ch10 reaction rates and equilibrium -...

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General, Organic, and Biological Chemistry: Structures of Life, 5/eKaren C. Timberlake

© 2016 Pearson Education, Inc.

Karen C. Timberlake

Lecture Presentation

Chapter 10

Reaction Rates and Chemical Equilibrium



A neonatal nurse works with newborns that are premature or have birth defects, cardiac malformations, and surgical problems.

Most neonatal nurses care for infants from the time of birth until the time they are discharged from the hospital.

Chapter 4 Reaction Rates and Chemical Equilibrium



Chapter 10 Readiness

Key Math Skills

• Solving Equations (1.4D)

• Writing Numbers in Scientific Notation (1.4F)

Core Chemistry Skills

• Using Significant Figures in Calculations (2.3)

• Balancing a Chemical Equation (7.1)

• Calculating Concentration (9.4)



10.1 Rates of Reactions

Reaction rates vary greatly for everyday processes. A banana ripens in a few days,

silver tarnishes in a few months, while the aging process of humans takes many years.

Learning Goal Describe how temperature, concentration, and catalysts affect the rate of a reaction.



10.1 Rates of Reactions

Reaction Rate:

• The speed at which a chemical reaction occurs

• Can be measured various ways

– The change in the concentration of a

reactant over time

– The change in the

concentration of a product over time



Collision Theory of Chemical Reactions

In order for a reaction to occur, reacting particles must collide, have a minimum amount of energy, and have the proper orientation to form products.

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Activation Energy

• Even when a collision has the proper orientation, there still must be sufficient energy to break the bonds between the atoms of the reactants.

Three Conditions Required for a Reaction to Occur

1. Collision The reactants must collide.

2. Orientation The reactants must align properly to break and form bonds.

3. Energy The collision must provide the energy of activation.



Activation Energy

• The activation energy is the minimum amount of energy required to break the bonds between atoms of the reactants.



Rate of Reaction

The rate (or speed) of a reaction is determined by measuring the amount of

• reactant used up in a certain period of time.

• product formed in a certain period of time.



Factors That Affect the Rate of a Reaction

Reactions with low activation energies go faster than reactions with high activation energies.

For any reaction, the rate is affected by

• changes in temperature.

• changes in reaction concentration.

• adding a catalyst.



Rate of Reaction: Effect of Temperature

At higher temperatures, molecules have more kinetic energy. The increase in kinetic energy of

the reactant molecules means

• they move faster, so

• they collide more often, and

• they collide with more energy.

For every 10 °C increase in temperature, most reaction rates approximately double.



Reaction Rate: Effect of Concentration

When there are more reacting

molecules, more collisions that form

products can occur, and the reaction

goes faster.

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Rate of Reaction: Catalysts

• Adding a catalyst speeds up the rate of the reaction by providing an alternative pathway that has a lower activation energy.

• When activation energy is lowered, more collisions provide sufficient energy for reactants to form product.

• During a reaction, a catalyst is not changed or consumed.



Factors That Affect Reaction Rates



Study Check

Indicate the effect of each factor listed on the rate of the reaction as (I) increases, (D) decreases, or

(N) no change.

2CO(g) + O2(g) � 2CO2 (g)

A. raising the temperature

B. removing O2

C. adding a catalyst

D. lowering the temperature



Solution

Indicate the effect of each factor listed on the rate of the reaction as (I) increases, (D) decreases, or

(N) no change.

2CO(g) + O2(g) � 2CO2 (g)

A. raising the temperature (I) increases

B. removing O2 (D) decreases

C. adding a catalyst (I) increases

D. lowering the temperature (D) decreases



Study Check

State the effect of each on the rate of reaction as (I) increases, (D) decreases, or (N) no change.

A. decreasing the temperature

B. removing one of the reactants

C. adding a catalyst

D. placing the reaction flask in ice

E. increasing the concentration of a reactant



Study Check

State the effect of each on the rate of reaction as (I) increases, (D) decreases, or (N) no change.

A. decreasing the temperature (D) decreases

B. removing one of the reactants (D) decreases

C. adding a catalyst (N) no change

D. placing the reaction flask in ice (D) decreases

E. increasing the concentration of areactant (I) increases

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Catalytic converters are used in automobile engines to reduce pollutants such as carbon monoxide (CO), hydrocarbons such as octane (C8H18), and nitrogen oxide (NO).

Chemistry Link to the Environment:Catalytic Converters



When pollutants pass through the surface, they react with the catalysts and are converted to CO2, N2, O2, and H2O.

