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Chapter 14 Lecture

© 2014 Pearson Education, Inc.

Sherril Soman

Grand Valley State University

Lecture Presentation

Chapter 14

Chemical

Equilibrium


Hemoglobin

• Hemoglobin is a protein (Hb), found in red blood cells, that reacts with O2.

– It enhances the amount of O2 that can be carried through the bloodstream.

Hb + O2 HbO2

– The is used to describe a process that is in dynamic equilibrium.


Hemoglobin Equilibrium System

Hb + O2 HbO2

• The concentrations of Hb, O2, and HbO2 are all

interdependent.

• The relative amounts of Hb, O2, and HbO2 at equilibrium

are related to a constant called the equilibrium

constant, K.

– A large value of K indicates a high concentration of

products at equilibrium.

• Changing the concentration of any one of these

necessitates changes the other concentrations to restore

equilibrium.


O2 Transport

• In the lungs:

– High concentration of O2

– The equilibrium shifts to the right

– Hb and O2 combine to make more HbO2

Insert cartoon at

top on page 650:

shifting of reaction

to the right


O2 Transport

• In the muscles:

– Low concentration of O2,

– The equilibrium shifts to

the right

– HbO2 breaks down

(dissociates) increasing

the amount of free O2.

Insert cartoon at

middle on page

650 : shifting of

reaction to the left


HbF

Hb

Fetal Hemoglobin, HbF HbF + O2 HbFO2

• Fetal hemoglobin’s equilibrium constant is larger than adult hemoglobin’s constant.

• Fetal hemoglobin is more efficient at binding O2.

• O2 is transferred to the fetal hemoglobin from the mother’s hemoglobin in the placenta.

Hb + O2 HbO2 O2 HbO2

O2

HbF + O2 HbFO2 HbFO2 O2

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Oxygen Exchange between Mother and Fetus


Arrow Conventions

• Chemists commonly use two kinds of arrows in reactions to indicate the degree of completion of the reactions.

• A single arrow indicates all the reactant molecules are converted to product molecules at the end.

• A double arrow indicates the reaction stops when only some of the reactant molecules have been converted into products.

– in these notes


Reaction Dynamics

• When a reaction starts, the reactants are consumed and products are made. – The reactant concentrations decrease and the product

concentrations increase.

– As reactant concentration decreases, the forward reaction rate decreases.

• Eventually, the products can react to re-form some of the reactants, assuming the products are not allowed to escape. – As product concentration increases, the reverse reaction rate

increases.

• Processes that proceed in both the forward and reverse direction are said to be reversible. reactants products


Dynamic Equilibrium

• As the forward reaction slows and the reverse reaction

accelerates, eventually they reach the same rate.

• Dynamic equilibrium is the condition wherein the rates of

the forward and reverse reactions are equal.

• Once the reaction reaches equilibrium, the concentrations

of all the chemicals remain constant because the

chemicals are being consumed and made at the

same rate.


H2(g) + I2(g) 2 HI(g)

At time 0, there are only reactants in the mixture,

so only the forward reaction can take place.

[H2] = 8, [I2] = 8, [HI] = 0


H2(g) + I2(g) 2 HI(g)

[H2] = 6, [I2] = 6, [HI] = 4

At time 16, there are both reactants and products in the

mixture, so both the forward reaction and reverse

reaction can take place.

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H2(g) + I2(g) 2 HI(g)

At time 32, there are now more products than reactants in

the mixture, the forward reaction has slowed down as the

reactants run out, and the reverse reaction accelerated.

[H2] = 4, [I2] = 4, [HI] = 8


H2(g) + I2(g) 2 HI(g)

At time 48, the amounts of products and reactants in the

mixture haven’t changed; the forward and reverse reactions

are proceeding at the same rate. It has reached equilibrium.


H2(g) + I2(g) 2 HI(g)

As the concentration of product increases and the

concentrations of reactants decrease, the rate of the

forward reaction slows down, and the rate of the reverse

reaction speeds up.


H2(g) + I2(g) 2 HI(g)

At dynamic equilibrium, the rate of the forward reaction is

equal to the rate of the reverse reaction.

The concentrations of reactants and products no longer

change.


Equilibrium Equal

• The rates of the forward and reverse reactions are equal

at equilibrium.

• But that does not mean the concentrations of reactants

and products are equal.

• Some reactions reach equilibrium only after almost all

the reactant molecules are consumed; we say the

position of equilibrium favors the products.

• Other reactions reach equilibrium when only a small

percentage of the reactant molecules are consumed; we

say the position of equilibrium favors the reactants.


When Country A citizens feel overcrowded,

some will emigrate to Country B .

An Analogy: Population Changes

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However, after a time, emigration will occur in

both directions at the same rate, leading to

populations in Country A and Country B that are

constant, but not necessarily equal.


Equilibrium Constant

• Even though the concentrations of reactants and products are not equal at equilibrium, there is a relationship between them.

• The relationship between the chemical equation and the concentrations of reactants and products is called the law of mass action.


Equilibrium Constant

• For the general equation aA + bB cC + dD, the law of mass action gives the relationship below. – The lowercase letters represent the coefficients of the

balanced chemical equation.

– Always products over reactants

• K is called the equilibrium constant.

