The document discusses chemical equilibrium for the reaction aA + bB ⇌ cC + dD. It provides the rate expressions for the forward and reverse reactions and shows that at equilibrium, these rates are equal. Graphs illustrate how the concentrations of reactants and products remain constant once equilibrium is reached, while the rates continue at the same value in both directions. The key concept covered is that chemical equilibrium is a dynamic state where the rates of the forward and reverse reactions are equal.

1. Chemical Equilibrium
k1
aA+bB cC+dD
k-1
rate of the forward reaction = k1 [A]a [B]b
rate of the reverse reaction = k -1 [C]c [D]d
At equilibrium:
rate of the forward reaction = rate of the reverse reaction
k1 [A]a [B]b = k -1 [C]c [D]d

2. Chemical Equilibrium
k1 [A]a [B]b  k -1 [C]c [D]d
k1 [C]c [D]d
 a b
k -1 [A] [B]
[C]c [D]d
K eq 
[A]a [B]b

3. Chemical Equilibrium
• When equilibrium, a dynamic state, is reached
the concentrations of A, B, C, and D are
invariant (i.e., no longer change)
• The reaction continues in a dynamic state, but
the concentrations remain the same.
• The rate of the forward reaction is equal to
the rate of the reverse reaction.

4. Chemical Equilibrium
Concentration versus time
1.2
at equilibrium
1
0.8 [C] and [D]
concentration, M
0.6
[A] and [B]
0.4
0.2
0
0 2 4 6 8 10 12 14 16 18 20
time (minutes)

5. Chemical Equilibrium
Rate versus time
1.4 at equilibrium
1.2
k1 [A]a [B]b
1
k1 [A]a [B]b = k -1 [C]c [D]d
0.8
Rate
0.6
0.4
k -1 [C]c [D]d
0.2
0
0 2 4 6 8 10
time (minutes) 12 14 16 18

6. Consider the Reaction between Hydrogen and Iodine
H2(g) + I2(g) 2 HI(g)
If the initial concentrations of hydrogen and iodine gases are 0.0175 M,
calculate the equilibrium concentrations of the reactants and products
if the equilibrium constant is 55.64 at 425oC.
2
 HI(g) 
  = 55.64
 H 2(g)   I2(g) 
  
0.0175  x 0.0175  x 2x
H2(g) + I2(g) 2 HI(g)

7. 2
 HI(g) 
  = 55.64
 H 2(g)   I2(g) 
  
 2x 
2
= 55.64
0.0175-x 0.0175-x 
4x 2
-4 2
= 55.64
3.06 x 10 - 0.035 x + x
4x 2 = 0.0170 - 1.95 x + 55.64 x 2
0 = 0.0170 - 1.95 x + 51.64 x 2
-b ± b2 -4ac
x=
2a
a = 51.64; b = -1.95; and c = 0.0170

8. -b ± b2 -4ac
x=
2a
1.95 ± 3.80-4 (51.64)(0.0170)
x=
2(51.64)
1.95 ± 0.54
x=
103.28
x = 1.37 x 10-2 M or 2.41 x 10-2 M If x = 1.37 x 10-2 M
HI(g)  = 2x = 2.73 x 10-2 M
 
H2(g)  = (0.0175 - x) M = (0.0175 - 0.0137) M = 3.8 x 10-3 M
 
I2(g)  = (0.0175 - x) M = (0.0175 - 0.0137) M = 3.8 x 10-3 M
 
x = 2.41 x 10-2 M is not possible, because it would lead to negative matter!

9. The equilibrium expression does not include solids
For example, the equilibrium expression for the following reaction:
1
S8 + O2 (g) SO2 (g)
8
SO2(g) 
  = K = 4.2 x 1025 at 25o C
O2(g) 
eq
 

10. The equilibrium expression when water is involved
For example,
- +
HCOOH(aq) + H2O(l) HCOO(aq) + H3O(aq)
 HCOO-(aq)   H3O(l) 
 
+
 = K H O  = K
 HCOOH (aq)   2 (l)  a
 

11. Kc versus Kp
N2(g) + 3 H2 (g) 2 NH3 (g)
2
 NH3 (g) 
 
Kc = = 3.5 x 108 at 25oC
 N 2 (g)   H 2 (g) 
   
2
PNH3 (g)
Kp = 2 2
= ? at 25o C
PN2 (g) PH2 (g)

12. From PV = nRT
n p
where = concentration =
V RT
K c = K p x (RT) 2
Kc
Kp =
(RT) 2
2
PNH3(g) 1
2
 NH3 (g) 
  (RT) 2 2
PNH3(g) (RT) 2
Kc = = 2 = 2 x
 N 2 (g)   H 2 (g) 
    PN2(g) 2
PH2(g) 2
PN2(g) x PH2(g) 1
2
x (RT) 4
(RT) (RT) 2
2
PNH3(g)
Kc = 2 2
x (RT) 2
PN2(g) x PH2(g)
K c = K P x (RT) 2