Chemistry Link to the Environment:Catalytic Converters



10.2 Chemical Equilibrium

In most chemical reactions, the reactants are not completely converted to products because a reverse

reaction takes place in which products collide to form the reactants.

Learning Goal Use the concept of reversible reactions to explain chemical equilibrium.



Reversible Reactions

When a reaction proceeds in both a forward and a reverse direction, it is said to be a reversible reaction.

As the reactants, H2 and I2, collide, the forward reaction begins.

HI molecules begin to form and collide with each other to form reactants in the reverse reaction. This reversible reaction is written with a double arrow.

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

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

reverse



Reversible Reactions

A reversible reaction

• occurs in both the forward and reverse direction at the same time.

forward

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

• has two rates, a rate for the forward reaction

and a rate for the reverse reaction.



Study Check

Write the forward and reverse reactions for the following:

CH4(g) + 2H2S(g) CS2(g) + 4H2(g)

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Solution

Write the forward and reverse reactions for the following:

CH4(g) + 2H2S(g) CS2(g) + 4H2(g)

The forward reaction isCH4(g) + 2H2S(g) CS2(g) + 4H2(g)

The reverse reaction isCS2(g) + 4H2(g) CH4(g) + 2H2S(g)



Rate of Reversible Reactions

As the reaction progresses, the rate of the forward reaction decreases and that of the reverse reaction increases. At equilibrium, the rates of the forward and reverse reactions are equal.



Equilibrium and Reversible Reactions

Equilibrium is reached when there are no further changes in the concentrations of reactants and products.



Rates of Forward and Reverse Reactions



Equilibrium

In the reaction

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

• the forward reaction is H2(g) + I2(g) 2HI(g).

• the reverse reaction is 2HI(g) H2(g) + I2(g).

As HI product builds up, the rate of the reverse reaction increases, while the rate of the forward reaction decreases.



Equilibrium

A reaction reaches equilibrium when no further changes take place in the concentration of the reactants and products.

At equilibrium,

• the rate of the forward reaction is equal to the rate of the reverse reaction.

• the forward and reverse reactions continue at the same rate.

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Study Check

Complete each of the following with is/are equal or is/are not equal, change(s) or do(es) not change.

A. Before equilibrium is reached, the concentrations

of the reactants and products ______.

B. At equilibrium, the rate of the forward reaction ______ to the rate of the reverse reaction.



Forward and Reverse Reactions

• If we start with reactants SO2 and O2, the reaction to form SO3 takes place until equilibrium is

reached.

• If we start with only the product SO3, the reaction to

form SO2 and O2 takes place until equilibrium is reached.



Study Check

Complete each sentence with

1) equal 2) not equal 3) forward 4) reverse 5) changes 6) does not change

A. Reactants form products in the _______ reaction.

B. At equilibrium, the reactant concentration _______.

C. Products form reactants in the _______ reaction.



Solution

Complete each sentence with

1) equal 2) not equal 3) forward 4) reverse 5) changes 6) does not change

A. Reactants form products in the forward reaction.

B. At equilibrium, the reactant concentration does not change.

C. Products form reactants in the reverse reaction.



10.3 Equilibrium Constants

At equilibrium, the number ofpeople riding up the lift and thenumber of people skiing down the slope are constant.

Learning Goal Calculate the equilibrium constant for a reversible reaction given the concentrations of reactants and products at equilibrium.



Equilibrium Constants

An equilibrium constant for a reversible chemical reaction

• multiplies the concentrations of the products together and divides by the concentrations of the reactants.

• raises the concentration (moles/liter) of each species to a power that is equal to its coefficient in the balanced chemical equation.

Core Chemistry Skill Writing the Equilibrium Constant Expression

aA + bB cC + dD

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Guide to Writing an Equilibrium Expression



Study Check

Write an equilibrium expression for the following reaction:

2CO(g) + O2(g) 2CO2(g)



Solution


2CO(g) + O2(g) 2CO2(g)

STEP 1 Write the balanced chemical equation. 2CO(g) + O2(g) 2CO2(g)

STEP 2 Write the concentrations of the products as the numerator and the reactants as thedenominator.



Solution


2CO(g) + O2(g) 2CO2(g)

STEP 3 Write any coefficient in the equation as an exponent.



Calculating Equilibrium Constants

The equilibrium constant, Kc, is the numerical value obtained by substituting experimentally measured molar concentrations at equilibrium into the expression.

For example,

• the Kc for the reaction of H2 and I2 is written as

• H2(g) + I2(g) 2HI(g)

• the Kc expression is written as

Core Chemistry Skill Calculating an Equilibrium Constant




For example,

• the concentrations of each species in Experiment 1 are[H2] = 0.10 M, [I2] = 0.20 M, and [HI] = 1.04 M, and the Kc is

• in additional Experiments 2 and 3, the mixtures have different equilibrium concentrations for the system at equilibrium at the same temperature, but they have the same value of Kc.

