© 2013 pearson education, inc. chapter 9, section 1 general, organic, and biological chemistry...
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© 2013 Pearson Education, Inc. Chapter 9, Section 1
General, Organic, and Biological Chemistry
Fourth EditionKaren Timberlake
Chapter 9
© 2013 Pearson Education, Inc.
Reaction Rates and Chemical Equilibrium
9.1Rates of Reactions
Lectures
© 2013 Pearson Education, Inc. Chapter 9, Section 1
Collision Theory of Reactions
A chemical reaction occurs when collisions between reactant molecules have sufficient
energy to break their bonds. molecules collide with the proper orientation. bonds between atoms of the reactants (N2 and O2)
are broken and new bonds (NO) form.
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© 2013 Pearson Education, Inc. Chapter 9, Section 1
Collision Theory of Reactions
A chemical reaction does not take place if the collisions between reactant molecules do not have
sufficient energy to break their bonds, or if molecules are not properly aligned.
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© 2013 Pearson Education, Inc. Chapter 9, Section 1
Activation Energy
Activation energy is the minimum energy needed for a reaction to take place or break the bonds between atoms of reactants.
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© 2013 Pearson Education, Inc. Chapter 9, Section 1
Reaction Rate
The rate of reaction is determined by measuring theamount of reactant used up or product formed in acertain period of time.
The rate or reaction is affected by temperature changes. changes in concentrations. the presence of a catalyst.
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© 2013 Pearson Education, Inc. Chapter 9, Section 1
Reaction Rate and Temperature
At higher temperatures, the increase in kinetic energy causes the reacting
molecules to move faster, more collisions occur, and more colliding molecules have sufficient energy to
react and form products.
For example, we refrigerate perishable foods to slowdown the reactions that cause spoilage.
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© 2013 Pearson Education, Inc. Chapter 9, Section 1
Reaction Rate and Concentration
Increasing theconcentration of reactants increases the number of
collisions and increases the reaction
rate.
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© 2013 Pearson Education, Inc. Chapter 9, Section 1
Reaction Rate and Catalysts
A catalyst speeds up the rate of a reaction. lowers the energy of activation. is not used up during the reaction.
Insert reaction rate and catalysis top pg 5
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© 2013 Pearson Education, Inc. Chapter 9, Section 1
Learning Check
State the effect of each change on the rate of reactionas increases, decreases or no change.
1. increasing the temperature
2. removing some of the reactants
3. adding a catalyst
4. placing the reaction flask in ice
5. increasing the concentration of a reactant
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© 2013 Pearson Education, Inc. Chapter 9, Section 1
Learning Check
Indicate the effect of each factor listed on the rate of thefollowing reaction as increases, decreases, or does not change.
A. raising the temperature
B. removing some O2
C. adding a catalyst
D. lowering the temperature
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© 2013 Pearson Education, Inc. Chapter 9, Section 2
General, Organic, and Biological Chemistry
Fourth EditionKaren Timberlake
9.2Chemical Equilibrium
Chapter 9Reaction Rates and
Chemical Equilibrium
© 2013 Pearson Education, Inc.Lectures
© 2013 Pearson Education, Inc. Chapter 9, Section 2
Reversible Reactions
A reversible reaction proceeds in both the forward andreverse directions. As a result there are two reactionrates: the rate of the forward reaction and the rate of thereverse reaction.
When molecules begin to react, the rate of the forward reaction is faster than the rate of the reverse reaction.
As reactants are consumed and products accumulate, the rate of the forward reaction decreases, whereas the rate of the reverse reaction increases.
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© 2013 Pearson Education, Inc. Chapter 9, Section 2 14
Reversible Reactions
Suppose SO2 and O2 are present initially. As they collide,the forward reaction begins.
2SO2(g) + O2(g) 2SO3(g)
As SO3 molecules form, they also collide in the reversereaction that forms reactants. This reversible reaction iswritten with a double arrow.
forward
2SO2(g) + O2(g) 2SO3(g) reverse
© 2013 Pearson Education, Inc. Chapter 9, Section 2
Chemical Equilibrium
At equilibrium, the rate of the forward
reaction becomes equal to the rate of the reverse reaction.
the forward and reverse reactions continue at equal rates.
no further changes occur in the concentration of reactants and products.
Insert bottom page 7 right column, reaction diagram.
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© 2013 Pearson Education, Inc. Chapter 9, Section 2
Equilibrium
Given the reaction, H2(g) + I2(g) 2HI(g) atequilibrium, the forward and reverse reactions occur atthe same time and are shown together using a doublearrow. Forward reaction: H2(g) + I2(g) 2HI(g)
Reverse reaction: 2HI(g) H2(g) + I2(g)
H2(g) + I2(g) 2HI(g)
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© 2013 Pearson Education, Inc. Chapter 9, Section 2
Equilibrium H2(g) + I2(g) 2HI(g)
Initially, the reaction flask contains only the reactants H2 and I2. The forward reaction begins to produce HI. As the reaction proceeds, there is less H2 and I2 and more HI, which
increases the rate of the reverse reaction. At equilibrium, the concentrations of reactants and product are constant. The reaction continues, with the rate of the forward reaction equal to the
rate of the reverse reaction.17
© 2013 Pearson Education, Inc. Chapter 9, Section 2 18
Learning Check
Write the forward and reverse reactions for the following.
