2. Equilibrium Concept
Most chemical reactions do not continue until all of the reactants are used up.
Most reactions are ongoing, reversible processes, preceding in both the forward direction to give products and in the reverse direction to give the original reactants.
We indicate an equilibrium reaction with a double arrow:
reactants products → ←
Chapter 16

3. Equilibrium Concept, continued
In an equilibrium reaction, initially the rate of the forward reaction is very fast.
As more products are formed, the rate of the reverse reaction speeds up.
When the rates of the forward and reverse reactions are the same, the system is at equilibrium.
reactants products
forward reaction
reverse reaction → ←
Chapter 16

4. Collision Theory Molecules must collide in order to react.
In a successful collision, existing bonds are broken as new bonds are formed and the reactants are transformed into products.
This is the collision theory of reactions.
Chapter 16

5. Factors in Successful Collisions
There are three factors that affect the rate of a chemical reaction.
Collision Frequency:
When we increase the frequency at which molecules collide, we increase the rate of reaction. The more collisions you have, the greater the odds that a collision will be successful.
Collision Energy:
For a reaction to occur, the molecules must collide with enough energy to form the new bonds.
Chapter 16

6. Factors in Successful Collisions
Collision Geometry:
For a reaction to occur, the molecules must be oriented in the proper geometry for the reaction to occur.
In (a), the reactants have the correct geometry and products are formed after the collision. In (b), the reactants do not have the correct geometry and they do not react.
Chapter 16

7. Factors That Affect Reaction Rate
There are three factors that affect the reaction rate:
Reactant Concentration:
As we increase the concentration of the reactant(s), the molecules are closer together and collide more frequently. The more collisions, the faster the reaction.
Reaction Temperature:
As we increase the temperature, we increase the energy of the reactants. As we increase the energy of the reactants, the rate of the reaction increases because of the increased collision frequency and the collision energy.
Chapter 16

8. Factors That Affect Reaction Rate
Reaction Concentration
Reaction Temperature
Addition of a Catalyst:
A catalyst increases the rate of a reaction. A catalyst increases the number of effective collisions by creating a more favorable collision geometry.
A catalyst is not consumed in a reaction.
Chapter 16

9. Energy Profiles of Reactions
For a chemical reaction to occur, the reactants must collide with sufficient energy to react.
This energy is required to achieve the transition state required to form the products (a).
Without sufficient energy, the reaction does not occur (b).
Chapter 16

10. Endothermic Reactions
An endothermic reaction absorbs heat as the reaction proceeds.
N2(g) + O2(g) + heat NO(g)
A reaction profile shows the energy of reactants and products during a reaction. → ←
The highest point on a reaction profile is the transition state.
Chapter 16

11. Reaction Profiles
The energy required for reactants to achieve the transition state is the activation energy, Eact.
The energy difference between reactants and products is the heat of reaction, DH.
The DH for an endothermic reaction is positive.
Chapter 16

12. NO(g) + O3(g) 2 NO2(g) + O2(g) + heat
Exothermic Reactions
An exothermic reaction releases heat as the reaction proceeds.
NO(g) + O3(g) NO2(g) + O2(g) + heat
The DH for an exothermic reaction is negative. → ←
Chapter 16

13. Effect of a Catalyst
A catalyst is a substance that allows a reaction to proceed faster by lowering the activation energy.
The reaction profile shows the effect of a catalyst on the reaction 2 H2(g) + O2(g) H2O(g) + heat → ←
A catalyst does not change DH for a reaction.
A catalyst speeds up both the forward and reverse reactions.
Chapter 16

14. Chemical Equilibrium Concept
A chemical change is a reversible process that can proceed simultaneously in both the forward and reverse directions.
When the rate of the forward and reverse reactions are proceeding at the same rate, the reaction is in a state of chemical equilibrium.
3 O2(g) O3(g)
At equilibrium,
ratef (O2 reaction) = rater (O3 reaction) → ←
Chapter 16

15. Rates and Equilibrium
The rate of reaction is the rate at which the concentrations of reactants decrease per unit time.
3 O2(g) O3(g)
Starting with only O2, as O2 is consumed, the rate of the forward reaction decreases.
As O3 is produced, the rate of the reverse reaction increases. When the rates are equal, equilibrium is achieved. → ←
Chapter 16

