1. Topics in CHM 1046 Intermolecular forces (IMF) Themodynamics
Chemical Kinetics
Chemical Equilibrium
Combination of above:
Unit 11 Intermolecular Forces, Liquids & Solids (T1)
Unit 12 Properties of Solutions (T1)
Unit 13 Thermochemistry (T1)
Unit 14 Chemical Thermodynamics (T2)
Unit 15 Chemical Kinetics (T2)
Unit 16 Chem. Equilibrium Gases & Heterogeneous (T3)
Unit 17.Acid Base Equilibria (T3)
Unit 18 Acid Base Equilibria Buffers & Hydrolysis (T3)
Unit 19 Acid Base Equilibria Titrations (T4)
Unit 20 Aqueous Equilibria Solubility Product (T4)
Unit 21 Electrochemistry (T4)

2. Unit 16 Chemical Equilibrium: Gases & Heterogeneous
CHM 1046: General Chemistry and Qualitative Analysis
Unit 16 Chemical Equilibrium: Gases & Heterogeneous
Dr. Jorge L. Alonso
Miami-Dade College – Kendall Campus
Miami, FL
Textbook Reference:
Chapter # 17
Module # 5 (& Appendix 2)

3. The Concept of Equilibrium
{Non-Equilibrium Reactions}: proceed in one direction, A  B
2 NO (g) + O2 (g) NO2 (g)
(clear gases)
(red-brown gas)
{Equilibrium Reactions}: proceed in both directions, A ↔ B 
N2O4 (g)
2 NO2 (g)
(clear gas)
(red-brown gas) 2X
Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.
This does not mean chemicals are found in same concentration!

4. The Equilibrium Constant (Keq)
N2O4 (g)
2 NO2 (g)
Therefore, at equilibrium
Ratef = Rater
kf [N2O4] = kr [NO2]2
Rewriting this, it becomes kf kr
[NO2]2
[N2O4] =
Keq =
The Equilibrium Constant is the ratio of the rate constants at a particular temperature.

5. The Equilibrium Constant (K)
To generalize this expression, consider the reaction
aA(aq) + bB(aq)
cC(aq) + dD(aq)
The equilibrium expression for this reaction would be
Kc =
[C]c[D]d
[A]a[B]b
where [X] = concentration of each chemical in moles/ liter (M).

6. What Does the Value of K Mean?
10 x
aA + bB
cC + dD
Kc =
[C]c[D]d
[A]a[B]b
Kc =10 = 10 1
If K >> 1, the reaction is product-favored; product predominates at equilibrium.
aA + bB
cC + dD
10 x
Kc =
[C]c[D]d
[A]a[B]b
Kc = 1 = 0.1 10
If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium.

7. The Equilibrium Constant for Gases
Kc =
[C]c[D]d
[A]a[B]b
aA(g) + bB(g)
cC(g) + dD(g)
Partial press: (75 torrs) (150 torrs) (300 torrs) (600 torrs) = total (1125 torrs)
Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written
Kp =
(PC)c (PD)d
(PA)a (PB)b
where Kp = equilibrium constant for gases, and Px = partial pressure of each gas.

8. Relationship between Kc and Kp
ideal gas law:
PV = nRT
Kc =
[C]c[D]d
[A]a[B]b
P = RT n V RT n V RT n V
( )c ( )d
( )a ( )b
Kp =
(PC)c (PD)d
(PA)a (PB)b = RT n V RT n V
Plugging this into the expression for Kp ☺
Kp = Kc (RT)n
Where
n = (moles of gaseous product) − (moles of gaseous reactant)

9. Equilibrium Can Be Reached from Either Direction
[NO2]2
[N2O4] = KC
The ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial concentrations of NO2 and N2O4 are.

