1. Equilibrium

2. Contents of Unit Power Point
Reversible reactions
Equilibrium Position
Le Chatelier Principle
K eq Equilibrium Constant
Q Reaction Quotient
K sp Solubulity Product Constant
(Just a special case of Keq)
Common ion Effect
(Just a special case of Le Chatelier Principle)

3. Reversible Reactions
It may seem that chemical reactions always happen in one direction. This is not true. Some reactions are reversible. A reversible reaction is one in which the conversion of reactants to products and the conversion of products to reactants occur at the same time.
2SO2(g) + O2(g)  2SO3(g)
2SO2(g) + O2(g)  2SO3(g)
forward reaction
reverse reaction

4. Reversible Reactions
The two equations can be combined into one using a double arrow.
2SO2(g) + O2(g) 2SO3(g)
The double arrow tells you that the reaction is reversible.
Molecules of SO2 and O2 react to give SO3.
Molecules of SO3 decompose to give SO2 and O2.

5. When the rates of the forward and reverse reactions are equal, the reaction has reached a state of balance called chemical equilibrium.
Reaction starting SO2 & O2, but no SO3.
Reaction starting with SO3, but no SO2 and O2.
Eventually, the concentrations remain constant.

6. Concentrations at Equilibrium
At chemical equilibrium, both the forward and reverse reactions continue, but because their rates are equal, no net change occurs in the concentrations of the reaction components. Although the rates of the forward and reverse reactions are equal at equilibrium, the concentrations of the components usually are not.
The relative concentrations of the reactants and
products at equilibrium mark the equilibrium position
of a reaction.

7. Equilibrium Position   Beginning: Blue squares represent reactant
Equilibrium reached:
75% orange (product);
25% blue (reactant)
The equilibrium mixture contains more orange squares than blue ones. “Position of equilibrium” is a way of expressing this. You can say things like:
"The position of equilibrium lies towards the orange."  
End result will be the same starting from product side.

8. A B Equilibrium Position
The equilibrium position tells you whether the forward or reverse reaction is more likely to happen.
Suppose a single reactant, A, forms a single product, B.
If the equilibrium mixture contains 1% A and 99% B, then the formation of B is said to be favored.
A B
1% %

9. Almost all reactions are reversible to some extent under the right conditions. In practice, one set of components is often so favored at equilibrium that the other set cannot be detected.
When no reactants can be detected, you can say that the reaction has gone to completion, or is irreversible.
When no products can be detected, you can say that no reaction has taken place.

10. Factors Affecting Equilibrium: Le Châtelier’s Principle
Le Châtelier’s principle: If a stress is applied to a system in dynamic equilibrium, the system changes in a way that relieves the stress.
Stresses that upset the equilibrium of a chemical system include:
changes in the concentration of reactants or products
changes in temperature
changes in pressure.

11. Le Châtelier’s Principle - Concentration
Changing the concentration of any reactant or product in a system at equilibrium disturbs the equilibrium. The system will adjust to minimize the effects of the change.
At equilibrium.
<1% carbonic acid
H2CO3(aq) CO2(aq) + H2O(l)
Now – add CO2 ; rate of the reverse reaction increases.
Remove CO2; rate of the reverse reaction decreases.

12. Le Châtelier’s Principle
Adding a product to a reaction at equilibrium pushes a reversible reaction in the direction of the reactants.
Removing a product always pulls a reversible reaction in the direction of the products.

13. Le Châtelier’s Principle - Temperature
Heat can be considered to be a product, just like NH3.
N2(g) + 3H2(g) NH3(g) + heat
Add heat
Direction of shift
Remove heat (cool)
Heating the reaction mixture at equilibrium pushes the equilibrium position to the left, which favors the reactants.
Cooling, or removing heat, pulls the equilibrium position to the right, and the product yield increases.

14. Summary of Temperature Effect on Equilibrium Position
Exothermic reaction (heat as a “product”)
adding heat pushes equilibrium position 
removing heat pushes equilibrium position
Endothermic reaction (heat as a “reactant”)
adding heat pushes equilibrium position 
removing heat pushes equilibrium position

15. Le Châtelier’s Principle: Pressure
A shift will occur only if there are an unequal number of moles of gas on each side of the equation
Predict equilibrium position shift by comparing the number of molecules of reactants and products.
N2(g) + 3H2(g) NH3(g)
Add pressure
Direction of shift
Reduce pressure 4
molecules
2 molecules
Added pressure shifts reaction to side with less particles

16. PCl5(g) + heat PCl3(g) + Cl2(g)
Applying Le Châtelier’s Principle
What effect will each of the following changes have on the equilibrium position for this reversible reaction?
PCl5(g) + heat PCl3(g) + Cl2(g)
a. Cl2 is added.
b. Pressure is increased.
c. Heat is removed.
d. PCl3 is removed as it forms.