13. K c = K P x (RT) 2
Kc
KP = 2
KP
(RT)
3.5 x 108
KP =
(0.08205 x 298) 2
3.5 x 105
KP =
(0.08205 x 298) 2
K P = 5.9 x 105

14. Consider the Following Reaction:
CaCO3 (s) CaO(s) +CO2 (g)
K c = CO2 (g) 
 
PCO2
K c = CO 2 (g)  =
  RT
PCO2 K p
Kc = 
RT RT
(RT) K c = K p
PCO2
K c = CO 2 (g)  =
  RT
PCO2 Kp
Kc = 
RT RT
(RT) K c = K p

15. Consider the Following Reaction:
Cu(NH3 )2(aq)
4 Cu 2+ + 4 NH3 (aq)
(aq)
4
 NH3 (aq)  Cu aq 
2+
Kc =    
Cu(NH3 )4(aq) 

2


16. Manipulating the Equilibrium Expression:
1
C(s) + O2 (g) CO (g)
2
CO (g) 
K1 =   = 4.6 x 1023 at 25o C
1
O2 (g)  2
 

17. Manipulating the Equilibrium Expression:
2 C(s) + O2 (g) 2CO (g)
2
CO (g) 
K2 =   = K 2 = 2.1 x 1047 at 25o C
O2 (g) 
1
 

18. Manipulating the Equilibrium Expression:
If Keq equals 3.5 x 108 for
N2 (g) + 3 H2 (g) 2NH3 (g)
Calculate Keq for
1 3
NH3 (g) N 2 (g) + H 2 (g)
2 2

19. Manipulating the Equilibrium Expression:
1 3
NH3 (g) N 2 (g) + H 2 (g)
2 2
1
= 5.3 x 10-5
3.5 x 108

20. Finding K for a Reaction from Other K values
AgCl (s) + 2 NH3 (aq) Ag(NH3 )2 (aq) + Cl-(aq)
+
 Ag(NH3 )2  Cl(aq) 
+ -
K eq =  
2

 NH3 (aq) 
 

21. Finding K for a Reaction from Other K values
AgCl (s) Ag +(aq) + Cl-(aq)
Ksp = Ag +(aq)  Cl(aq)  = 1.8 x 10-10
 
-

+ +
Ag(aq) + 2 NH3 (aq) Ag(NH3 )2 (aq)
 Ag(NH3 ) + 
 2
Kf = 2
= 1.6 x 107
 Ag (aq)   NH3 (aq) 

+
 

22. Finding K for a Reaction from Other K values
 Ag(NH3 ) 2 

+
  Ag(NH 3 ) 2  Cl(aq) 
+ -
2
x  Ag (aq)  Cl(aq)  = 

+

-


2

 Ag (aq)   NH 3 (aq) 

+
   NH 3 (aq) 
 
 Ag(NH3 ) 2  Cl(aq) 
+ -
K f x K sp =  
2

 NH3 (aq) 
 
 Ag(NH 3 ) 2  Cl(aq) 
+ -
1.6 x 107 x 1.8 x 10-10 =  
2

 NH3 (aq) 
 
 Ag(NH3 ) 2  Cl(aq) 
+ -
2.9 x 10-3 =  
2
 = K eq
 NH 3 (aq) 
 

23. Calculate Kp at 226.8oC for Reaction (4) given Reactions
(1), (2), and (3):
(1) H 2 (g) + Br2 (g) 2 HBr (g) K P = 7.9 x 1011
(2) H 2 (g) 2 H (g) K P = 4.8 x 10-41
(3) Br2 (g) 2 Br (g) K P = 2.2 x 10-15
(4) H (g) + Br (g) HBr (g) KP = ?

24. Kp at 226.8oC for
H(g) + Br (g) HBr (g)
(1) H 2 (g) + Br2 (g) 2 HBr (g) K P1 = 7.9 x 1011
(2) H 2 (g) 2 H (g) K P2 = 4.8 x 10-41
(3) Br2 (g) 2 Br (g) K P3 = 2.2 x 10-15
(4) H (g) + Br (g) HBr (g) K P4 = ?

