explanation of Solution

Explanation

Given

The reaction is given as,

XY(g)X(g)+Y(g)

The mass of the sample is 2g . Molar mass of the sample XY is 165g/mol . The temperature is remains constant that is 25°C . The pressure is 0.967atm . The dissociative mole percent is 35% . Ideal behavior is to be assumed.

The concentration of the gaseous sample is determined by the formula,

n=mM

Where,

• n is the number of moles.
• m is the given mass.
• M is the molar mass.

Substitute the values of masses in the above formula.

n=mMn=2g165g/moln=0.0121mol

This is the initial value of moles. When the piston is moved the dissociation in moles is 35% therefore, the number of moles on dissociation is given as,

Dissociativemoles=35%of0.0121=35100×0.0121=0.00424mol

The dissociative moles are the change in moles. To determine the equilibrium moles the ICE-table is given as,

XY(g)X(g)+Y(g)Initialmol0.012100Changeinmol0.00424+0.00424+0.00424equilibriummol0.00790.004240.00424

Total moles at equilibrium is the sum of all the moles at equilibrium and is given as,

Totalmoles=0.0079+0.00424+0.00424=0.0164moles

The ideal gas equation is given as,

PV=nRT

Where,

• P is the pressure of the gas.
• V is the volume of the gas.
• n is the number of moles of the gas.
• R is the universal gas constant.
• T is the temperature.

At constant temperature and pressure the ideal gas equation becomes as,

V=n(RTP)Vn

The above expression for two different volume is given as,

V=k×nVfinal=k×nfinalVinitial=k×ninitial

Where,

k is proportionality constant.

For initial and final volume conditions the expression is as,

VfinalVinitial=k×nfinalk×ninitial

Substitute the initial and final values of moles in the above expression.

VfinalVinitial=nfinalninitialVfinalVinitial=0.01640.0121VfinalVinitial=1.36 (1)

For the given value of initial moles, temperature and pressure the initial volume is calculated by the formula,

Vinitial=ninitial(RTP)Vinitial=0

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