– Unitless


Writing Equilibrium Constant Expressions

• So, for the reaction

2 N2O5(g) 4 NO2(g) + O2(g)

the equilibrium constant expression is as follows:


What Does the Value of Keq Imply?

• When the value of Keq >> 1, when the reaction

reaches equilibrium there will be many more

product molecules present than reactant

molecules.

• The position of equilibrium favors products.


What Does the Value of Keq Imply?

• When the value of Keq << 1, when the reaction

reaches equilibrium there will be many more

reactant molecules present than product

molecules.

• The position of equilibrium favors reactants.

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A Large Equilibrium Constant


A Small Equilibrium Constant


The equilibrium between reactants and products may

be disturbed in three ways:

(1) by changing the temperature

(2) by changing the concentration of a reactant

(3) by changing the volume (for systems involving

gases)

A change in any of these factors will cause a system

at equilibrium to shift back towards a state of

equilibrium.

This statement is often referred to as Le Chatelier’s

principle.

Disturbing a Chemical Equilibrium


Disturbing and Restoring Equilibrium

• Once a reaction is at equilibrium, the concentrations of all the reactants and products remain the same.

• However, if the conditions are changed, the concentrations of all the chemicals will change until equilibrium is restored.

• The new concentrations will be different, but the equilibrium constant will be the same, unless you change the temperature.


Le Châtelier’s Principle

• Le Châtelier's principle guides us in predicting the effect

various changes in conditions have on the position of

equilibrium.

• It says that if a system at equilibrium is disturbed, the

position of equilibrium will shift to minimize the disturbance.

– Disturbances all involve making the system open.



When the populations of Country A and Country B

are in equilibrium, the emigration rates between the

two countries are equal so the populations stay constant.

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When an influx of population enters Country B

from somewhere outside Country A, it disturbs the

equilibrium established between Country A and Country B.


The result will be people moving from Country B into

Country A faster than people moving from Country A into

Country B.

This will continue until a new equilibrium between the

populations is established; the new populations will have

different numbers of people than the old ones.



Disturbing Equilibrium: Adding or

Removing Reactants

• After equilibrium is established, a reactant is added,

as long as the added reactant is included in the

equilibrium constant expression. • That is, not a solid or liquid

• How will this affect the rate of the forward reaction?

• How will it affect the rate of the reverse reaction?

• How will it affect the value of K?

© 2014 Pearson Education, Inc. 34


• Changes in Concentration continued

Change Shifts the Equilibrium

Increase concentration of product(s) left

Decrease concentration of product(s) right

Decrease concentration of reactant(s)

Increase concentration of reactant(s) right

left

aA + bB cC + dD

Add Add Remove Remove


The Effect of Concentration Changes on

Equilibrium

When NO2 is added, some of it

combines to make more N2O4.


The Effect of Concentration Changes

on Equilibrium

When N2O4 is added, some of it

decomposes to make more NO2.

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The Effect of Concentration Changes

on Equilibrium

When N2O4 is added, some of it

decomposes to make more NO2.


The Effect of Adding a Gas to a

Gas Phase Reaction at Equilibrium

• Adding a gaseous reactant increases its partial pressure, causing the equilibrium to shift to the right.

– Increasing its partial pressure increases its concentration.

– It does not increase the partial pressure of the other gases in the mixture.

• Adding an inert gas to the mixture has no effect on the position of equilibrium.

– It does not affect the partial pressures of the gases in the reaction.

© 2014 Pearson Education, Inc. 39


• Changes in Volume and Pressure

A (g) + B (g) C (g)

Change Shifts the Equilibrium

Increase pressure Side with fewest moles of gas

Decrease pressure Side with most moles of gas

Decrease volume

Increase volume Side with most moles of gas

Side with fewest moles of gas

• When the number of gas moles on either side is

the same, there is no effect.


Effect of Volume Change on Equilibrium

• Decreasing the volume of the container increases the

concentration of all the gases in the container.

– It increases their partial pressures.

– It does not change the concentrations of solutions!

• If their partial pressures increase, then the total

pressure in the container will increase.

• According to Le Châtelier’s Principle, the equilibrium

should shift to remove that pressure.

• The way the system reduces the pressure is to reduce

the number of gas molecules in the container.

• When the volume decreases, the equilibrium shifts

to the side with fewer gas molecules.


Disturbing Equilibrium: Changing the

Volume

• After equilibrium is established, the container volume is

decreased.

• How will it affect the concentration of solids, liquid,

solutions, and gases?

• How will this affect the total pressure of solids, liquid,

and gases?

• How will it affect the value of K?


Disturbing Equilibrium: Reducing the

Volume

• Decreasing the container volume will increase the total pressure. – Boyle’s law

– If the total pressure increases, the partial pressures of all the gases will increase—Dalton’s law of partial pressures.

• Because the total pressure increases, the position of equilibrium will shift to decrease the pressure by removing gas molecules. – Shift toward the side with fewer gas molecules

• At the new equilibrium position, the partial pressures of gaseous reactants and products will be such that the value of the equilibrium constant is the same.

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The Effect of Volume Changes

on Equilibrium

Because there are more

gas molecules on the

reactants side of the

reaction, when the pressure

is increased, the position of

equilibrium shifts toward the

side with fewer molecules

to decrease the pressure.