CH4(g) + 2H2S(g) CS2(g) + 4H2(g)
© 2013 Pearson Education, Inc. Chapter 9, Section 2 19
Learning Check
Complete with equal, not equal, forward, reverse,changes, or does not change.1. Reactants form products in the ________ reaction.
2. At equilibrium, the reactant concentration _______.
3. When products form reactants, it is the _______ reaction.
4. At equilibrium, the rate of the forward reaction is ______ to the rate of the reverse reaction.
5. If the forward reaction is faster than the reverse, the amount of products ________.
© 2013 Pearson Education, Inc. Chapter 9, Section 3
General, Organic, and Biological Chemistry
Fourth EditionKaren Timberlake
9.3Equilibrium Constants
Chapter 9Reaction Rates and
Chemical Equilibrium
© 2013 Pearson Education, Inc.Lectures
© 2013 Pearson Education, Inc. Chapter 9, Section 3
For the reaction,
the equilibrium constant expression, Kc, gives the
concentrations of the reactants and products at equilibrium,
the square brackets indicate the moles/liter of each substance, and
the coefficients b and a are written as superscripts that raise the moles/liter to a specific power.
Equilibrium Constants
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© 2013 Pearson Education, Inc. Chapter 9, Section 3
Writing a Kc Expression
Write the Kc expression for the following reaction.
Step 1 Write the balanced chemical equation.
Step 2 Write the concentrations of the products as the numerator and the reactants as the denominator.
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© 2013 Pearson Education, Inc. Chapter 9, Section 3
Writing a Kc Expression
Write the Kc expression for the following reaction.
Step 3 Write any coefficient in the equation as an exponent.
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© 2013 Pearson Education, Inc. Chapter 9, Section 3
Learning Check
Which of the following is the correctly written Kc expression for the reaction shown below?
A. B.
C. D.
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© 2013 Pearson Education, Inc. Chapter 9, Section 3
Homogeneous, Heterogeneous Equilibrium In a homogeneous equilibrium, all reactants and
products in the reaction are in the same physical state.
In a heterogeneous equilibrium, the reactants and products are in two or more physical states.
The concentration of solids and liquids are constant,and therefore omitted from the equilibrium constantexpression.
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© 2013 Pearson Education, Inc. Chapter 9, Section 3
Learning Check
Write the equilibrium constant expression for thefollowing reaction at equilibrium.
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© 2013 Pearson Education, Inc. Chapter 9, Section 3
Calculating Equilibrium Constants
What is the Kc for the following reaction at 427 ˚C ifthe equilibrium concentrations are [H2] = 0.20 M,
[I2] = 0.20 M and [HI] = 1.47 M?
Step 1 Write the Kc expression for the equilibrium.
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© 2013 Pearson Education, Inc. Chapter 9, Section 3
Calculating Equilibrium Constants
What is the Kc for the following reaction at 427 ˚C ifthe equilibrium concentrations are [H2] = 0.20 M,
[I2] = 0.20 M and [HI] = 1.47 M?
Step 2 Substitute equilibrium (molar) concentrations and calculate Kc.
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© 2013 Pearson Education, Inc. Chapter 9, Section 3
Learning Check
Calculate the Kc for the reaction,
with the following equilibrium concentrations,
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© 2013 Pearson Education, Inc. Chapter 9, Section 4 32
General, Organic, and Biological Chemistry
Fourth EditionKaren Timberlake
9.4Using Equilibrium
Constants
Chapter 9Reaction Rates and
Chemical Equilibrium
© 2013 Pearson Education, Inc.Lectures
© 2013 Pearson Education, Inc. Chapter 9, Section 4 33
Reaching Chemical Equilibrium
The values of equilibrium constants can be large orsmall.
The size of the Kc depends on whether equilibrium isreached with more products than reactants (Kc is large), or
more reactants than products (Kc is small).
© 2013 Pearson Education, Inc. Chapter 9, Section 4 34
Kc is Large.
When a reaction has a large equilibrium constant, itmeans the forward reaction produced a large amount of
products when equilibrium was reached. the equilibrium mixture contains mostly products.
© 2013 Pearson Education, Inc. Chapter 9, Section 4 36
Kc is Small.
When a reaction has a small equilibrium constant, it means
the reverse reaction converted a large amount of products back to reactants.
the equilibrium mixture contains mostly reactants.
© 2013 Pearson Education, Inc. Chapter 9, Section 4 38
Summary of Kc Values
A reaction that favors products has a large Kc. with about equal concentrations of products and
reactants has a Kc close to 1. that favors reactants has a small Kc.
© 2013 Pearson Education, Inc. Chapter 9, Section 4 40
Learning Check
For each Kc, indicate whether the reaction at equilibrium
contains mostly reactants or products.