Thus, a reaction at a specific temperature can have only one value for the equilibrium constant.

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The units of Kc depend on the specific equation. In this example, the units of [M2]/[M2] cancel out to give a value of 54. In this text, the numerical value will be given without any units.



Guide to Calculating the Kc



Study Check

The decomposition of dinitrogen tetroxide forms nitrogen dioxide.

N2O4(g) 2NO2(g)

What is the numerical value of Kc at 100 °C if a reaction mixture at equilibrium contains 0.45 M N2O4 and 0.31 M NO2?



Solution


N2O4(g) 2NO2(g)

What is the numerical value of Kc at 100 °C if a reaction mixture at equilibrium contains 0.45 M N2O4 and 0.31 M NO2?

STEP 1 State the given and needed quantities.

ANALYZE Given NeedTHE PROBLEM 0.45 M N2O4 Kc

0.31 M NO2

Equation

N2O4(g) 2NO2(g)

ANALYZE Given NeedTHE PROBLEM 0.45 M N2O4 Kc

0.31 M NO2

Equation

N2O4(g) 2NO2(g)



Solution


N2O4(g) 2NO2(g)

What is the numerical value of Kc at 100 °C if a reactionmixture at equilibrium contains 0.45 M N2O4 and 0.31 M NO2?

STEP 2 Write the Kc expression for the equilibrium.

STEP 3 Substitute equilibrium (molar)concentrations and calculate Kc.



Study Check

Given the following chemical reaction,

H2(g) + I2(g) 2HI(g)what is the numerical value of Kc at 443 °C if the equilibrium concentrations are as follows?

[H2] = 1.2 M [I2] = 1.2 M [HI] = 0.35 M

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Solution



[H2] = 1.2 M [I2] = 1.2 M [HI] = 0.35 M


ANALYZE Given NeedTHE PROBLEM 1.2 M H2, 1.2 M I2, Kc

0.35 M HI

EquationH2(g) + I2(g) 2HI(g)

ANALYZE Given NeedTHE PROBLEM 1.2 M H2, 1.2 M I2, Kc

0.35 M HI

EquationH2(g) + I2(g) 2HI(g)



Solution



[H2] = 1.2 M [I2] = 1.2 M [HI] = 0.35 M

STEP 2 Write the Kc expression for equilibrium.



Solution



[H2] = 1.2 M [I2] = 1.2 M [HI] = 0.35 M

STEP 3 Substitute equilibrium (molar) concentrationsand calculate Kc.



In the reaction of SO2(g) and O2(g), the equilibrium mixture contains mostly product SO3(g), which results in a large Kc.

Learning Goal Use an equilibrium constant to predict the extent of the reaction and to calculate equilibrium concentrations.

10.4 Using Equilibrium Constants



The values of Kc can be large or small, depending on whether equilibrium is reached with

• more products than reactants.

• more reactants than products.

However, the size of the equilibrium constant does not affect how fast equilibrium is reached.

Using Equilibrium Constants



Reactions with a large Kc have large amounts of products produced from the forward reaction at equilibrium.

The equilibrium constant for the reaction of SO2 and O2

has a large Kc. At equilibrium, the reaction mixture contains mostly product and few reactants.

2SO2(g) + O2(g) 2SO3(g)

Equilibrium with a Large Kc

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Reactions with a small Kc have an equilibrium mixture with a low concentration of products and high concentration of reactants.

The equilibrium constant for the reaction of N2 and O2 has a small Kc. At equilibrium, the reaction mixture contains mostly reactants and few products.

N2(g) + O2(g) 2NO(g)

Equilibrium with a Small Kc



The equilibrium mixture contains a very small amount of the product NO and a large amount of the reactants N2 and O2, which results in a small Kc.

Equilibrium with a Small Kc



A few reactions have equilibrium constants close to 1, which means they have about equal concentrations of reactants and products at equilibrium.

At equilibrium, a reaction with a large Kc contains mostly products, whereas a reaction with a small Kc contains mostly reactants.

Equilibrium with a Kc close to 1



Equilibrium Constants: Kc



Guide to Using the Equilibrium Constant

Core Chemistry Skill Calculating Equilibrium Concentrations



For the reaction of carbon dioxide and hydrogen, the equilibrium concentrations are 0.25 M CO2, 0.80 M H2, and 0.50 M H2O. What is the equilibrium concentration of CO(g)?