Left side of

figure

14.11


The Effect of Volume Changes

on Equilibrium

When the pressure is

decreased by increasing the

volume, the position of

equilibrium shifts toward the

side with the greater

number of molecules—the

reactant side.

Right side

of figure

14.11


The Effect of Temperature Changes on

Equilibrium Position

• Exothermic reactions release energy and endothermic reactions absorb energy

• Writing heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, helps us use Le Châtelier’s principle to predict the effect of temperature changes, even though heat is not matter and not written in a proper equation.



Equilibrium for Exothermic Reactions

• For an exothermic reaction, heat is a product.

• Increasing the temperature is like adding heat.

• According to Le Châtelier’s principle, the equilibrium will shift away from the added heat.



Equilibrium for Exothermic Reactions

• Adding heat to an exothermic reaction will decrease the concentrations of products and increase the concentrations of reactants.

• Adding heat to an exothermic reaction will decrease the value of K.

• How will decreasing the temperature affect the system?



Equilibrium for Endothermic Reactions

• For an endothermic reaction, heat is a reactant

• Increasing the temperature is like adding heat

• According to Le Châtelier’s Principle, the equilibrium will shift away from the added heat

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Equilibrium for Endothermic Reactions

• Adding heat to an endothermic reaction will decrease the concentrations of reactants and increase the concentrations of products.

• Adding heat to an endothermic reaction will increase the value of K.

• How will decreasing the temperature affect the system?



Equilibrium


Effect of Temperature Changes on Gas-Phase

Equilibria

Conclusion:

• Increasing the temperature of an endothermic

reaction favors the products, equilibrium shifts to

the right.

• Increasing the temperature of an exothermic

reaction favors the reactants, equilibrium shifts to

the left.

• Lowering temperature results in the reverse effects.

Disturbing a Chemical Equilibrium


Not Changing the Position of Equilibrium:

The Effect of Catalysts

• Catalysts provide an alternative, more efficient mechanism.

• Catalysts work for both forward and reverse reactions.

• Catalysts affect the rate of the forward and reverse reactions by the same factor.

• Therefore, catalysts do not affect the position of equilibrium.


Relationships between K and Chemical

Equations

• When the reaction is written backward, the

equilibrium constant is inverted.

For the reaction aA + bB cC + dD

the equilibrium constant expression

is as follows:

For the reaction cC + dD aA + bB

the equilibrium constant expression

is as follows:



Equations

• When the coefficients of an equation are multiplied

by a factor, the equilibrium constant is raised to

that factor.

For the reaction aA + bB cC

the equilibrium constant

expression is as follows:

For the reaction 2aA + 2bB 2cC



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Equations

• When you add equations to get a new equation, the equilibrium constant of the new equation is the product of the equilibrium constants of the old equations.

For the reactions (1) aA bB

and (2) bB cC the equilibrium

constant expressions are as

follows:

For the reaction aA cC



Chapter 14 Lecture


Solution The equilibrium constant is the equilibrium concentrations of the products raised to their stoichiometric coefficients

divided by the equilibrium concentrations of the reactants raised to their stoichiometric coefficients.

For Practice 14.1 Express the equilibrium constant for the combustion of propane as shown by the balanced chemical equation:

Express the equilibrium constant for the chemical equation:

Example 14.1 Expressing Equilibrium Constants for Chemical

Equations

Chapter 14 Lecture


Solution You want to manipulate the given reaction and value of K to obtain the desired reaction and value of K. You can see

that the given reaction is the reverse of the desired reaction, and its coefficients are twice those of the desired reaction.

Begin by reversing the given reaction and taking the inverse of the value of K.

Next, multiply the reaction by and raise the equilibrium constant to the power.

Consider the chemical equation and equilibrium constant for the synthesis of ammonia at 25 °C:

Calculate the equilibrium constant for the following reaction at 25 ° C:

Example 14.2 Manipulating the Equilibrium Constant to Reflect

Changes in the Chemical Equation

Chapter 14 Lecture


Calculate the value of K′.

For Practice 14.2 Consider the following chemical equation and equilibrium constant at 25 ° C:

Calculate the equilibrium constant for the following reaction at 25 ° C:

For More Practice 14.2 Predict the equilibrium constant for the first reaction shown here given the equilibrium constants for the second

and third reactions:

Continued

Example 14.2 Manipulating the Equilibrium Constant to Reflect

Changes in the Chemical Equation


Equilibrium Constants for Reactions

Involving Gases

• The concentration of a gas in a mixture is proportional to

its partial pressure.

• Therefore, the equilibrium constant can be expressed as

the ratio of the partial pressures of the gases.

• For aA(g) + bB(g) cC(g) + dD(g) the equilibrium

constant expressions are as follows:

or


Kc and Kp

• In calculating Kp, the partial pressures are always in atm.

• The values of Kp and Kc are not necessarily the same because of the difference in units.

– Kp = Kc when Dn = 0

• The relationship between them is as follows:

Dn is the difference between the number of

moles of reactants and moles of products.

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Deriving the Relationship between Kp

and Kc


Deriving the Relationship between Kp

and Kc

for aA(g) + bB(g) cC(g) + dD(g)

substituting


Heterogeneous Equilibria

• The concentrations of pure solids and pure liquids do not change during the course of a reaction.

