Chapter 13, Problem 116IP

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

Chapter
Section

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

# Given K = 3.50 at 45°C for the reaction A ( g ) + B ( g ) ⇌ C ( g ) and K = 7.10 at 45°C for the reaction 2 A ( g ) + D ( g ) ⇌ C ( g ) what is the value of K at the same temperature for the reaction C ( g ) + D ( g ) ⇌ 2 B ( g ) What is the value of Kp, at 45°c for the reaction? Starting with 1.50 atm partial pressures of both C and D, what is the mole fraction of B once equilibrium is reached?

Interpretation Introduction

Interpretation: In terms of partial pressure and in terms of molar concentration the equilibrium constant for the given reaction is to be calculated. The mole fraction of B 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.

The equilibrium constant Kc 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.

Law of mass action states that at a given temperature the equilibrium constant is equal to the partial pressure of the products to the reactants raised the power of stoichiometric coefficient.

To determine: The equilibrium constants for the given reaction in terms of partial pressure and molar concentration; the mole fraction of reactant B at equilibrium.

Explanation

Explanation

Given

The reactions are given as,

A(s)+B(g)C(g)......(1)2A(s)+D(g)C(g)......(2)C(s)+D(g)2B(g)......(3)

At 45°C for the first reaction the equilibrium constant =3.50

At 45°C for the second reaction the equilibrium constant =7.10

To obtain third reaction multiplies first reaction with 2 then reverses and then adds with second reaction as,

2A(s)+2B(g)2C(g)2C(g)2A(g)+2B(g)2A(g)+D(g)C(g)C(g)+D(g)2B(g)

The equilibrium constant follow the same changes as in the respective reaction occur.

In the addition of two reactions the new equilibrium constant is the multiplication of the equilibrium constant of two reactions.

Therefore the equilibrium constant K for the reverse reaction is calculated as,

Kreverse=1(K)2

Where,

• Kreverse is the equilibrium constant for the reverse reaction.
• K is the equilibrium constant of the first reaction.

Substitute the given values to calculate the Kreverse as,

Kreverse=1(K)2=1(3.50)2=0.0816

The equilibrium constant for the second reaction is represents as K2 .

The equilibrium constant for the resultant reaction in terms of molar concentration is given as,

K=Kreverse×K2

Where,

• K2 is the equilibrium constant for the second reaction.

Substitute the values of Kreverse and K2 in the above equation.

K=Kreverse×K2=0.0816×7.10=0.58_

The relation between the equilibrium constant in terms of partial pressure and in terms of molar concentration is given as,

Kp=Kc(RT)Δn

Where,

• Kp is the equilibrium constant in terms of partial pressure.
• Kc is the equilibrium constant in terms of molar concentration.
• R is the universal gas constant.
• T is the temperature.
• Δn is the difference in the number of moles of product to the number of moles of reactants.

For the given reaction,

C(g)+D(g)2B(g)

The difference in moles is zero.

Substitute the values of Kc, and Δn the calculated value of Kp is given as,

Kp=Kc(RT)Δn=0.58(RT)0=0.58×1=0.58_

The mole fraction of B at equilibrium is 0.273_

The initial partial pressure is 1.50atm and the change in partial pressure is assumed to be x . The ICE table for the given reaction is,

C(g)+D(g)2B(g)Initialpressure1

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