1. Equilibrium is obtained when the rate of the forward reaction is equal to the rate of the reverse reaction.
CO (g) H2O (g) CO2 (g) H2 (g)
The arrows pointing in opposite directions indicates that the reaction can easily proceed in either direction.
By convention, the arrow pointing to the right is the forward reaction and the arrow pointing to the left is the reverse reaction.
What are the consequences of being at equilibrium?
Or in other words, what would a system at equilibrium look like?
Or, how could we tell if a system is at equilibrium?

2. The forward reaction has a maximum rate at the start and the reverse reaction has a minimum rate at the start. As the reaction proceeds, the rate of the forward reaction eventually equals the rate of the reverse reaction.

3. Why does the rate of the forward reaction decrease as the reaction proceeds?
As the concentrations of the reactants decrease, the forward reaction slows down-fewer collisions.
Why does the rate of the reverse reaction increase as the reaction proceeds?
As the concentrations of the products increase, the reverse reaction speeds up-more collisions.

4. At the start of the reaction:
H2O (g) CO (g)
Only reactants are present.
After the reaction has been started:
As the reaction proceeds, the amount of reactants decrease while the amount of products increases.
H2O (g) CO (g) CO2 (g) H2 (g)
What happens when equilibrium has been reached?
The amounts of reactants and the amounts of products no longer changes.

5. H2O (g) + CO (g) CO2 (g) + H2 (g)
a) Composition at the start of the reaction.
b) Composition after a short period of time-composition is different from the starting composition.
c) Composition after a longer period of time-composition is different from the that in figure b.
d) Composition after an even longer period of time-composition is the same as it was in figure c.
Did the reaction reach equilibrium? If so, when?

6. A picture of equilibrium from the viewpoint of concentrations of reactant and products.
H2 (g) + N2 (g) NH3 (g)
When the concentrations stop changing, the reaction has reached equilibrium.

7. Equilibrium Constant (Keq)
a A (g) b B (g) c C (g) d D (g)
Keq =
[C]c[D]d
[A]a[B]b
For a given reaction at the same temperature, Keq will have the same value regardless of how much starting materials are used.
What does it mean if the value of Keq is greater than 1?
What does it mean if the value of Keq is less than 1?
When all reactants and products are in the same phase (gaseous or aqueous), the equilibrium is called a homogeneous equilibrium.

8. Write the equilibrium constant equation for the following reaction:
H2 (g) I2 (g) HI (g)
Keq =
[C]c[D]d
[A]a[B]b
Keq1 =
[HI]2
[H2] [I2]
What would the equilibrium constant equation be for the reverse reaction?
2 HI (g) H2 (g) I2 (g)
Keq2 =
[HI]2
[H2] [I2]
What is the relationship between Keq1 and Keq2?
Keq1 =
Keq2 1
Keq1 = (Keq2)-1 or

9. Only reactants and products that can change concentration during a reaction may appear in the equilibrium expression.
Only gases and aqueous solutions can change concentration during a reaction, therefore only these phases may appear in an equilibrium expression.
What would the equilibrium expression be for the equilibrium reaction below?
C (s) O2 (g) CO2 (g)
Keq =
[CO2]
[O2]
Notice that the C (s) does not appear in the expression!
Pure solids and pure liquids will not appear in equilibrium expressions.
Note that aqueous does not indicate a pure liquid, so it will appear.
When reactants or products are in different phases than each other, the equilibrium is called a heterogeneous equilibrium.

10. Write Equilibrium Constant Expressions for each of the reactions below.
CHO2CH3 (aq) H2O (l) ↔ CHO2H (aq) CH3OH (aq)
2 NO (g) O2 (g) ↔ NO2 (g)
S8 (s) O2 (g) ↔8 SO3 (g)
N2 (g) H2 (g) ↔ NH3 (g)
2 C5H10 (g) ↔ C10H20 (l)

11. A reaction at equilibrium may be thought of as being in balance.
If anything causes it to become out of balance, the reaction will “shift” itself until balance is obtained again.
A “stress” is any change that causes a reaction at equilibrium to go out of balance.
Anything that causes the concentration of a reactant or product to change will be a stress.
Adding or removing a reactant or a product is a stress.
Changing the pressure or volume for a gaseous reaction is a stress.
Changing the temperature is a stress.

