by Steven S. Zumdahl & Don J. DeCoste University of Illinois

by Steven S. Zumdahl & Don J. DeCoste

University of Illinois

Introductory Chemistry: A Foundation, 6th Ed.

Introductory Chemistry, 6th Ed.

Basic Chemistry, 6th Ed.

Chapter 17

Equilibrium

Copyright © Houghton Mifflin Company. All rights reserved. 17 | 3

Collision Theory of Kinetics

• Kinetics is the study of the factors that affect the speed of a reaction and the mechanism by which a reaction proceeds.

• In order for a reaction to take place, the reacting molecules must collide into each other.


Collision Theory of Kinetics (cont.)

• Once molecules collide they may react together or they may not, depending on two factors:1. Whether the reacting molecules collide in the

proper orientation for new bonds to form

2. Whether the collision has enough energy to "break the bonds holding reactant molecules together"


Effective Collisions

• Collisions in which these two conditions are met (and the reaction occurs) are called effective collisions.

• The higher the frequency of effective collisions the faster the reaction rate.


Effective Collisions (cont.)

• When two molecules have an effective collision, a temporary, high energy (unstable) chemical species called an activated complex is formed.– Not a true molecule because its bonds

aren’t complete


Activated Complex

• The difference in potential energy between the reactant molecules and the activated complex is called the activation energy, Ea

• The larger the activation energy, the slower the reaction


Activated Complex (cont.)

• The energy to overcome the activation energy comes from the kinetic energy of the collision being converted into potential energy, or from energy available in the environment, i.e. heat.

• Different reactions have different activated complexes and therefore different activation energies.


Reactions and Activated Complex


Reactions and Activated Complex (cont.)


Factors Affecting Reaction Rate

• The kind of molecules and what condition the reactants are in

• Increasing temperature always increases reaction rate


Factors Affecting Reaction Rate (cont.)


• The larger the concentration of reactant molecules, the faster the reaction will go.– Increases the frequency of reactant molecule

collisions



• Catalysts are substances that affect the speed of a reaction without being consumed.

• Most catalysts are used to speed up a reaction.

• Homogeneous = catalyst present in same phase

• Heterogeneous = catalyst present in different phase



Catalysts Speed Up Reactions

• Catalysts work by providing a pathway for the reaction with a lower activation energy.

• Enzymes are biochemical catalysts.


Catalysts Speed Up Reactions (cont.)


• Increasing the temperature raises the kinetic energy of the reactant molecules, causing them to collide more frequently and to have more collisions with sufficient energy to form the activated complex.

Molecular Interpretationof Factors Affecting Rate


• The larger the concentration of reactant molecules, the more frequently they collide and the larger the number of effective collisions will be.

• Therefore the speed will increase.

Molecular Interpretationof Factors Affecting Rate (cont.)


• Catalysts work by providing an alternative pathway for the reaction with a lower activation energy.

• Lowering the activation energy means more molecules have enough kinetic energy so that when they collide they can form the activated complex.

• The result is the reaction goes faster.

Molecular Interpretationof Factors Affecting Rate (cont.)


Reaction Dynamics

• If the products of a reaction are removed from the system as they are made, then a chemical reaction will proceed until the limiting reactants are used up.

• However, if the products are allowed to accumulate they will start reacting together to form the original reactants - called the reverse reaction.


Reaction Dynamics (cont.)

• The forward reaction slows down as the amounts of reactants decreases because the reactant concentrations are decreasing, while the reverse reaction speeds up as the concentration of the products increases.

• Eventually rateforward = ratereverse


Reaction Dynamics (cont.)


Chemical Equilibrium

• Equilibrium only occurs in a closed system.

• When a system reaches equilibrium, the amounts of reactants and products in the system stay constant.


Chemical Equilibrium (cont.)


Equilibrium Constant

• Law Of Chemical Equilibrium

• In this expression, K is a number called the equilibrium constant.

aA + bB cC + dD = [C]c[D]d

[A]a[B]bKeq


Equilibrium Constant (cont.)

• Do not include solids or liquids, only solutions and gases.

• For a reaction, the value of K for a reaction depends on the temperature.

• K is independent of the amounts of reactants and products you start with.

aA + bB cC + dD = [C]c[D]d

[A]a[B]bKeq


• The relative concentrations of reactants and products when a reaction reaches equilibrium is called the position of equilibrium.

• Different initial amounts of reactants (and or products) will result in different equilibrium concentrations but the same equilibrium constant.

Position of Equilibrium


• If K is large, then there will be a larger concentration of products at equilibrium than of reactants, and we say the position of equilibrium favors the products.