16. The Equilibrium Process
Initially, the container has only O2 gas (a).
As the reaction proceeds, O3 is formed (b).
At equilibrium, the forward and reverse reactions occur at the same rate (c).
Later, the amounts of O2 and O3 are unchanged (d).
Chapter 16

17. Chemistry Connection: Ozone Hole
The ozone layer is a region of the atmosphere that contains ozone, O3.
Ozone absorbs harmful UV radiation and prevents it from reaching the Earth’s surface.
Molecules called chloroflurorcarbons (CFCs) were found to cause the destruction of ozone.
Many nations have stopped using CFCs, and the size of the ozone hole is starting to shrink.
Chapter 16

18. Law of Chemical Equilibrium
Consider the following general reaction:
a A + b B c C + d D
The law of chemical equilibrium states that the molar concentrations of the products (raised to the powers c and d), divided by the molar concentrations of the reactants (raised to the powers a and b), equals a constant. → ←
Chapter 16

19. Equilibrium Constant Keq
Mathematically, we express the law of chemical equilibrium as follows:
Keq =
[C]c[D]d
[A]a[B]b
The constant, Keq, is the general equilibrium constant.
The value of Keq varies with temperature. So a given value of Keq is valid only for a specific temperature.
Chapter 16

20. Writing Equilibrium Constants
Let’s write the equilibrium constant expression for the reaction 2A B.
Recall, Keq is product(s) over reactant(s), each raised to its coefficient in the balanced reaction. Recall that square brackets represent the molar concentration of a species.
The equilibrium constant expression is: → ←
Keq =
[B]
[A]2
Chapter 16

21. Homogeneous Equilibria
A homogeneous equilibrium is a reaction where all of the products and reactants are in the same physical state.
What is the equilibrium constant expression for the homogeneous equilibrium:
2 SO2(g) + O2(g) SO3(g) → ←
Keq =
[SO3]2
[SO2]2[O2]
Chapter 16

22. Heterogeneous Equilibria
A heterogeneous equilibrium is a reaction where one of the substances is in a different physical state.
C(s) + H2O(g) CO(g) + H2(g)
The concentrations of liquids and solids do not change, and they are therefore omitted from equilibrium constant expressions: → ←
Keq =
[CO][H2]
[H2O]
Chapter 16

23. Equilibrium Constant Expressions
So when we write equilibrium constant expressions, we only include the concentrations of substances in the gas or aqueous state.
What is the equilibrium constant expression for the following reaction?
NH4NO3(s) N2O(g) + 2 H2O(g)
Keq = [N2O][H2O]2 → ←
Chapter 16

24. Experimental Determination of Keq
We can calculate the numerical value of Keq if we know the concentrations of all the species in the reaction.
H2(g) + I2(g) HI(g)
If the concentrations at equilibrium are [H2] = M, [I2] = M, and [HI] = M, what is Keq? → ←
Keq =
[HI]2
[H2][I2]
(1.576)2
(0.212)(0.212) =
= 55.3
Chapter 16

25. Values of Keq
It doesn’t matter how much of each substance we start with, the value of Keq is always 55.3.
Chapter 16

26. Shifts in Gaseous Equilibria
Le Chatelier’s principle states that when a reversible reaction at equilibrium is stressed by a change in concentration, temperature, or pressure, the equilibrium shifts to relieve the stress.
Let’s look at the equilibrium between colorless N2O4 and brown NO2:
N2O4(g) NO2(g)
If we increase the amount of N2O4, the reaction shifts to the right to produce more NO2.
If we increase the amount of NO2, the reaction shifts to the left to produce more N2O4. → ←
Chapter 16

27. Effect of Temperature The reaction is endothermic:
N2O4(g) + heat NO2(g)
If we lower the temperature, the reaction shifts to produce more colorless N2O4.
If we raise the temperature, the reaction shifts to produce more brown NO2. → ←
Chapter 16