12. A system at equilibrium
A system immediately after a reactant has been added.
Which direction does the reaction need to shift to bring the system back into balance?
Or, which do you need more of-product or reactant?
If the reaction shifts to the right, more product will be made and this will bring the system back into balance.
A system immediately after a product has been added.
If the reaction shifts to the left, more reactant will be made and this will bring the system back into balance.

13. A system immediately after a reactant has been removed.
If the reaction shifts to the right, more product will be made and this will bring the system back into balance.
A system immediately after a product has been removed.
If the reaction shifts to the left, more reactant will be made and this will bring the system back into balance.

14. N2 (g) H2 (g) NH3 (g)
(a) An Equilibrium Mixture of N2, H2, and NH3 (b) Addition of N2 causes an imbalance (too much N2). (c) The reaction must shift to the right to consume some N2.

15. For a reaction containing gases, changing the volume of the container changes the concentration of the gases and the pressure of the gasses.
Since the concentration changes, changing the volume is a stress.
Decreasing volume increases pressure, so the system will shift toward the side that has the fewest number of gas particles in order to reduce the pressure.
Increasing volume decreases pressure, so the system will shift toward the side that has the greater number of gas particles in order to increase the pressure.

16. N2 (g) + 3 H2 (g) 2 NH3 (g) 4 moles total 2 moles total
(a) A Mixture of NH3(g), N2(g), and H2(g) at Equilibrium (b) The volume is suddenly decreased which increases the pressure and concentration of all of the gases. (c) The reaction will shift to the right to reduce the number of particles is the container.

17. A + B C + D + Heat Exothermic A + B + Heat C + D Endothermic
Remember that heat can be a reactant (endothermic reaction) or a product (exothermic
reaction). Therefore, changing the temperature is like adding or removing a reactant or a
product (heat). Changing the temperature is a stress.
A B C D Heat
Exothermic
A B Heat C D
Endothermic
Increasing the temperature for an exothermic reaction is like adding a product.
Increasing the temperature for an endothermic reaction is like adding a reactant.
Decreasing the temperature for an exothermic reaction is like removing a product.
Decreasing the temperature for an endothermic reaction is like removing a reactant.

18. Cooling the reaction down caused a shift to the
NO2 is dark brown. N2O4 is colorless.
Cooling the reaction down caused a shift to the
right (more N2O4).
2 NO2 (g) N2O4 (g)
Heat
Is the reaction endothermic or exothermic?

19. [D] = ((Keq[A]2[B])/[C])½
2 A B C D
If [A] = 2.0 M Keq = 3.7
[B] = 1.5 M
[C] = 0.82 M
[D] = ?
Keq =
[C] [D]2
[A]2[B]
(##) ½ means take
the square root!
Solve for [D]:
[D] = ((Keq[A]2[B])/[C])½
Plug in the values:
[D] = ((3.7*(2.0)2(1.5))/0.82)½
[D] = 5.2
Practice: Find the value for [R] at equilibrium given the information below.
3 S R T V
If [S] = M Keq = 0.76
[R] = ? M
[T] = M
[V] = M

20. Notice that even though each of the three reactions started with completely different amounts of materials in the flask the same value is obtained for Keq.

21. Solubility
When a solid dissolves, the process becomes an equilibrium situation when the solution becomes saturated.
At any time before the solution becomes saturated, the rate of dissolving is greater than the rate of crystallizing.
At the moment the solution becomes saturated, the rate of crystallizing becomes equal to the rate of dissolving.
For an ionic solid:
McXd (s) c M+ (aq) d X- (aq)
Ksp = [M+]c[X-]d
Ksp is called the solubility product
Example:
PbCl2 (s) Pb+2 (aq) Cl- (aq)
Ksp = [Pb+2][Cl-]2
= 1.7X10-5