17. The equilibrium shifts to the left.
a. Cl2 is added.
Cl2 is a product; Increasing the concentration of a product shifts the equilibrium to the left.
b. Pressure is increased. 1 molecule on left; 2 on right.
The equilibrium shifts to the left.
Heat is removed
Heat is a “reactant”.
The removal of heat causes the equilibrium to shift to the left.
d. PCl3 is removed as it forms.
PCl3 is a product.
Removal of a product as it forms causes the equilibrium to shift to the right.

18. 4HCl(g) + O2(g) 2Cl2(g) +2H2O(g)
In the following equilibrium reaction, in which direction would the equilibrium position shift with an increase in pressure?
4HCl(g) + O2(g) Cl2(g) +2H2O(g)
Reducing the number of molecules that are gases decreases the pressure.
The equilibrium will shift to the right.

19. Equilibrium Constants
Equilibrium follows the Law of Mass Action:
Which states that the relative concentrations of reactants and products can be expressed in terms of a mathematical constant, Keq
For the equation aA + bB cC + dD
where the coefficients a, b, c, and d represent the number of moles.
The equilibrium constant (Keq) is the ratio of [product]
to [reactant] at equilibrium. Each concentration is
raised to a power equal to the coefficient.

20. Equilibrium Constants
General equation for Keq Keq represents the equilibrium position
Keq =
[C]c x [D]d
[A]a x [B]b

21. Equilibrium Constants
The value of Keq depends only on the temperature of the reaction, NOT on the concentrations.
The size of the equilibrium constant indicates whether reactants or products are more common at equilibrium.
Keq greater than 1 products are favored at equilibrium.
Keq less than 1 reactants are favored at equilibrium.

22. A homogeneous equilibrium has everything present in the same phase
A homogeneous equilibrium has everything present in the same phase. The usual examples include reactions where everything is a gas, or everything is present in the same solution.
A heterogeneous equilibrium has things present in more than one phase. The usual examples include reactions involving solids and gases, or solids and liquids.
Solids and pure liquids will always be excluded from the equilibrium constant. It is assumed that these concentrations are constant.

23. Equilibrium Position versus Equilibrium Constant
Each set of equilibrium concentrations is called an equilibrium position. The equilibrium constant and equilibrium positions for a reaction are NOT the same thing. There is only one equilibrium constant for a reaction at a particular temperature, but there are an infinite number of equilibrium positions. A specific equilibrium position depends on initial concentrations, the equilibrium constant does not.

24. Equilibrium Position versus Equilibrium Constant
Example, for a reaction with 2 products and 2 reactants,
and a Keq of 15, there are an infinite number of
combinations of concentrations equal to 15.
(15)(1) (15)(3) (7.5)(4) (30)(1)
(2)(0.5) (2)(1.5) (1)(2) (4)(0.5) etc.
The concentration combinations are the equilibrium
positions (infinite). But there can only be one Keq at any specific temperature

25. Expressing and Calculating Keq
The colorless gas dinitrogen tetroxide (N2O4) and the brown gas nitrogen dioxide (NO2) exist in equilibrium with each other.
N2O4(g) NO2(g)
A liter of the gas mixture at equilibrium contains mol of N2O4 and mol of NO2 at 10oC.
Write the expression for the equilibrium constant (Keq) and calculate the value of the constant for the reaction.