CO2(g) + H2(g) CO(g) + H2O(g) Kc = 0.11


Calculating Concentrations at Equilibrium

ANALYZE Given NeedTHE 0.25 M CO2, 0.80 M H2, [CO]

PROBLEM 0.50 M H2OEquationCO2(g) + H2(g) CO(g) + H2O(g) Kc= 0.11

ANALYZE Given NeedTHE 0.25 M CO2, 0.80 M H2, [CO]

PROBLEM 0.50 M H2OEquationCO2(g) + H2(g) CO(g) + H2O(g) Kc= 0.11

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CO2(g) + H2(g) CO(g) + H2O(g) Kc = 0.11

STEP 2 Write the Kc expression for the equilibrium and solve for the needed concentration.





CO2(g) + H2(g) CO(g) + H2O(g) Kc = 0.11

STEP 3 Substitute the equilibrium (molar) concentrations and calculate the needed concentration.




For each Kc, indicate whether the reaction mixture at equilibrium contains mostly reactants or products.

A. H2(g) + F2(g) 2HF(g) Kc = 1 × 1095

B. 3O2(g) 2O3(g) Kc = 1.8 × 10−7

Study Check



For each Kc, indicate whether the reaction mixture at equilibrium contains mostly reactants or products.

A. H2(g) + F2(g) 2HF(g) Kc = 1 × 1095

Kc is large; products are favored at equilibrium.

B. 3O2(g) 2O3(g) Kc = 1.8 × 10−7

Kc is small; reactants are favored at equilibrium.

Solution



At equilibrium, the reaction

PCl5(g) PCl3(g) + Cl2(g) Kc = 4.2 × 10−2

has the concentrations [PCl3] = [Cl2] = 0.10 M. What is the concentration of PCl5 at equilibrium?

Study Check




PCl5(g) PCl3(g) + Cl2(g) Kc = 4.2 × 10−2



Solution

ANALYZE Given NeedTHE [PCl3] = [Cl2] = 0.10 M [PCl5]

PROBLEMEquationPCl5(g) PCl3(g) + Cl2(g) Kc = 4.2 ×

10−2

ANALYZE Given NeedTHE [PCl3] = [Cl2] = 0.10 M [PCl5]

PROBLEMEquationPCl5(g) PCl3(g) + Cl2(g) Kc = 4.2 ×

10−2

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PCl5(g) PCl3(g) + Cl2(g) Kc = 4.2 × 10−2


STEP 2 Write the Kc expression for the equilibrium and solve for the needed concentration.

Solution




PCl5(g) PCl3(g) + Cl2(g) Kc = 4.2 × 10−2


STEP 3 Substitute the equilibrium (molar)concentrations and calculate the needed concentration.

Solution



10.5 Changing Equilibrium Conditions: Le Châtelier’s Principle

The transport of oxygen involves an equilibrium between hemoglobin (Hb), oxygen, and oxyhemoglobin (HbO2).

Hb(aq) + O2(g) HbO2(aq)

For a person living at higher altitudes, hemoglobin concentration increases, causing a shift in the equilibrium back in the direction of HbO2 product.

Learning Goal Use Le Châtelier’s Principle to describe the

changes made in equilibrium concentrations when the reaction conditions change.

Hypoxia may occur at high altitudes where the oxygen concentration is lower.



Le Châtelier’s Principle

When the conditions of a reaction at equilibrium are changed, the forward and reverse reactions will no longer be equal.

Le Châtelier’s principle states that when a

stress is placed on a reaction at equilibrium, the system responds by changing the rate of the

forward or reverse reaction in the direction that relieves that stress.



Le Châtelier’s Principle

The stress of adding water to tank A increases the rate of the forward direction to reestablish equal water levels and equilibrium.

Core Chemistry Skill Using Le Châtelier’s Principle



Effect of Concentration Change on Equilibrium: Adding Reactant

Consider the following reaction at equilibrium:

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

If more reactant (H2 or I2) is added,

• the rate of the forward reaction increases to form more product until the system is again at equilibrium.

• the equilibrium shifts toward the products.

Add H2Add H2

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Effect of Concentration Change on Equilibrium



Effect of Concentration Change on Equilibrium: Removing Reactant


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

If some of a reactant (H2 or I2) is removed,

• the rate of the reverse reaction increases to form

more reactant until the equilibrium is reached.

• the equilibrium shifts toward the reactants.

Remove H2Remove H2



Effect of Concentration Change on Equilibrium: Adding Product


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

If more product (HI) is added,

• the rate of the reverse reaction increases to form more H2 and I2 reactants.

• the equilibrium shifts toward the reactants.