10. Equilibrium Can Be Reached from Either Direction
N2O4 (g)
2 NO2 (g)

11. Manipulating Equilibrium Constants
RULE #1: Reciprocal Rule
The equilibrium constant of a reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.
KI = = at 100C
[NO2]2
[N2O4]
N2O4 (g)
2 NO2(g)
N2O4(g)
2 NO2 (g) 1 =
0.212
KII =
[N2O4]
[NO2]2
= at 100C

12. Manipulating Equilibrium Constants
If the coefficients of a chemical equation are changed (increased or decreased) by a factor n, then the value of K is raised to that power.
RULE #2: Coefficient Rule
KI = = at 100C
[NO2]2
[N2O4]
N2O4 (g)
2 NO2 (g)
KIII = = (0.212)2 at 100C
[NO2]4
[N2O4]2
2 N2O4 (g)
4 NO2 (g)
KIV =
½N2O4(g)
NO2(g)

13. Manipulating Equilibrium Constants
RULE #3: Multiple Equilibria Rule
When two or more equations are added to obtain a final resultant equation, the equilibrium constant for the resultant equation is product of the constants of the added equations.
2 A(aq) + B(aq) ↔ 4 C(aq)
4 C(aq) + E(aq) ↔ 2 F(aq)
2 A(aq) + B(aq) + E ↔ 2 F(aq)

14. Homogeneous vs. Heterogeneous Equilibria
Reactions involving only gasses or only solutions
[NO2]2
[N2O4]
N2O4 (g)
2 NO2(g)
Kc = = at 100C
Homogeneous vs. Heterogeneous Equilibria
Reactions involving different phases (s, l, g, aq) of matter
CO2 (g) + CaO(s)
CaCO3 (s)

15. The Concentrations of Solids and Liquids Are Essentially Constant
PbCl2 (s)
Pb2+ (aq) + 2 Cl−(aq)
Write the equilibrium expression!
Kc = [Pb2+] [Cl−]2
The concentrations of solids and liquids do not appear in equilibrium expressions because their concentration does not change.
Concentration of both solids and liquids can be obtained by dividing the density of the substance by its molar mass—and both of these are constants at constant temperature.

16. What are the Equilibrium Expressions for these Heterogeneous Equilibria?

17. CaCO3 (s)
CO2 (g) + CaO(s)
How does the concentration of CO2 differ in the two bell jars?
As long as some CaCO3 or CaO remain in the system, the amount of CO2 above the solid will remain the same.

18. Equilibrium Calculations
(1) Calculating Kc from experimental data: [ ]i & [ ]e
(2) Calculating Equilibrium Concentrations [ ]e when Kc & [ ]i are known
[A] + [B]
[C] + [D]
ICE Tables:
Initial I
[A]i
[B]i C -
change
- change
+ change E
[A]e
[B]e
[C]e
[D]e
Change
Equilibrium

19. (1) Calculating Kc from experimental data: [ ]i & [ ]e
[HI]2
[H2] [I2]
H2 (g) + I2 (g)
2 HI (g)
A closed system initially containing x 10−3 M H2 and x 10−3 M I2 at 448C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10−3 M. Calculate Kc at 448C for the reaction taking place, which is
[H2], M +
[I2], M 
2 [HI], M
Initially
1.000 x 10-3
2.000 x 10-3
Change
-9.35 x 10-4
+1.87 x 10-3
At Equilibrium
6.5 x 10-5
1.065 x 10-3
1.87 x 10-3
[HI] Increases by 1.87 x 10-3 M
Stoichiometry tells us [H2] and [I2] decrease by half as much

20. …and, therefore, the equilibrium constant
[H2], M +
[I2], M 
[HI], M
Initially
1.000 x 10-3
2.000 x 10-3
Change
-9.35 x 10-4
+1.87 x 10-3
At Equilibrium
6.5 x 10-5
1.065 x 10-3
1.87 x 10-3
= 51 =
(1.87 x 10-3)2
(6.5 x 10-5)(1.065 x 10-3)
Kc =
[HI]2
[H2] [I2]

21. Calculating Kc from experimental data: [ ]i & [ ]e
N2O4 (g)
2 NO2 (g)
[NO2]2
[N2O4] = I
0.0
0.0200 C
+ x
- 2x E
0.0014 x Kc
The ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial concentrations of NO2 and N2O4 are.