26. Solve
Put concentration of the product in the numerator and concentration of the reactant in the denominator.
Keq =
[NO2]2
[N2O2]
Raise each concentration to the power equal to its coefficient in the chemical equation.
Keq =
(0.030 mol/L)2
( mol/L) =
(0.030 mol/L x mol/L)
Keq = 0.20 mol/L = 0.20

27. X = 0.78 moles Finding the Equilibrium Constant
One mole of H2 (g) and one mole I2 (g) react in a sealed 1-L flask at 450oC. At equilibrium, 1.56 mol of HI (g) is present, with some of the reactant gases. Calculate Keq for the reaction.
H2(g) + I2(g) HI(g)
Use an ICE chart to calculate all amounts at equilibrium H2 I2 HI
Initial
1 mole
0 moles
Change
- x 2x
Equilibrium
0.22 moles
1.56 moles
X = 0.78 moles

28. Solve [HI]2 Keq = [H2] x [I2] Write equilibrium expression
Substitute values at equilibrium and solve
Keq =
(1.56 mol/L)2
0.22 mol/L x 0.22 mol/L
1.56 mol/L x 1.56 mol/L
Keq = 5.0 x 101

29. Finding Concentrations at Equilibrium
Bromine chloride (BrCl) decomposes to form bromine and chlorine.
2BrCl(g) Br2(g) + Cl2(g)
At a certain temperature, Keq is 11.1.
BrCl is placed in a 1-L container and decomposes.
At equilibrium, the reaction mixture contains 4.00 mol Cl2. What are the equilibrium concentrations of Br2 and BrCl?
Concentrations are not given; only moles of Cl2
However, the volume of the container is 1 L,
so [Cl2] = 4 moles of Cl2 in 1 L or 4 moles/L
From the balanced equation, [Br2] = [Cl2] .

30. Solve [Br2] x [Cl2] Keq = [BrCl]2
Write the equilibrium expression for the reaction.
Keq =
[BrCl]2
[BrCl]2 =
Keq
[Br2] x [Cl2]
Rearrange the equation to solve
for [BrCl]2.
Substitute the known values
for Keq, [Br2], and [Cl2].
[BrCl]2 =
11.1
4.00 mol/L x 4.00 mol/L
= 1.44 mol2/L2
[BrCl] =1.20 mol/L

31. HCl is formed when H2 and Cl2 react at high temperatures.
H2(g) + Cl2(g)  2HCl(g)
At equilibrium, [HCl] = 1.76 x 10–2 mol/L,
and [H2] = [Cl2] = 1.60 x 10–3 mol/L.
What is the value of the equilibrium constant?

32. HCl is formed when H2 and Cl2 react at high temperatures.
H2(g) + Cl2(g)  2HCl(g)
At equilibrium, [HCl] = 1.76 x 10–2 mol/L, and [H2] = [Cl2] = 1.60 x 10–3 mol/L.
What is the value of the equilibrium constant?
[HCl]2
(1.76 x 10–2 mol/L)2
Keq = =
[H2] x [Cl2]
(1.60 x 10–3 mol/L) x (1.60 x 10–3 mol/L)
Keq = 121

33. Catalysts and Equilibrium
Equilibrium constants aren't changed if you add (or change) a catalyst. The only thing that changes an equilibrium constant is a change of temperature.
The position of equilibrium is not changed if you add (or change) a catalyst.
Explanation
A catalyst speeds up both the forward and back reactions by exactly the same amount. Dynamic equilibrium is established when the rates of the forward and back reactions become equal. If a catalyst speeds up both reactions to the same extent, then they will remain equal without any need for a shift in position of equilibrium.

34. Reaction Quotient, Q Many reaction systems are not at equilibrium.
Q, reaction quotient, is a mass action expression
used under any set of conditions, not just
equilibrium.
Q is used to determine which direction a
Reaction will shift to reach equilibrium.
If K > Q, a reaction will 
If K < Q, the reaction will 
If Q = K, system is at equilibrium.

35. Calculating & Using Reaction Quotient, Q
0.035 moles of SO2, moles of SO2Cl2, and moles of Cl2 in a 5.00 L flask at 100oC. What is Q before the reaction begins? Which direction will the reaction proceed in order to establish equilibrium?
SO2Cl2(g)  SO2(g) + Cl2(g) Kc = at 100oC
Q = [SO2] [Cl2] mol/5L = mol/L SO2
[SO2Cl2] mol/5L = mol/L SO2Cl2
0.080 mol/5L = mol/L Cl2
Substitute and solve
Q =[0.007] [0.016]= (which is less than 0.078)
[0.100]
Since K >Q, rxn 

36. Calculating & Using Reaction Quotient, Q
Substitute the values in to the expression and solve for Q. Compare the answer to the value for the equilibrium constant and predict the shift (K) > (Q) Since K >Q, the reaction will proceed in the forward direction in order to increase the concentrations of both SO2 and Cl2 and decrease that of SO2Cl2 until Q = K.