28. Effect of Pressure
In a gaseous equilibrium, increasing the pressure will shift the reaction to the side with fewer gas molecules.
In the reaction N2O4(g) NO2(g), increasing the pressure will shift the reaction to the left, producing more N2O4. → ←
Chapter 16

29. Effect of an Inert Gas
If we add an inert gas to a gaseous reaction at equilibrium, what will happen?
The volume of the container does not change, therefore the concentration (and the partial pressure) of the substances do not change.
Adding an inert gas has no effect on a system at equilibrium.
Chapter 16

30. Ionization Equilibrium Constant, Ki
The equilibrium constant for the ionization of a weak acid or base is the ionization equilibrium constant, Ki.
What is Ki constant for the ionization of hydrofluoric acid?
HF(aq) H+(aq) + F– (aq) → ←
Ki =
[H+][F-]
[HF]
Chapter 16

31. Ionization of a Weak Base
We can also write a Ki expression for the ionization of ammonium hydroxide:
NH4OH(aq) NH4+(aq) + OH– (aq) → ←
Ki =
[NH4+][OH-]
[NH4OH]
Chapter 16

32. Experimental Determination of Ki
We can calculate the numerical value of Ki if we know the concentrations of all the species in the reaction.
HC2H3O2(aq) H+(aq) + C2H3O2– (aq)
If the concentrations at equilibrium are [HC2H3O2] = M, [H+] = M, and [C2H3O2–] = M, what is Ki? → ←
Ki =
[H+][C2H3O2–]
[HC2H3O2]
( )( )
(0.100) =
= 1.80 × 10-5
Chapter 16

33. Weak Acid-Base Equilibria Shifts
Let’s look at the following weak acid equilibrium:
HF(aq) H+(aq) + F– (aq)
If we increase the amount of HF, the reaction shifts to the right to produce more H+ and F-.
If we raise the pH (by adding base, for example), we decrease [H+], and the reaction shifts to the right.
If we add some soluble NaF, we increase the [F-], and the reaction shifts to the left.
If we add some soluble NaCl, nothing happens. → ←
Chapter 16

34. Solubility Product Equilibria
Insoluble salts are really very slightly soluble.
If we add Ag2SO4 to water, some of the slightly soluble Ag2SO4 dissolves:
Ag2SO4(s) Ag+(aq) + SO42-(aq)
We can write the solubility product equilibrium constant, Ksp, for the reaction:
Ksp = [Ag+]2 [SO42-]
Recall, we don’t include pure solids or liquids in equilibrium constant expressions. → ←
Chapter 16

35. Experimental Determination of Ksp
We can calculate the numerical value of Ksp if we know the concentrations of all the species in the reaction.
Mg(OH)2(s) Mg2+(aq) + 2 OH-(aq)
If the concentrations at equilibrium are [Mg2+] = M, and [OH-] = M, what is Ksp?
Ksp = [Mg2+][OH-]2 = ( )( )2
Ksp = 1.6 × 10-11 → ←
Chapter 16

36. Solubility Equilibria Shifts
Let’s look at the following solubility equilibrium:
AgCl(s) Ag+(aq) + Cl-(aq)
Ksp = [Ag+][Cl-]
What happens if we add more AgCl?
Nothing, since AgCl does not appear in Ksp.
What happens if we add some soluble NaCl?
We increase [Cl-], and the equilibrium shifts to the left producing more solid AgCl. → ←
Chapter 16

37. Chapter Summary
According to collision theory we can speed up a reaction in three ways:
increasing the concentration of reactants
raising the temperature of the reaction
adding a catalyst
An endothermic reaction absorbs heat energy, and an exothermic reaction releases heat energy.
Chapter 16

38. Chapter Summary, continued
The amount of energy necessary to achieve the transition state is the activation energy, Eact.
The difference in the energy of the reactants and products is the heat of reaction, DH.
A catalyst speeds up a reaction by lowering the activation energy.
A catalyst speeds up both the forward and reverse reactions.
Chapter 16

39. Chapter Summary, continued
We can write an equilibrium expression for reactions at equilibrium.
Chapter 16

40. Chapter Summary, continued
According to Le Chatelier’s principle, a reaction at equilibrium shifts in order to relieve a stress.
Chapter 16