• Because their concentration doesn’t change, solids and liquids are not included in the equilibrium constant expression.

• For the reaction the equilibrium constant expression is as follows:


Heterogeneous Equilibria

The amount of C is

different, but the

amounts of CO and

CO2 remain the same.

Therefore, the amount

of C has no effect on

the position of

equilibrium.


Calculating Equilibrium Constants from

Measured Equilibrium Concentrations

• The most direct way of finding the equilibrium constant is to

measure the amounts of reactants and products in a

mixture at equilibrium.

• The equilibrium mixture may have different amounts of

reactants and products, but the value of the equilibrium

constant will always be the same, as long as the

temperature is kept constant.

– The value of the equilibrium constant is independent of

the initial amounts of reactants and products.

Chapter 14 Lecture


Sort You are given Kp for the reaction and asked to find Kc.

Given: Kp = 2.2 × 1012

Find: Kc

Strategize Use Equation 14.2 to relate Kp and Kc.

Equation Kp = Kc(RT)Δn

Solve Solve the equation for Kc.

Calculate n.

Substitute the required quantities to calculate Kc. The temperature must be in kelvins. The units are dropped when

reporting Kc as described previously.

Nitrogen monoxide, a pollutant in automobile exhaust, is oxidized to nitrogen dioxide in the atmosphere according

to the equation:

Find Kc for this reaction.

Example 14.3 Relating Kp and Kc

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Chapter 14 Lecture


Solution

Check The easiest way to check this answer is to substitute it back into Equation 14.2 and confirm that you get the original

value for Kp.

Continued


Chapter 14 Lecture


For Practice 14.3 Consider the following reaction and corresponding value of Kc:

What is the value of Kp at this temperature?

Continued


Chapter 14 Lecture


Solution Since CaCO3(s) and CaO(s) are both solids, omit them from the equilibrium expression.

Kc = [CO2]

For Practice 14.4 Write an equilibrium expression (Kc) for the equation:

Write an expression for the equilibrium constant (Kc) for this chemical equation:

Example 14.4 Writing Equilibrium Expressions for Reactions

Involving a Solid or a Liquid

Chapter 14 Lecture


Consider the following reaction:

A reaction mixture at 780 °C initially contains [CO] = 0.500 M and [H2] = 1.00 M. At equilibrium, the CO

concentration is found to be 0.15 M. What is the value of the equilibrium constant?

Procedure For… Finding Equilibrium Constants from Experimental Concentration Measurements

Step 1 Using the balanced equation as a guide, prepare an ICE table showing the known initial

concentrations and equilibrium concentrations of the reactants and products. Leave space in

the middle of the table for determining the changes in concentration that occur during the reaction.

Example 14.5 Finding Equilibrium Constants from Experimental

Concentration Measurements

Chapter 14 Lecture


Step 2 For the reactant or product whose concentration is known both initially and at equilibrium,

calculate the change in concentration that occurs.

Step 3 Use the change calculated in step 2 and the stoichiometric relationships from the balanced

chemical equation to determine the changes in concentration of all other reactants and products.

Since reactants are consumed during the reaction, the changes in their concentrations are negative.

Since products are formed, the changes in their concentrations are positive.

Continued



Chapter 14 Lecture


Step 4 Sum each column for each reactant and product to determine the equilibrium concentrations.

Step 5 Use the balanced equation to write an expression for the equilibrium constant and substitute the

equilibrium concentrations to calculate K.

For Practice 14.5 The reaction in Example 14.5 between CO and H2 is carried out at a different temperature with initial concentrations

of [CO] = 0.27 M and [H2] = 0.49 M. At equilibrium, the concentration of CH3OH is 0.11 M. Find the equilibrium

constant at this temperature.

Continued



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Initial and Equilibrium Concentrations for

H2(g) + I2(g) 2HI(g) at 445 °C


Calculating Equilibrium Concentrations

• Stoichiometry can be used to determine the equilibrium

concentrations of all reactants and products if you know

initial concentrations and one equilibrium concentration.

• Use the change in the concentration of the material that

you know to determine the change in the other

chemicals in the reaction.


The Reaction Quotient

• If a reaction mixture containing both reactants and products is not at equilibrium, how can we determine in which direction it will proceed?

• The answer is to compare the current concentration ratios to the equilibrium constant.

• The concentration ratio of the products (raised to the power of their coefficients) to the reactants (raised to the power of their coefficients) is called the reaction quotient, Q.


The Reaction Quotient

For the gas phase reaction

aA + bB cC + dD

the reaction quotient is as follows:


The Reaction Quotient: Predicting the

Direction of Change

• If Q > K, the reaction will proceed fastest in the reverse direction.

– The products will decrease and reactants will increase.



Direction of Change

• If Q < K, the reaction will proceed fastest in the forward direction.

– The products will increase and reactants will decrease.

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Direction of Change

• If Q = K, the reaction is at equilibrium

– The products and reactants will not change.

• If a reaction mixture contains just reactants, then Q = 0, and the reaction will proceed in the forward direction.

• If a reaction mixture contains just products, then Q = ∞, and the reaction will proceed in the reverse direction.


Finding Equilibrium Concentrations

When Given the Equilibrium Constant

and Initial Concentrations or Pressures

Step 1: Decide in which direction the reaction will proceed. – Compare Q to K.

