1. CHEMICAL EQUILIBRIUM Chapter 16
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2. Properties of an Equilibrium
Equilibrium systems are
DYNAMIC (in constant motion)
REVERSIBLE
can be approached from either direction

3. Properties of an Equilibrium
Equilibrium systems are
DYNAMIC (in constant motion)
REVERSIBLE
can be approached from either direction
Pink to blue
Co(H2O)6Cl2 ---> Co(H2O)4Cl H2O

4. Properties of an Equilibrium
Equilibrium systems are
DYNAMIC (in constant motion)
REVERSIBLE
can be approached from either direction
Pink to blue
Co(H2O)6Cl2 ---> Co(H2O)4Cl H2O
Blue to pink
Co(H2O)4Cl H2O ---> Co(H2O)6Cl2

5. Chemical Equilibrium Fe3+ + SCN- FeSCN2+
Fe(H2O)63+
+ SCN-
Fe(SCN)(H2O)53+
+ H2O

6. Chemical Equilibrium Fe3+ + SCN- FeSCN2+
After a period of time, the concentrations of reactants and products are constant.
The forward and reverse reactions continue after equilibrium is attained.

7. Examples of Chemical Equilibria
Phase changes such as H2O(s) H2O(liq)

8. Examples of Chemical Equilibria
Formation of stalactites and stalagmites
CaCO3(s) + H2O(liq) + CO2(g) Ca2+(aq) + 2 HCO3-(aq)

9. Chemical Equilibria
CaCO3(s) + H2O(liq) + CO2(g) Ca2+(aq) + 2 HCO3-(aq)
At a given T and P of CO2, [Ca2+] and [HCO3-] can be found from the EQUILIBRIUM CONSTANT.

10. THE EQUILIBRIUM CONSTANT
For any type of chemical equilibrium of the type
a A + b B c C + d D
the following is a CONSTANT (at a given T)

11. THE EQUILIBRIUM CONSTANT
For any type of chemical equilibrium of the type
a A + b B c C + d D
the following is a CONSTANT (at a given T)
If K is known, then we can predict concs. of products or reactants.

12. Determining K 2 NOCl(g) 2 NO(g) + Cl2(g)
Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K.
Solution
Set of a table of concentrations
[NOCl] [NO] [Cl2]
Before
Change
Equilibrium

13. Determining K 2 NOCl(g) 2 NO(g) + Cl2(g)
Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K.
Solution
Set of a table of concentrations
[NOCl] [NO] [Cl2]
Before
Change
Equilibrium

14. Determining K 2 NOCl(g) 2 NO(g) + Cl2(g) [NOCl] [NO] [Cl2]
Before
Change
Equilibrium

15. Determining K 2 NOCl(g) 2 NO(g) + Cl2(g) [NOCl] [NO] [Cl2]
Before
Change
Equilibrium

16. Writing and Manipulating K Expressions
Solids and liquids NEVER appear in equilibrium expressions.
S(s) + O2(g) SO2(g)

17. Writing and Manipulating K Expressions
Solids and liquids NEVER appear in equilibrium expressions.
S(s) + O2(g) SO2(g)

18. Writing and Manipulating K Expressions
Solids and liquids NEVER appear in equilibrium expressions.
NH3(aq) + H2O(liq) NH4+(aq) + OH-(aq)

19. Writing and Manipulating K Expressions
Solids and liquids NEVER appear in equilibrium expressions.
NH3(aq) + H2O(liq) NH4+(aq) + OH-(aq)

20. Writing and Manipulating K Expressions
Adding equations for reactions
S(s) + O2(g) SO2(g) K1 = [SO2] / [O2]
SO2(g) + 1/2 O2(g) SO3(g)
K2 = [SO3] / [SO2][O2]1/2
NET EQUATION
S(s) /2 O2(g) SO3(g)

21. Writing and Manipulating K Expressions
Adding equations for reactions
S(s) + O2(g) SO2(g) K1 = [SO2] / [O2]
SO2(g) + 1/2 O2(g) SO3(g)
K2 = [SO3] / [SO2][O2]1/2
NET EQUATION
S(s) /2 O2(g) SO3(g)

22. Writing and Manipulating K Expressions
Changing coefficients
S(s) /2 O2(g) SO3(g)
2 S(s) O2(g) SO3(g)

23. Writing and Manipulating K Expressions
Changing coefficients
S(s) /2 O2(g) SO3(g)
2 S(s) O2(g) SO3(g)

24. Writing and Manipulating K Expressions
Changing coefficients
S(s) /2 O2(g) SO3(g)
2 S(s) O2(g) SO3(g)

25. Writing and Manipulating K Expressions
Changing direction
S(s) + O2(g) SO2(g)
SO2(g) S(s) + O2(g)

26. Writing and Manipulating K Expressions
Changing direction
S(s) + O2(g) SO2(g)
SO2(g) S(s) + O2(g)

27. Writing and Manipulating K Expressions
Changing direction
S(s) + O2(g) SO2(g)
SO2(g) S(s) + O2(g)

28. Writing and Manipulating K Expressions
Concentration Units
We have been writing K in terms of mol/L.
These are designated by Kc
But with gases, P = (n/V)•RT = conc • RT
P is proportional to concentration, so we can write K in terms of P. These are designated by Kp.
Kc and Kp may or may not be the same.

