   Chapter 13, Problem 113CP

Chapter
Section
Textbook Problem

# A sample of gaseous nitrosyl bromide (NOBr) was placed in a container tiued with a frictionless, massless piston, where it decomposed at 25°C according to the following equation: 2 NOBr ( g ) ⇌ 2 NO ( g ) + Br 2 ( g ) The initial density of the system was recorded as 4.495 g/L. After equilibrium was reached, the density was noted to be 4.086 g/L.a. Determine the value of the equilibrium constant K for the reaction.b. If Ar(g) is added to the system at equilibrium at constant temperature, what will happen to the equilibrium position? What happens to the value of K? Explain each answer

(a)

Interpretation Introduction

Interpretation: The equilibrium constant for the given reaction is to be determined. If Argon is added to the system at constant temperature then the effect on the equilibrium position is to be determined. The effect on the value of the K is to be stated.

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.

At equilibrium sum of all the partial pressure is equal to the total pressure.

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.

Every chemical reaction follows the concept of the conservation of mass.

To determine: The equilibrium constant for the given reaction

Explanation

Explanation

The value of equilibrium constant K is 2.33×10-4_ .

Given

The reactions are given as,

2NOBr(g)2NO(s)+2Br2(g)

At 25°C the initial density is 4.495g/L

At equilibrium density is 4.086g/L

In the given chemical reaction the density of the system decreases while the mass remains constant. As the number of moles of gas increases, the volume of the system increases.

The density of the given reaction is represents as,

(Density)i(Density)e=m/(V)im/(V)e

Where,

• (Density)i is the initial density of the system.
• (Density)e is the density at equilibrium.
• m is the mass that is remain constant.
• Vi is the initial volume.
• Ve is the final volume.

Substitute the given values in the above equation.

(Density)i(Density)e=4.495g/L4.086g/L(m/V)i(m/V)e=4.495g/L4.086g/L(V)e(V)i=1.1000

The volume is proportional to number of moles at constant temperature and pressure.

Simplify the above expression.

(V)e(V)i=1.100neni=1.100 (1)

The initial volume of NOBr is assumed to be one liter.

The number of moles of NOBr is calculate by the expression,

Numberofmoles=GivenmassMolarmassMolesofNOBr=4.495g109.91g/mol=0.04090mol

The ICE-table is given as,

2NOBr2NO+Br2Initial0.049000Change2x+2x+2xEquilibrium0.04902x2x2x

Substitute the values of number of moles from ICE table in equation (1)

neni=1

(b)

Interpretation Introduction

Interpretation: The equilibrium constant for the given reaction is to be determined. If Argon is added to the system at constant temperature then the effect on the equilibrium position is to be determined. The effect on the value of the K is to be stated.

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.

At equilibrium sum of all the partial pressure is equal to the total pressure.

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 sat constant temperature.

Every chemical reaction follows the concept of the conservation of mass.

To determine: The effect on equilibrium position and on equilibrium constant if the Argon is added to the system.

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 