22. If Ksp is known for an ionic compound, then the solubility for a compound can be found.
For lead (II) chloride, Ksp = 1.7X10-5
What is the solubility of lead (II) chloride?
Solubility (s) will be expressed in moles of the solute per L of solution.
First write the dissociation equilibrium expression:
PbCl2 (s) Pb+2 (aq) Cl- (aq)
Then write the solubility product expression
Ksp = [Pb+2][Cl-]2
For every one PbCl2, you get one Pb+2 and two Cl- ions.
Therefore for “x” PbCl2, you get “x” Pb+2 and “2x” Cl- ions.
Now, Ksp = [Pb+2][Cl-]2 = (x)(2x)2
Ksp = (x)(4x2) = (4x3)
x = (Ksp/4)1/3
x = ((1.7X10-5)/4)1/3
x = = 0.16 mol/L
x is the concentration of Pb+2 ions and 2x is the concentration of Cl- ions.

23. In the previous example, we found “x” which was the concentration of Pb+2 ion.
Since for “x” PbCl2, we get “x” Pb+2 ions, the concentration of Pb+2 ions will also be solubility of the lead (II) chloride.
Therefore, s = x = 0.16 mol/L
However, for more complicated ionic compounds, s may not be equal to the concentration of any individual ion.
Al2(SO4)3 (s) Al+3 (aq) SO4-2 (aq)
Ksp = [Al+3]2[SO4-2] 3
For one aluminum sulfate (“x”) you get 2x aluminum ions and 3x sulfate ions, so
Ksp = (2x)2(3x) 3
If Ksp = 2.7X10-47, then what is the solubility of aluminum sulfate?
2.7X = (2x)2(3x)3 = (4x2)(27x3) = 108x5
x = ((2.7X10-47)/108)1/5
x = 1.9X10-10 mol/L
Note that x is the solubility, but it is not the concentration of aluminum or sulfate ions.

24. Predicting Precipitates
A student mixes 45.0 mL of M Pb(NO3)2 solution with mL of M NaF solution. Does a precipitate form?
Step 1: Identify the possible precipitate (i.e. the chemical formula of the precipitate).
Step 2: Write the dissolution equation for the precipitate.
Step 3: Write the Ksp equation for the precipitate.
Step 4: Calculate the concentrations of the important ions in the new solution.
Step 5: Plug the concentrations of the ions into the Qsp equation.
Step 6: Compare Qsp to Ksp
If Qsp < Ksp a precipitate will NOT form.
If Qsp > Ksp a precipitate will form.
If Qsp = Ksp a precipitate must b e present because the solution is saturated.

25. Pb(NO3)2 (aq) + NaF (aq) PbCl2 (s) + 2 NaNO3 (aq)
A student mixes 45.0 mL of M Pb(NO3)2 solution with mL of M NaF solution. Does a precipitate form?
Pb(NO3)2 (aq) + NaF (aq) PbCl2 (s) NaNO3 (aq)
PbF2 (s) Pb+2 (aq) F- (aq)
Ksp = [Pb+2][F-]2
= 3.3X10-8
From Chapter 15: M1V1 = M2V2 you need to find M2 for Pb+2 and for F-
M2 for Pb+2 = (0.0045)(45.0)/(205.0) = 9.9X10-4 M
M2 for F- = ( )(160.0)/(205.0) = 5.7X10-4 M
Qsp = [Pb+2][F-]2
= (9.9X10-4)(5.7X10-4)2 = 3.2X10-10
Qsp < Ksp so a precipitate will NOT form.

26. What is the solubility of AgCl? Ksp = 1.8X10-10
Common Ion Effect
The presence of an ion in a solution from another source will effect the solubility of a substance that contains that ion.
What is the solubility of AgCl? Ksp = 1.8X10-10
Ksp = [Ag+][Cl-] = 1.8X10-10
(x)(x) = x2 = 1.8X10-10
x = 1.3X10-5 M
What is the solubility of AgCl in 0.50 M NaCl solution?
Now [Cl-] = x
Ksp = (x)(x ) = x x = 1.8X10-10
Either solve the quadratic equation for x or
assume that x is small compared to 0.50
0.50x = 1.8X and x = 3.6X10-10