Step 2: Define the changes of all materials in terms of x. – Use the coefficient from the chemical equation as the

coefficient of x.

– The x change is + for materials on the side the reaction is proceeding toward.

– The x change is for materials on the side the reaction is proceeding away from.

Step 3: Solve for x. – For second order equations, take square roots of both sides or use

the quadratic formula.

– Simplify and approximate answer for very large or small equilibrium constants, if possible.


Approximations to Simplify the Math

• When the equilibrium constant is very small, the

position of equilibrium favors the reactants.

• For relatively large initial concentrations of

reactants, the reactant concentration will not change

significantly when it reaches equilibrium.

– assuming the reaction is proceeding forward

– The [X]equilibrium = ([X]initial ax) [X]initial

• We are approximating the equilibrium concentration of

reactant to be the same as the initial concentration.


Checking the Approximation and Refining

as Necessary

• We can check our approximation by

comparing the approximate value of x to the

initial concentration.

• If the approximate value of x is less than 5%

of the initial concentration, the approximation

is valid.

Chapter 14 Lecture


Solution Since CaCO3(s) and CaO(s) are both solids, omit them from the equilibrium expression.

Kc = [CO2]

For Practice 14.4 Write an equilibrium expression (Kc) for the equation:

Write an expression for the equilibrium constant (Kc) for this chemical equation:

Example 14.4 Writing Equilibrium Expressions for Reactions

Involving a Solid or a Liquid

Chapter 14 Lecture



A reaction mixture at 780 °C initially contains [CO] = 0.500 M and [H2] = 1.00 M. At equilibrium, the CO

concentration is found to be 0.15 M. What is the value of the equilibrium constant?







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Chapter 14 Lecture








Continued



Chapter 14 Lecture





For Practice 14.5 The reaction in Example 14.5 between CO and H2 is carried out at a different temperature with initial concentrations

of [CO] = 0.27 M and [H2] = 0.49 M. At equilibrium, the concentration of CH3OH is 0.11 M. Find the equilibrium

constant at this temperature.

Continued



Chapter 14 Lecture



A reaction mixture at 1700 °C initially contains [CH4] = 0.115 M. At equilibrium, the mixture contains

[C2H2] = 0.035 M. What is the value of the equilibrium constant?







Chapter 14 Lecture








Continued



Chapter 14 Lecture





For Practice 14.6 The reaction of CH4 in Example 14.6 is carried out at a different temperature with an initial concentration of

[CH4] = 0.087 M. At equilibrium, the concentration of H2 is 0.012 M. Find the equilibrium constant at this

temperature.

Continued



Chapter 14 Lecture


Solution To determine the progress of the reaction relative to the equilibrium state, first calculate Q.

Compare Q to K.

Qp = 10.8; Kp = 81.9

Since Qp < Kp, the reaction is not at equilibrium and will proceed to the right.

Consider the reaction and its equilibrium constant:

A reaction mixture contains = 0.114 atm, = 0.102 atm, and PICl = 0.355 atm. Is the reaction mixture at

equilibrium? If not, in which direction will the reaction proceed?

Example 14.7 Predicting the Direction of a Reaction by Comparing

Q and K

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Chapter 14 Lecture


For Practice 14.7 Consider the reaction and its equilibrium constant:

A reaction mixture contains [NO2] = 0.0255 M and [N2O4] = 0.0331 M. Calculate Qc and determine the direction in

which the reaction will proceed.

Continued

Example 14.7 Predicting the Direction of a Reaction by Comparing

Q and K

Chapter 14 Lecture



In an equilibrium mixture, the concentration of COF2 is 0.255 M and the concentration of CF4 is 0.118 M. What is

the equilibrium concentration of CO2?

Example 14.8 Finding Equilibrium Concentrations When You

Know the Equilibrium Constant and All but One of

the Equilibrium Concentrations of the Reactants

and Products

Sort You are given the equilibrium constant of a chemical reaction, together with the equilibrium concentrations of the

reactant and one product. You are asked to find the equilibrium concentration of the other product.

Given:

Find: [CO2]

Strategize You can calculate the concentration of the product using the given quantities and the expression for Kc.

Chapter 14 Lecture


Continued




and Products

Conceptual Plan

Solve Solve the equilibrium expression for [CO2] and then substitute in the appropriate values to calculate it.

Solution

Chapter 14 Lecture


Continued




and Products

Check Check your answer by mentally substituting the given values of [COF2] and [CF4] as well as your calculated value

for CO2 back into the equilibrium expression.

[CO2] was found to be roughly equal to 1. [COF2]2 ≈ 0.06 and [CF4] ≈ 0.12. Therefore Kc is approximately 2, as

given in the problem.

For Practice 14.8 Diatomic iodine [I2] decomposes at high temperature to form I atoms according to the reaction:

In an equilibrium mixture, the concentration of I2 is 0.10 M. What is the equilibrium concentration of I?

Chapter 14 Lecture


Consider the reaction:

A reaction mixture at 2000 °C initially contains [N2] = 0.200 M and [O2] = 0.200 M. Find the equilibrium

concentrations of the reactants and product at this temperature.

Procedure For… Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Constant

Step 1 Using the balanced equation as a guide, prepare a table showing the known initial concentrations

of the reactants and products. Leave room in the table for the changes in concentrations and for the

equilibrium concentrations.