© 2013 Pearson Education, Inc. Chapter 9, Section 4 42
At equilibrium, the reaction, has a Kc of and containsWhat is the equilibrium concentration of PCl5? Problem Facts
Calculating Equilibrium Concentration
© 2013 Pearson Education, Inc. Chapter 9, Section 4 43
At equilibrium, the reaction, has a Kc of 4.2 x 10–2 and containsWhat is the equilibrium concentration of PCl5?
Step 1 Write the Kc expression for the equilibrium equation.
Step 2 Solve the Kc expression for the unknown concentration.
Calculating Equilibrium Concentration
© 2013 Pearson Education, Inc. Chapter 9, Section 4 44
Calculating Equilibrium Concentration
At equilibrium, the reaction, has a Kc of and containsWhat is the equilibrium concentration of PCl5?
Step 3 Substitute the known values into the rearranged Kc expression and calculate.
© 2013 Pearson Education, Inc. Chapter 9, Section 4 45
Learning Check
The Kc is 2.0 for the reaction,
If the equilibrium concentrations are and [ what is the equilibrium concentration of Br2?
A. 0.39 M
B. 0.78 M
C. 1.3 M
© 2013 Pearson Education, Inc. Chapter 9, Section 5 46
General, Organic, and Biological Chemistry
Fourth EditionKaren Timberlake
9.5Changing Equilibrium
Conditions:Le Châtelier’s Principle
Chapter 9Reaction Rates and
Chemical Equilibrium
© 2013 Pearson Education, Inc.Lectures
© 2013 Pearson Education, Inc. Chapter 9, Section 5 47
Le Châtelier’s Principle
Chemical equilibrium can be disturbed by a change inconcentration, volume, or temperature. Altering any ofthese conditions puts the system under stress.
Le Châtelier’s principle states that when a system at equilibrium is disturbed, the system
will shift in the direction that will reduce that stress. there will be a change in the rate of the forward or
reverse reaction to return the system to equilibrium.
© 2013 Pearson Education, Inc. Chapter 9, Section 5 48
Changing Concentrations
For the following reaction at equilibrium,
the rate of the forward reaction is increased to relieve the stress when more H2 or I2 is added, or HI is removed.
the rate of the reverse reaction is increased to relieve the stress when H2 or I2 is removed or more HI is added.
© 2013 Pearson Education, Inc. Chapter 9, Section 5 49
Adding Reactant or Product
The equilibrium shifts toward products when H2(g) or I2(g) is added.
reactants when HI(g) is added.
Add H2 or I2
Add HI
© 2013 Pearson Education, Inc. Chapter 9, Section 5 50
Removing Reactant or Product
The system shifts toward
a reverse reaction when H2 or I2 is removed.
a forward reaction when HI(g) is removed.
© 2013 Pearson Education, Inc. Chapter 9, Section 5 51
Concentration Changes and Equilibrium
(a) The addition of H2 places stress on the equilibrium system ofH2 (b) To relieve the stress, the forwardreaction converts some reactants, to product, HI(g).(c) A new equilibrium is established when the rates of the forwardreaction and the reverse reaction become equal.
© 2013 Pearson Education, Inc. Chapter 9, Section 5 52
Effect of a Catalyst
Adding a catalyst lowers the activation energy of the forward reaction. increases the rate of the forward reaction. lowers the activation energy of the reverse reaction. increases the rate of the reverse reaction. decreases the time to reach equilibrium. has no effect on the concentrations at equilibrium.
© 2013 Pearson Education, Inc. Chapter 9, Section 5 53
Learning Check
Predict any shift in the forward or reverse reactions foreach of the following changes on the reaction.
1. H2S(g) is added.
2. NH3(g) is removed.
3. A catalyst is added.
© 2013 Pearson Education, Inc. Chapter 9, Section 5 54
Effect of Volume Change
Changing the volume of a gas mixture at equilibrium willchange the concentrations of gases in the mixture,upsetting the equilibrium. Decreasing the mixture volume will increase the
concentration of gases. Increasing the mixture volume will decrease the
concentration of gases.
© 2013 Pearson Education, Inc. Chapter 9, Section 5 55
Effect of Volume Change
Decreasing the volume of the gas mixture shifts theequilibrium towards the fewer number of moles.
Increasing the volume of the gas mixture shifts theequilibrium toward the larger number of moles.
© 2013 Pearson Education, Inc. Chapter 9, Section 5 56
Volume Decrease and Equilibrium
A decrease in the volume of the container causes the system to shift in the direction of fewer moles of gas.
An increase in the volume of the container causes the system to shift in the direction of more moles of gas.
© 2013 Pearson Education, Inc. Chapter 9, Section 5 57
Temperature Change and Endothermic ReactionsFor an endothermic reaction at equilibrium, heat is areactant. A decrease in temperature (T) removes heat, shifting
the equilibrium toward the reactants. An increase in temperature adds heat, shifting the
equilibrium toward the products.
© 2013 Pearson Education, Inc. Chapter 9, Section 5 58
Temperature Change and Exothermic ReactionsFor an exothermic reaction at equilibrium, an increase in temperature adds heat and the
equilibrium shifts toward the reactants. a decrease in temperature (T) removes heat and the
equilibrium shifts toward the products.