Position of Equilibrium (cont.)


• If K is small then there will be a larger concentration of reactants at equilibrium than of products, and we say the position of equilibrium favors the reactants.

• The position of equilibrium is not affected by adding a catalyst.

Position of Equilibrium (cont.)


Determine the value of the equilibrium constant for the reaction 2 SO2 + O2 2 SO3

Example #1:


• Determine the equilibrium expression

• Plug the equilibrium concentrations into the equilibrium expression

• Solve the equation3.503.00SO3

1.251.50O2

1.502.00SO2

[Equilibrium][Initial]Chemical

4.36 1.25)((1.50)

(3.50)

]O[][SO

][SO K

2

2

22

2

23

===

Example #1 (cont.)


Le Châtelier’s Principle

• Le Châtelier's Principle guides us in predicting the effect various changes have on the position of equilibrium.

• When a change is imposed on a system at equilibrium, the position of equilibrium will shift in the direction that will reduce the effect of that change.


• The position of equilibrium can be affected without changing the equilibrium constant.

• Adding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found.

Concentration Changesand Le Châtelier’s Principle


Concentration Changesand Le Châtelier’s Principle (cont.)


• Changing the volume of a gas is like changing its concentration.– Has the same effect as changing the

concentration on the position of equilibrium

• Decreasing the volume of the system increases its pressure.– Increasing the pressure on the system causes the

position of equilibrium to shift toward the side of the reaction with the fewer gas molecules

Changing Volume and Le Châtelier’s Principle


Changing Volume and Le Châtelier’s Principle (cont.)


• The equilibrium constant will change if the temperature changes.

• For exothermic reactions, heating the system decreases K.– Think of heat as a product of the reaction– Therefore shift the position of equilibrium

toward the reactants

Changing Temperatureand Le Châtelier’s Principle


Changing Temperatureand Le Châtelier’s Principle (cont.)

• For endothermic reactions, heating the system increases K.– Think of heat as a reactant– The position of equilibrium will shift toward

the products

• Cooling an exothermic or endothermic reaction will have the opposite effects on K and equilibrium position.


Example #2:

If the value of the equilibrium constant for the reaction 2 SO2 + O2 2 SO3 is 4.36, determine the equilibrium concentration of SO3


Example #2 (cont.)

• Determine the equilibrium expression

• Plug the equilibrium concentrations and equilibrium constant into the equilibrium expression

• Solve the equation

?3.00SO3

1.251.50O2

1.502.00SO2

[Equilibrium][Initial]Chemical


Example #2 (cont.)

3.51 12.3 ][SO

12.3 (1.25) x (1.50) x 4.36 ][SO

4.36 1.25)((1.50)

][SO

]O[][SO

][SO K

3

223

2

23

22

2

23

==

==

===


Solubility & Solubility Product

• Even “insoluble” salts dissolve somewhat in water– Insoluble = less than 0.1 g per 100 g H2O

• The solubility of insoluble salts is described in terms of equilibrium between undissolved solid and aqueous ions produced.

AnXm(s) n A+(aq) + m Y-(aq)


Solubility and Solubility Product (cont.)

• Equilibrium constant called solubility product.

Ksp = [A+]n[Y-]m

• If undissolved solid is in equilibrium with the solution, the solution is saturated.

• Larger K = more soluble– For salts that produce same the number of ions


Example #3:

Calculate the solubility of AgI in water at 25°C if the value of Ksp = 1.5 x 10-16.


Example #3 (cont.)

• Determine the balanced equation for the dissociation of the salt

AgI(s) Ag+(aq) + I-(aq)

• Determine the expression for the solubility product– Same as the equilibrium constant expression

Ksp = [Ag+][I-]


• Define the concentrations of dissolved ions in terms of x

AgI(s) Ag+(aq) + I-(aq)Stoichiometry tells us that we get 1 mole of Ag+ and

1 mol I- for each mole of AgI dissolved.Let x = [Ag+], then [I-] = x

• Plug the ion concentrations into the expression for the solubility product and solve for Ksp

[Ag+] = [I-] = x

Example #3 (cont.)


8-16-

16-2

2--16sp

10 x 2.110 x 5.1

10 x 5.1

)( x )( ][I][Ag 10 x 5.1 K

==

=

==== +

x

x

xxx

[Ag+] = 1.2 x 10-8 mol/L = [AgI]The solubility of AgI = 1.2 x 10-8 M

Example #3 (cont.)

by Steven S. Zumdahl & Don J. DeCoste University of Illinois

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