Example 14.9 Finding Equilibrium Concentrations from Initial

Concentrations and the Equilibrium Constant

Chapter 14 Lecture


Step 2 Use the initial concentrations to calculate the reaction quotient (Q) for the initial concentrations.

Compare Q to K to predict the direction in which the reaction will proceed.

Q < K; therefore, the reaction will proceed to the right.

Step 3 Represent the change in the concentration of one of the reactants or products with the variable

x. Define the changes in the concentrations of the other reactants or products in terms of x. It is

usually most convenient to let x represent the change in concentration of the reactant or product with

the smallest stoichiometric coefficient.

Continued



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Chapter 14 Lecture


Step 4 Sum each column for each reactant and each product to determine the equilibrium concentrations

in terms of the initial concentrations and the variable x.

Step 5 Substitute the expressions for the equilibrium concentrations (from step 4) into the expression for

the equilibrium constant. Using the given value of the equilibrium constant, solve the expression

for the variable x. In some cases, such as Example 14.9, you can take the square root of both sides of

the expression to solve for x. In other cases, such as Example 14.10, you must solve a quadratic

equation to find x.

Remember the quadratic formula:

Continued



Chapter 14 Lecture


Continued



Chapter 14 Lecture


Step 6 Substitute x into the expressions for the equilibrium concentrations of the reactants and products

(from step 4) and calculate the concentrations. In cases where you solved a quadratic and have two

values for x, choose the value for x that gives a physically realistic answer. For example, reject the value

of x that results in any negative concentrations.

Step 7 Check your answer by substituting the calculated equilibrium values into the equilibrium

expression. The calculated value of K should match the given value of K. Note that rounding errors

could cause a difference in the least significant digit when comparing values of the equilibrium constant.

Continued



Chapter 14 Lecture


Since the calculated value of Kc matches the given value (to within one digit in the least significant

figure), the answer is valid.

For Practice 14.9 The reaction in Example 14.9 is carried out at a different temperature at which Kc = 0.055. This time, however, the

reaction mixture starts with only the product, [NO] = 0.0100 M, and no reactants. Find the equilibrium concentrations

of N2, O2, and NO at equilibrium.

Continued



Chapter 14 Lecture



A reaction mixture at 100 °C initially contains [NO2] = 0.100 M. Find the equilibrium concentrations of NO2 and

N2O4 at this temperature.

Procedure For… Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Constant

Step 1 Using the balanced equation as a guide, prepare a table showing the known initial concentrations

of the reactants and products. Leave room in the table for the changes in concentrations and for the

equilibrium concentrations.



Chapter 14 Lecture


Step 2 Use the initial concentrations to calculate the reaction quotient (Q) for the initial concentrations.

Compare Q to K to predict the direction in which the reaction will proceed.

Q > K; therefore, the reaction will proceed to the left.

Step 3 Represent the change in the concentration of one of the reactants or products with the variable

x. Define the changes in the concentrations of the other reactants or products in terms of x. It is

usually most convenient to let x represent the change in concentration of the reactant or product with

the smallest stoichiometric coefficient.

Continued



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Chapter 14 Lecture


Step 4 Sum each column for each reactant and each product to determine the equilibrium concentrations

in terms of the initial concentrations and the variable x.

Step 5 Substitute the expressions for the equilibrium concentrations (from step 4) into the expression for

the equilibrium constant. Using the given value of the equilibrium constant, solve the expression

for the variable x. In some cases, such as Example 14.9, you can take the square root of both sides of

the expression to solve for x. In other cases, such as Example 14.10, you must solve a quadratic

equation to find x.

Remember the quadratic formula:

Continued



Chapter 14 Lecture


Continued



Chapter 14 Lecture


Step 6 Substitute x into the expressions for the equilibrium concentrations of the reactants and products

(from step 4) and calculate the concentrations. In cases where you solved a quadratic and have two

values for x, choose the value for x that gives a physically realistic answer. For example, reject the value

of x that results in any negative concentrations.

We reject the root x = 0.176 because it gives a negative concentration for NO2. Using x = 0.014, we get

the following concentrations:

Step 7 Check your answer by substituting the calculated equilibrium values into the equilibrium

expression. The calculated value of K should match the given value of K. Note that rounding errors

could cause a difference in the least significant digit when comparing values of the equilibrium constant.

Continued



Chapter 14 Lecture


Since the calculated value of Kc matches the given value (to within one digit in the least significant

figure), the answer is valid.

For Practice 14.10 The reaction in Example 14.10 is carried out at the same temperature, but this time the reaction mixture initially

contains only the reactant, [N2O4] = 0.0250 M, and no NO2. Find the equilibrium concentrations of N2O4 and NO2.

Continued



Chapter 14 Lecture


Solution Follow the procedure used in Example 14.5 and 14.6 (using partial pressures in place of concentrations) to solve the

problem.

1. Using the balanced equation as a guide, prepare a table showing the known initial partial pressures of the reactants

and products.


A reaction mixture at 25 °C initially contains = 0.100 atm, = 0.100 atm, and PICl = 0.100 atm. Find the

equilibrium partial pressures of I2, Cl2, and ICl at this temperature.