22. 2006B Q5
Practice Problem
{ }

25. Equilibrium Calculations
(1) Calculating Kc from experimental data: [ ]i & [ ]e
[H2], M +
[I2], M 
[HI], M
Initially
1.000 x 10-3
2.000 x 10-3
Change
At Equilibrium
1.87 x 10-3
(2) Calculating Equilibrium Concentrations
when K & [ ]i are known
H2(g) I2(g) ↔ 2HI(g) I
0.0224 C
- x
+ 2x E x x x
51 =

26. (2) Calculating Equilibrium Concentrations [ ]e when K & [ ]i are known
aA + bB
cC + dD I
[A]i
[B]i C
- ax
- bx
+ cx
+ dx E
[A]i- ax
[B]i- bx cx dx
Solve for x by using the quadratic equation:

27. Calculating Equilibrium Concentrations when K & [ ]i are known
Problem: for the reaction H2(g) + I2(g) HI(g) , K= 430ºC. Suppose that the initial concentrations of H2, I2, and HI are M, M, and M, respectively. Calculate the concentrations of these species at equilibrium.
H2(g) I2(g) HI(g) I
0.0224 C
- x
+ 2x E x x x
Deciding if x is negligible:

28. Problem: for the reaction H2(g) + I2(g) 2HI(g) , K= 54. 3 @ 430ºC
Problem: for the reaction H2(g) + I2(g) HI(g) , K= 430ºC. Suppose that the initial concentrations of H2, I2, and HI are M, M, and M, respectively. Calculate the concentrations of these species at equilibrium.
Deciding if x is negligible:
Take the largest [A]i and multiply it by 100 and also divide it by 100.
Compare the answers of above to the value of K
If comparison are not significant as compared to K, then x is negligible and you can delete the x that is subtracted from [A]I
If these calc values are significant compared to K, then x is not negligible and you must use the quadratic to solve for x.
YOU MUST USE QUADRATIC!

29. Calculating Equilibrium Concentrations when K & [ ]i are known
[H2]i = E x x x
[I2]i =
First answer is physically impossible, since conc. of H2 and I2 would be more than original conc.
x calc without quadratic:
[H2]e = – ) = M
[I2]e = ( – ) = M
[HI]e = ( x ) = M

30. Equilibrium Calculations
(1) Calculating Kc from experimental data: [ ]i & [ ]e
[H2], M +
[I2], M 
2 [HI], M
Initially
1.000 x 10-3
2.000 x 10-3
Change
At Equilibrium
1.87 x 10-3
(2) Calculating Equilibrium Concentrations
when K & [ ]i are known
H2(g) I2(g) HI(g)
51 = I
0.0224 C
- x
+ 2x E x x x
Deciding if x is negligible:

31. Le Châtelier’s Principle
{Equi.Intro1}
{Equi.Intro2}
“If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.”
[N2O4]
[NO2]2 = KC
The effect of (1) Concentration
Henri Louis Le Chatelier
2 NO2 (g)
N2O4 (g) 
(red-brown gas)
(clear gas)
The effect of (2) Pressure & Volume
(3) Temperature
{L.C.&PressVol}

32. Le Châtelier’s Principle
The effect of (3) Temperature
2 NO2 (g)
N2O4 (g)
+ HEAT 
H = kJ
(red-brown gas)
(clear gas)
Why does heat favor the formation of NO2 over N2O4?
{Le Châtelier’s&Temp}
{Molecular Explanation}
Keq at different Temp:

33. The Effect of Temperature Changes
Add heat (), which way will the equlibrium be shifted? 
Co(H2O)62+(aq) + 4 Cl(aq)
CoCl4 (aq) + 6 H2O (l)
H= +
(pink)
+ HEAT
(blue) 
{CoCl}

34. Briggs-Rauscher reaction
{ClockReaction}
IO3− + 2H2O2 + CH2(COOH)2 + H+ → ICH(COOH)2 + 2O2 + 3H2O

35. Le Châtelier’s Principle & the Haber Process
It was not until the early 20th century that this method was developed to harness the atmospheric abundance of nitrogen to create ammonia, which can then be oxidized to make the nitrates and nitrites essential for the production of fertilizers and ammunitions.
∆H = kJ/n @ 250 C N
NH3 + HNO3  NH4NO3 (explosives & fertilizers)