37. Solubility Many ionic compounds are soluble in water.
But, some ionic compounds do not dissolve very well in water.
For example, most sulfide compounds are considered insoluble.
However, most insoluble ionic compounds will actually dissolve to some extent in water. These compounds are said to be slightly soluble in water.

38. The Solubility Product Constant
When the “insoluble” compound silver chloride is mixed with water, a very small amount of silver chloride dissolves in the water.
An equilibrium is established between the solid and the dissolved ions in the saturated solution.
AgCl(s) Ag+(aq) + Cl–(aq)
You can write an equilibrium-constant expression for this process.
Keq =
[Ag+]  [Cl–]
[AgCl]

39. The Solubility Product Constant (Ksp)
To compare the solubility of salts, it is useful to have a constant that tells us only how much salt has dissolved into ions.
Salt AB  aA bB- salt dissociation
This is the solubility product constant (Ksp). It is equal
to the product of the concentrations of the ions each
raised to a power equal to the coefficient.
Ksp = [A]a  [B]b
Smaller Ksp means lower solubility

40. Finding the Ion Concentrations in a Saturated Solution
What is the concentration of lead ions and chromate ions in a saturated solution of lead(II) chromate at 25oC? (Ksp = 1.8  10–14)
PbCrO4(s) Pb2+(aq) + CrO42–(aq)
Ksp = [A]a  [B]b
Ksp = [Pb2+]  [CrO42–] = 1.8  10–14
[Pb2+] = [CrO42–] = 1.3  10–7 M

41. Which of the following compounds is least soluble at 25oC?
A. PbF2
B. ZnS
C. SrCO3
D. CaSO4
Ksp at 25oC
Ionic compound
Ksp
CaSO4
2.4  10–5
PbF2
3.6  10–8
SrCO3
9.3  10–10
ZnS
3.0  10–23

42. PbCrO4(s) Pb2+(aq) + CrO42–(aq)
The Common Ion Effect
In a saturated solution of lead(II) chromate, an equilibrium is established between the solid lead(II) chromate and its ions in solution.
PbCrO4(s) Pb2+(aq) + CrO42–(aq)
What would happen if you added some Pb(NO3)2 (s)?
Pb(NO3)2 is soluble in water, so [lead ion] increases.
Applying Le Châtelier’s principle, adding product shifts eq. position to reactant side. PbCrO4 will precipitate.
NO3 is not part of the reaction and has no effect.

43. The Common Ion Effect In this example, the lead ion is a common ion.
A common ion is an ion that is found in both ionic compounds in a solution.
The lowering of the solubility of an ionic compound as a result of the addition of a common ion is called the common ion effect.

44. The Ksp of BaSO4 is 1.1  10–10. Why can a patient ingest the toxic BaSO4 without being harmed?
Barium sulfate is not very soluble. Therefore, very little of it can dissolve in bodily fluids and be absorbed as a toxin.

45. Finding Equilibrium Ion Concentrations in the Presence of a Common Ion
Small amounts of silver bromide can be added to the lenses used for eyeglasses. The silver bromide causes the lenses to darken in the presence of large amounts of UV light.
AgBr Ksp = 5.0  10–13.
What is the concentration of bromide ion in a 1.00-L saturated solution of AgBr to which mol of AgNO3 is added?
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

46. Solve Before AgNO3 added, [Ag+] = [Br–] = x
After AgNO3 added, x = [Ag+]
Ksp = [Ag+]  [Br–] = (x ) x
substitute values
5.0  10–13 = (x ) x
solve for x
Based on the small value of Ksp, you can assume that x will be very small compared to Thus, [Ag+] ≈ M.
5.0  10–13 = (0.020) x
[Br–] = 2.5  10–11 M

47. The Common Ion Effect
A precipitate will form if the product of the concentrations of two ions in the mixture is greater than the Ksp value for the compound formed from the ions.
As solutions of barium nitrate and sodium sulfate are mixed, a precipitate of BaSO4 is formed. Does a precipitate form when 1 L of M Na2SO4 is mixed with 1 L of M Ba(NO3)2? The Ksp for BaSO4 is 1.1  10–10.
[Ba2+]  [SO42–] = ( M)  (0.014 M) = 8.4  10–6
The result is larger than the Ksp value for BaSO4, so a precipitate will form.