Example 14.11 Finding Equilibrium Partial Pressures When You Are

Given the Equilibrium Constant and Initial Partial

Pressures

Chapter 14 Lecture


2. Use the initial partial pressures to calculate the reaction quotient (Q). Compare Q to K to predict the direction in

which the reaction will proceed.

Q < K; therefore, the reaction will proceed to the right.

3. Represent the change in the partial pressure of one of the reactants or products with the variable x. Define the

changes in the partial pressures of the other reactants or products in terms of x.

Continued



Pressures

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Chapter 14 Lecture


4. Sum each column for each reactant and product to determine the equilibrium partial pressures in terms of the initial

partial pressures and the variable x.

5. Substitute the expressions for the equilibrium partial pressures (from step 4) into the expression for the equilibrium

constant. Use the given value of the equilibrium constant to solve the expression for the variable x.

Continued



Pressures

Chapter 14 Lecture


6. Substitute x into the expressions for the equilibrium partial pressures of the reactants and products (from step 4)

and calculate the partial pressures.

7. Check your answer by substituting the calculated equilibrium partial pressures into the equilibrium expression.

The calculated value of K should match the given value of K.

Since the calculated value of Kp matches the given value (within the uncertainty indicated by the significant

figures), the answer is valid.

Continued



Pressures

Chapter 14 Lecture


For Practice 14.11 The reaction between I2 and Cl2 in Example 14.11 is carried out at the same temperature, but with these initial

partial pressures: = 0.150 atm, = 0.150 atm, PICl = 0.00 atm. Find the equilibrium partial pressures of all

three substances.

Continued



Pressures

Chapter 14 Lecture


Consider the reaction for the decomposition of hydrogen disulfide:

A 0.500 L reaction vessel initially contains 0.0125 mol of H2S at 800 °C. Find the equilibrium concentrations of

H2 and S2.

Procedure For… Finding Equilibrium Concentrations from Initial Concentrations in Cases with a Small Equilibrium Constant

Step 1 Using the balanced equation as a guide, prepare a table showing the known initial concentrations of the

reactants and products. (In these examples, you must first calculate the concentration of H2S from the

given number of moles and volume.)


Concentrations in Cases with a Small Equilibrium

Constant

Chapter 14 Lecture


Step 2 Use the initial concentrations to calculate the reaction quotient (Q). Compare Q to K to predict the

direction in which the reaction will proceed.

By inspection, Q = 0; the reaction will proceed to the right.

Step 3 Represent the change in the concentration of one of the reactants or products with the variable x.

Define the changes in the concentrations of the other reactants or products with respect to x.

Continued



Constant

Chapter 14 Lecture


Step 4 Sum each column for each reactant and product to determine the equilibrium concentrations in terms

of the initial concentrations and the variable x.

Step 5 Substitute the expressions for the equilibrium concentrations (from step 4) into the expression for the

equilibrium constant. Use the given value of the equilibrium constant to solve the resulting equation

for the variable x.

In this case, the resulting equation is cubic in x. Although cubic equations can be solved, the solutions

are not usually simple. However, since the equilibrium constant is small, we know that the reaction

does not proceed very far to the right. Therefore, x will be a small number and can be dropped from

any quantities in which it is added to or subtracted from another number (as long as the number itself

is not too small).

Continued



Constant

2/24/2014

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Chapter 14 Lecture


Continued



Constant

Chapter 14 Lecture


Check whether your approximation was valid by comparing the calculated value of x to the number it

was added to or subtracted from. The ratio of the two numbers should be less than 0.05 (or 5%) for the

approximation to be valid. If approximation is not valid, proceed to step 5a.

Checking the x is small approximation:

The x is small approximation is valid, proceed to step 6.

Step 5a If the approximation is not valid, you can either solve the equation exactly (by hand or with your

calculator), or use the method of successive approximations. In this case, we use the method of

successive approximations.

Substitute the value obtained for x in step 5 back into the original cubic equation, but only at the exact

spot where x was assumed to be negligible, and then solve the equation for x again. Continue this

procedure until the value of x obtained from solving the equation is the same as the one that is

substituted into the equation.

Continued



Constant

Chapter 14 Lecture


Step 6 Substitute x into the expressions for the equilibrium concentrations of the reactants and products (from

step 4) and calculate the concentrations.

Step 7 Check your answer by substituting the calculated equilibrium values into the equilibrium expression.

The calculated value of K should match the given value of K. Note that the approximation method and

rounding errors could cause a difference of up to about 5% when comparing values of the equilibrium

constant.

The calculated value of K is close enough to the given value when we consider the uncertainty

introduced by the approximation. Therefore the answer is valid.

Continued



Constant

Chapter 14 Lecture


For Practice 14.12 The reaction in Example 14.12 is carried out at the same temperature with the following initial concentrations:

[H2S] = 0.100 M, [H2] = 0.100 M, and [S2] = 0.00 M. Find the equilibrium concentration of [S2].

Continued



Constant

Chapter 14 Lecture


Consider the reaction for the decomposition of hydrogen disulfide:

A 0.500 L reaction vessel initially contains 1.25 × 10–4 mol of H2S at 800 °C. Find the equilibrium concentrations

of H2 and S2.