36. Le Châtelier’s Principle & the Haber Process
Shift equilibrium to the right by increasing press & conc. of N2 + H2 and by removing NH3 from system Fe
N2 (g) + 3H2 (g) NH3 (g)
This apparatus helps push the equilibrium to the right by removing the ammonia (NH3) from the system as a liquid
Removed by liquefaction
Subs
b. pt.
NH3
-33°C N2
-196°C H2
-253°C Fe

37. Increase the rate of both the forward and reverse reactions.
The Effect of Catalysts on Equilibrium
How does the addition of a catalyst affect the equilibrium between subs. A and B?
catalyst
Increase the rate of both the forward and reverse reactions.
Equilibrium is achieved faster, but the equilibrium composition remains unaltered.

38. Equilibrium Constant (K) vs. Reaction Quotient (Q)
aA(aq) + bB(aq)
cC(aq) + dD(aq)
At Equilibrium:
At Non-Equilibrium:
Q =
[C]ic[D]id
[A]ia[B]ib
K =
[C]c[D]d
[A]a[B]b
To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression.
Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.

39. If Q = K, the system is at equilibrium.
aA(aq) + bB(aq)
cC(aq) + dD(aq)
K =
[C]c[D]d
[A]a[B]b
Q =
[C]ic[D]id
[A]ia[B]ib
If Q = K, the system is at equilibrium.
If Q < K, there is too much reactant, and the equilibrium shifts to the right.
If Q < K?
If Q > K, there is too much product and the equilibrium shifts to the left.
If Q > K?

40. Kc, Q and Spontaneity (ΔG)
[NH3]2eq
[N2]eq[H2]eq
Q =
[NH3]i2
[N2]i[H2]i
N2(g) + 3H2(g)
2NH3(g)
ΔG= -
Reverse Process ΔG = +
ΔG= -
Reverse Process ΔG = +
@ Equilibrium K = Q, ΔG = 0

41. Free Energy (G) & Reaction Quotient (Q) under non-Standard Conditions☺
G = G + RT ln Q = G + RT log Q
Non-standard
conditions
Non-standard conditions
G = - (spontaneity), implies COMPLETE conversion of all reactants to →
products @ standard conditions (25ºC, 1 atm, 1M).
G implies DIFFERENT CONCENTRATIONS of both reactants ↔ products and under non-standard conditions of concentration, temperature and pressure.
If [Reactants]↑, then G = -
If [Products]↑, then G = +
If at Equilibrium , then G = 0

42. Free Energy (G) & Equilibrium Constant (K) at Equilibrium
At equilibrium, G = 0 and Q = K so……
G = G + RT ln Q
0 = G + RT ln K
K = eG/RT
Rearranging …… ☺
G = - RT ln K
= RT log K
G K
Product formation –
> 1
Equilibrium
= 1
[Products] = Equilibrium (RARE) +
< 1
Equilibrium

43. Free Energy (ΔG), the Equilibrium Constant (ΔK),and Temperature
2 NO2 (g)
N2O4 (g)
Low Temp.
How would T ↓or ↑ affect K?
H = kJ
High Temp.
(red-brown gas)
(clear gas)
@ T ↓, K ↑
G°= -
G = - RT ln K
@ T ↑, K ↓
G°= +
Low Temperature:
High Temperature:
G°= -
G°= +
Q = K
Reactants only
Equilibrium mixture
Products only
{G, Q & T↑NO2 ← N2O4}HighT
{G, Q & T↓ NO2 → N2O4}LowT

44. 2003B Q1
2 HIH2 + I2

45. Kp = Kc(RT)Δn
2003B Q1

46. 2000A Q1

47. 2000A Q1

50. 2004B Q1

52. 2007B Q1

53. 2000 1

56. 2003 B

58. 2004 B

60. 2006 A

62. 2006 (B)

69. 2007 (B)