29. The Meaning of K
1. Can tell if a reaction is product-favored or reactant-favored.
For N2(g) H2(g) NH3(g)

30. The Meaning of K
1. Can tell if a reaction is product-favored or reactant-favored.
For N2(g) H2(g) NH3(g)

31. The Meaning of K
1. Can tell if a reaction is product-favored or reactant-favored.
For N2(g) H2(g) NH3(g)
Conc. of products is much greater than that of reactants at equilibrium.

32. The Meaning of K
1. Can tell if a reaction is product-favored or reactant-favored.
For N2(g) H2(g) NH3(g)
Conc. of products is much greater than that of reactants at equilibrium.
The reaction is strongly product-favored.

33. The Meaning of K For AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 1.8 x 10-5
Conc. of products is much less than that of reactants at equilibrium.
The reaction is strongly reactant-favored.
Ag+(aq) + Cl-(aq)
AgCl(s)
is product-favored.

34. The Meaning of K
2. Can tell if a reaction is at equilibrium. If not, which way it moves to approach equilibrium.

35. The Meaning of K
If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium?
Which way does the reaction “shift” to approach equilibrium?
See Screen 16.9.

36. The Meaning of K
In general, all reacting chemical systems are characterized by their REACTION QUOTIENT, Q.
If Q = K, then system is at equilibrium.

37. The Meaning of K
In general, all reacting chemical systems are characterized by their REACTION QUOTIENT, Q.
If Q = K, then system is at equilibrium.
If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium?
Q (2.3) < K (2.5).

38. The Meaning of K
In general, all reacting chemical systems are characterized by their REACTION QUOTIENT, Q.
If Q = K, then system is at equilibrium.
Q (2.33) < K (2.5).
Reaction is NOT at equilibrium, so [Iso] must become ________ and [n] must ____________.

39. Typical Calculations H2(g) + I2(g) ¸ 2 HI(g)
PROBLEM: Place 1.00 mol each of H2 and I2 in a 1.00 L flask. Calc. equilibrium concentrations.
H2(g) + I2(g) ¸ 2 HI(g)

40. H2(g) + I2(g) HI(g), Kc = 55.3
Step 1. Set up table to define EQUILIBRIUM concentrations.
[H2] [I2] [HI]
Initial
Change
Equilib

41. H2(g) + I2(g) HI(g), Kc = 55.3
Step 1. Set up table to define EQUILIBRIUM concentrations.
[H2] [I2] [HI]
Initial
Change -x -x +2x
Equilib 1.00-x 1.00-x 2x
where x is defined as am’t of H2 and I2 consumed on approaching equilibrium.

42. H2(g) + I2(g) HI(g), Kc = 55.3
Step 2. Put equilibrium concentrations into Kc expression.

43. H2(g) + I2(g) HI(g), Kc = 55.3
Step 3. Solve Kc expression - take square root of both sides.

44. H2(g) + I2(g) HI(g), Kc = 55.3
Step 3. Solve Kc expression - take square root of both sides.

45. H2(g) + I2(g) HI(g), Kc = 55.3
Step 3. Solve Kc expression - take square root of both sides.
x = 0.79

46. H2(g) + I2(g) HI(g), Kc = 55.3
Step 3. Solve Kc expression - take square root of both sides.
x = 0.79
Therefore, at equilibrium
[H2] = [I2] = x = M

47. H2(g) + I2(g) HI(g), Kc = 55.3
Step 3. Solve Kc expression - take square root of both sides.
x = 0.79
Therefore, at equilibrium
[H2] = [I2] = x = M
[HI] = 2x = M

48. Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)

49. Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)
If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations?
Step 1. Set up an equilibrium table
[N2O4] [NO2]
Initial
Change
Equilib

50. Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)
If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations?
Step 1. Set up an equilibrium table
[N2O4] [NO2]
Initial
Change -x +2x
Equilib x 2x

51. Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)
Step 2. Substitute into Kc expression and solve.
Rearrange: ( x) = 4x2
x = 4x2
4x x = 0
This is a QUADRATIC EQUATION
ax bx c = 0
a = 4 b = c =

52. Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)
Solve the quadratic equation for x.
ax bx c = 0
a = 4 b = c =

53. Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)
Solve the quadratic equation for x.
ax bx c = 0
a = 4 b = c =

54. Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)
Solve the quadratic equation for x.
ax bx c = 0
a = 4 b = c =
x = ± 1/8(0.046)1/2 = ± 0.027

55. Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)
x = or
But a negative value is not reasonable.
Conclusion
[N2O4] = x = M
[NO2] = 2x = M