Procedure For… Finding Equilibrium Concentrations from Initial Concentrations in Cases with a Small Equilibrium Constant

Step 1 Using the balanced equation as a guide, prepare a table showing the known initial concentrations of the

reactants and products. (In these examples, you must first calculate the concentration of H2S from the

given number of moles and volume.)



Constant

Chapter 14 Lecture


Step 2 Use the initial concentrations to calculate the reaction quotient (Q). Compare Q to K to predict the

direction in which the reaction will proceed.

By inspection, Q = 0; the reaction will proceed to the right.

Step 3 Represent the change in the concentration of one of the reactants or products with the variable x.

Define the changes in the concentrations of the other reactants or products with respect to x.

Continued



Constant

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Chapter 14 Lecture


Step 4 Sum each column for each reactant and product to determine the equilibrium concentrations in terms

of the initial concentrations and the variable x.

Step 5 Substitute the expressions for the equilibrium concentrations (from step 4) into the expression for the

equilibrium constant. Use the given value of the equilibrium constant to solve the resulting equation

for the variable x.

In this case, the resulting equation is cubic in x. Although cubic equations can be solved, the solutions

are not usually simple. However, since the equilibrium constant is small, we know that the reaction

does not proceed very far to the right. Therefore, x will be a small number and can be dropped from

any quantities in which it is added to or subtracted from another number (as long as the number itself

is not too small).

Continued



Constant

Chapter 14 Lecture


Continued



Constant

Chapter 14 Lecture


Check whether your approximation was valid by comparing the calculated value of x to the number it

was added to or subtracted from. The ratio of the two numbers should be less than 0.05 (or 5%) for the

approximation to be valid. If approximation is not valid, proceed to step 5a.

Checking the x is small approximation:

The approximation does not satisfy the <5% rule (although it is close).

Step 5a If the approximation is not valid, you can either solve the equation exactly (by hand or with your

calculator), or use the method of successive approximations. In this case, we use the method of

successive approximations.

Substitute the value obtained for x in step 5 back into the original cubic equation, but only at the exact

spot where x was assumed to be negligible, and then solve the equation for x again. Continue this

procedure until the value of x obtained from solving the equation is the same as the one that is

substituted into the equation.

Continued



Constant

Chapter 14 Lecture


If we substitute this value of x back into the cubic equation and solve it, we get x = 1.28 × 10–5, which

is nearly identical to 1.27 × 10–5. Therefore, we have arrived at the best approximation for x.

Step 6 Substitute x into the expressions for the equilibrium concentrations of the reactants and products (from

step 4) and calculate the concentrations.

Continued



Constant

Chapter 14 Lecture


Step 7 Check your answer by substituting the calculated equilibrium values into the equilibrium expression.

The calculated value of K should match the given value of K. Note that the approximation method and

rounding errors could cause a difference of up to about 5% when comparing values of the equilibrium

constant.

The calculated value of K is equal to the given value. Therefore the answer is valid.

For Practice 14.13 The reaction in Example 14.13 is carried out at the same temperature with the following initial concentrations:

[H2S] = 1.00 × 10–4 M, [H2] = 0.00 M, and [S2] = 0.00 M. Find the equilibrium concentration of [S2].

Continued



Constant

Chapter 14 Lecture


Consider the following reaction at equilibrium:

What is the effect of adding additional CO2 to the reaction mixture? What is the effect of adding additional CaCO3?

Example 14.14 The Effect of a Concentration Change on Equilibrium

Solution Adding additional CO2 increases the concentration of CO2 and causes the reaction to shift to the left. Adding additional

CaCO3, however, does not increase the concentration of CaCO3 because CaCO3 is a solid and therefore has a constant

concentration. Thus, adding additional CaCO3 has no effect on the position of the equilibrium. (Note that, as we saw in

Section 14.5, solids are not included in the equilibrium expression.)

For Practice 14.14 Consider the following reaction in chemical equilibrium:

What is the effect of adding additional Br2 to the reaction mixture? What is the effect of adding additional BrNO?

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Chapter 14 Lecture


Consider the following reaction at chemical equilibrium:

What is the effect of decreasing the volume of the reaction mixture? Increasing the volume of the reaction mixture?

Adding an inert gas at constant volume?

Example 14.15 The Effect of a Volume Change on Equilibrium

Solution The chemical equation has 3 mol of gas on the right and zero moles of gas on the left. Decreasing the volume of the

reaction mixture increases the pressure and causes the reaction to shift to the left (toward the side with fewer moles of

gas particles). Increasing the volume of the reaction mixture decreases the pressure and causes the reaction to shift to

the right (toward the side with more moles of gas particles.) Adding an inert gas has no effect.

For Practice 14.15 Consider the following reaction at chemical equilibrium:

What is the effect of decreasing the volume of the reaction mixture? Increasing the volume of the reaction mixture?

Chapter 14 Lecture


The following reaction is endothermic:

What is the effect of increasing the temperature of the reaction mixture? Decreasing the temperature?

Solution Since the reaction is endothermic, we can think of heat as a reactant:

Raising the temperature is equivalent to adding a reactant, causing the reaction to shift to the right. Lowering the

temperature is equivalent to removing a reactant, causing the reaction to shift to the left.

For Practice 14.16 The following reaction is exothermic:

What is the effect of increasing the temperature of the reaction mixture? Decreasing the temperature?

Example 14.16 The Effect of a Temperature Change on Equilibrium

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