Add HIAdd HI



Effect of Concentration Change on Equilibrium: Removing Product


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

When some of the product (HI) is removed,

• there is an decrease in collisions of HI molecules.

• the rate of the forward reaction increases and forms more product (HI).

Remove HIRemove HI



Effect of Concentration Change on Equilibrium



Effect of a Catalyst on Equilibrium

• Adding a catalyst speeds up a reaction by lowering the activation energy, thus increasing the rate of the forward and reverse reactions.

• The time to reach equilibrium is shorter; however, the same ratios of reactants and products are present.

• The addition of a catalyst does not change the equilibrium mixture.

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Effect of Volume Change on Equilibrium: Decreasing Volume

A change in the volume of a gas mixture at equilibrium will change the concentration of the gases in the mixture.

2CO(g) + O2(g) 2CO2(g)

Decreasing the volume increases the concentration of the gases, and the system shifts in the direction of the smaller number of moles to compensate.

Decrease VolumeDecrease Volume



Effect of Volume Change on Equilibrium: Increasing Volume

A change in the volume of a gas mixture at equilibrium will change the concentration of the gases in the mixture.

2CO(g) + O2(g) 2CO2(g)

Increasing the volume decreases the concentration of the gases, and the system shifts in the direction of the larger number of moles to compensate.

Increase volumeIncrease volume



Effect of Volume Change on Equilibrium

(a) A decrease in the volume of the container causes the system to shift in the direction of fewer moles of gas. (b) An increase in the volume of the container causes the system to shift in the direction of more moles of gas.



Chemistry Link to Health: Hb

Oxygen transport involves an equilibrium between hemoglobin (Hb), oxygen, and oxyhemoglobin (HbO2).


• When there is a high concentration of O2 in the alveoli of the lungs, the reaction shifts in the direction of oxyhemoglobin.

• When the concentration of O2 is low in the tissues, the reverse reaction releases O2 from oxyhemoglobin.




Given the reaction of hemoglobin,


• At normal atmospheric pressure, oxygen diffuses into the blood because the partial pressure of oxygen in the alveoli is higher than that in the blood.

• At altitudes above 8000 ft, a decrease in atmospheric pressure results in a lower pressure of O2.




At an altitude of 18,000 ft, a person will obtain 29% less oxygen and may experience hypoxia.

Hypoxia may occur at high altitudes where the oxygen concentration is lower.

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According to Le Châtelier’s principle, a decrease in oxygen

• shifts the equilibrium in the direction of the reactants.

• depletes the concentration of HbO2, causing hypoxia.


Remove O2Remove O2



Effect of Temperature Change on Equilibrium: Decreasing Temperature

Decreasing the temperature of an endothermic reaction

• causes the system to respond by shifting the reaction toward more heat.

• shifts the reaction toward the reactants, increasing heat in system.

N2(g) + O2(g) + heat 2NO(g)

Decrease TemperatureDecrease Temperature



Effect of Temperature Change on Equilibrium: Increasing Temperature

Increasing the temperature of an endothermic reaction

• causes the system to respond by shifting the reaction to remove heat.

• shifts the reaction toward the products, using up the heat.

H2(g) + O2(g) + heat 2NO(g)

Increase Temperature



Effect of Temperature Change on Equilibrium: Decreasing Temperature

Decreasing the temperature of an exothermic reaction

• causes the system to respond by shifting the reaction toward more heat.

• shifts the reaction toward the products, increasing heat in system.

2SO2(g) + O2(g) 2SO3(g) + heat

Decrease TemperatureDecrease Temperature



Effect of Temperature Change on Equilibrium: Increasing Temperature

Increasing the temperature of an exothermic reaction

• causes the system to respond by shifting the reaction toward removing heat.

• shifts the reaction toward the reactants, decreasing heat in system.

2SO2(g) + O2(g) 2SO3(g) + heat

Increase Temperature



Effects of Changing Conditions on Equilibrium

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Study Check

Indicate the shift in equilibrium of each change.

2NO2(g) + heat 2NO(g) + O2(g)

1) toward products 2) toward reactants

A. adding more NO

B. decreasing the temperature

C. removing some O2

D. increasing the volume

E. removing some NO



Study Check

Indicate the shift in equilibrium of each change.

2NO2(g) + heat 2NO(g) + O2(g)

1) toward products 2) toward reactants

A. adding more NO 2) toward reactants

B. decreasing the temperature 2) toward reactants

C. removing some O2 1) toward products

D. increasing the volume 1) toward products

E. removing some NO 1) toward products



Concept Map

ch10 reaction rates and equilibrium -...

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