   # For the reaction PCl 5 ( g ) ⇌ PCl 3 ( g ) + Cl 2 ( g ) at 600. K, the equilibrium constant, K p , is 11.5. Suppose that 2.450 g PCl 5 is placed in an evacuated 500.-mL bulb, which is then heated to 600. K. a . What would be the pressure of PCl 5 if it did not dissociate? b . What is the partial pressure of PCl 5 at equilibrium? c . What is the total pressure in the bulb at equilibrium? d . What is the percent dissociation of PCl 5 at equilibrium? ### 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 78AE
Textbook Problem
61 views

## For the reaction PCl 5 ( g ) ⇌ PCl 3 ( g ) + Cl 2 ( g ) at 600. K, the equilibrium constant, Kp, is 11.5. Suppose that 2.450 g PCl5 is placed in an evacuated 500.-mL bulb, which is then heated to 600. K.a. What would be the pressure of PCl5 if it did not dissociate?b. What is the partial pressure of PCl5 at equilibrium?c. What is the total pressure in the bulb at equilibrium?d. What is the percent dissociation of PCl5 at equilibrium?

(a)

Interpretation Introduction

Interpretation: The pressure of PCl5 in undissociated form is to be calculated. The partial pressure, the total pressure and the percent dissociation of PCl5 at equilibrium is to be calculated.

Concept introduction: The equilibrium constant Kp describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure..

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 molar mass of PCl5=208.23gmol-1 .

Dissociation is also known as degree of dissociation which tells that dissociated amount of the species at a particular time.

To determine: The pressure of PCl5 in undissociated form.

### Explanation of Solution

Explanation

The pressure of PCl5 in undissociated form is 1.16atm_ .

Given

The reactions are given as,

PCl5(s)PCl5(g)+Cl2(g)

The equilibrium constant Kp=11.5 at 600K .

Mass of PCl5=2.450g .

Volume of the PCl5 =500ml .

Moles of the PCl5 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 in the above equation.

n=mM=2.450g208

(b)

Interpretation Introduction

Interpretation: The pressure of PCl5 in undissociated form is to be calculated. The partial pressure, the total pressure and the percent dissociation of PCl5 at equilibrium is to be calculated.

Concept introduction: The equilibrium constant Kp describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure..

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 molar mass of PCl5=208.23gmol-1 .

Dissociation is also known as degree of dissociation which tells that dissociated amount of the species at a particular time.

To determine: The partial pressure of PCl5 at equilibrium.

(c)

Interpretation Introduction

Interpretation: The pressure of PCl5 in undissociated form is to be calculated. The partial pressure, the total pressure and the percent dissociation of PCl5 at equilibrium is to be calculated.

Concept introduction: The equilibrium constant Kp describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure..

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 molar mass of PCl5=208.23gmol-1 .

Dissociation is also known as degree of dissociation which tells that dissociated amount of the species at a particular time.

To determine: The total pressure in the bulb at equilibrium.

(d)

Interpretation Introduction

Interpretation: The pressure of PCl5 in undissociated form is to be calculated. The partial pressure, the total pressure and the percent dissociation of PCl5 at equilibrium is to be calculated.

Concept introduction: The equilibrium constant Kp describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure..

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 molar mass of PCl5=208.23gmol-1 .

Dissociation is also known as degree of dissociation which tells that dissociated amount of the species at a particular time.

To determine: The percent dissociation of PCl5 at equilibrium.

### 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

Find more solutions based on key concepts
What is the difference between a molecular formula and a structural formula?

General Chemistry - Standalone book (MindTap Course List)

Which of the following is a physical consequence of fasting? a. loss of lean body tissues b. lasting weight los...

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

How many centimeters are there in a kilometer?

An Introduction to Physical Science

You throw a ball vertically upward so that it leaves the ground with velocity +5.00 m/s. (a) What is its veloci...

Physics for Scientists and Engineers, Technology Update (No access codes included)

Why do hot stars look bluer than cool stars?

Foundations of Astronomy (MindTap Course List) 