25. Kp at 226.8oC for
H(g) + Br (g) HBr (g)
2 2
 HBr(g) 
   H 2 (g) 
  x  Br2 (g)  =
   HBr(g) 
 
x
 H 2 (g)   Br2 (g) 
    H (g) 
2
 Br (g) 
2 2
 H (g)   Br (g) 
2
       
1 1
K P1 x x = K 24
P
K P2 K P3
2
 HBr(g) 
   HBr(g) 
 
= = K 24 = K P4
 H (g)   Br (g) 
2 2 P
 H (g)   Br (g) 
      

26. Kp at 226.8oC for
H(g) + Br (g) HBr (g)
7.9 x 1011
= K P4 = 2.7 x 1033
4.8 x 10-41 x 2.2 x 10-15

27. The Reaction Quotient, Q
• Q is obtained by using a similar formula as K, but
Q may not be at equilibrium. The reaction is at
dynamic equilibrium when Q = Keq
• If Q is greater than Keq , then the reaction is not at
equilibrium and must adjust so that product(s)
will go to reactant(s) until a state of dynamic
equilibrium is reached
• If Q is less than Keq , then the reaction is not at
equilibrium and must adjust so that more
reactant(s) will go to product(s) until a state of
dynamic equilibrium is reached

28. Application of Q
For the following equilibrium:
CH3CH2CH2CH3 (CH3 )2CHCH3
Keq = 2.5 at 25oC is the reaction at equilibrium when
[CH3CH2CH2CH3] = 0.97 M and [(CH3)2CHCH3] = 2.18M?
If not, calculate the equilibrium concentrations.

29. Application of Q
(CH3 )2CHCH3  =Q
CH3CH 2CH 2CH3 
2.18 M
=Q
0.97 M
2.25 = Q
Therefore, Q is less than K eq

30. Application of Q
Since Q (2.25) is less than Keq (2.5)
0.97 M - x 2.18 + x
CH3CH2CH2CH3 (CH3)2CHCH3
2.18 + x
= 2.5
0.97 - x
x = 0.07
Therefore, the equilibrium concentration of
CH3CH2CH2CH3 is 0.90 M (0.97 – 0.07) and the equilibrium
concentration of (CH3)2CHCH3 is 2.25M (2.18 + 0.07)

31. Another Application of Q
For the following equilibrium:
N2 (g) + O2 (g) 2 NO (g)
Keq = 1.7 x 10-3 is the reaction at equilibrium when
[N2] = 0.50 M , [O2] = 0.25 M, and [NO] = 0.0042 M?
If not, calculate the equilibrium concentrations.

32. Another Application of Q
For the following equilibrium:
 NO
2
Q=
 N 2 O2 
0.0042
2
Q= = 1.41 x 10-4
0.500.25
Q (0.000141) is less than Keq (0.0017)

33. Another Application of Q
0.50 M - x 0.25 - x 0.0042+2x
N 2 (g) + O 2 (g) 2 NO (g)
(0.0042+2x)2
K eq = 1.7 x 10-3 =
(0.50 -x)(0.25 - x)
x = 5.1 x 10-3

34. Another Application of Q
x = 5.1 x 10-3
(0.0042+2 x 5.1 x 10-3 ) 2
=
(0.50 -0.0051)(0.25 - 0.0051)
(0.0144) 2
= 1.7x10-3
(0.495)(0.245)

35. Let’s calculate the equilibrium constant for the
following reaction at 1000 K:
2 SO2 (g) + O2 (g) 2 SO3 (g)
1.00 mol of SO2 and 1.00 mol of O2
are placed in a 1.00 L flask. When equilibrium
is achieved 0.925 mol of SO2 is formed.

36. Let’s calculate the KP for the following reaction at 800 K:
2 H2S (g) 2 H2 (g) + S2 (g)
A tank contains hydrogen sulfide gas at 10.00 atm at 800 K.
When equilibrium is achieved, the partial pressure of S2
is 2.0 x 10-2 atm.
What is Kc for this reaction?

37. Let’s calculate the Kc for the following reaction:
H
I
I + I2
H
If 0.050 mol of trans-1,2-diodocyclohexane is dissolved
in carbon tetrachloride to make a 1.00 L solution.
At equilibrium the concentration of I2 is 0.035 M.
What is Kp for this reaction?

38. Kc for the following reaction is 1.0 x 10-5 at 1227oC:
N2 (g) + O2 (g) 2 NO(g)
Calculate the equilibrium concentrations if 0.80 mole of N2
and 0.20 mole of O2 are placed initially in a 2.00 L flask at 1227oC.
Calculate Kp at 1227oC, and calculate the equilibrium
pressures at N2, O2, and NO at 1227oC.

39. Le Chatelier’s Principle
When a system in dynamic equilibrium
encounters stress, the system will adjust itself
in order to relieve the stress.

40. The Stresses
• Change the temperature
• Change the concentration of reactants or
products
• Change the volume
• Change the pressure (only when there is a
difference between the number of moles of
gaseous reactants and products)

41. Changing the Temperature
• Adjustment in the equilibrium is determined
by whether the reaction is endothermic or
exothermic.
• If the reaction is exothermic, an increase in
temperature shifts the equilibrium toward the
reactants. A decrease in temperature shifts
the reaction toward the products.

42. Changing the Temperature
• If the reaction is endothermic, an increase in
temperature shifts the equilibrium toward the
products. A decrease in temperature shifts
the reaction toward the reactants.
• Changes in temperature result in changes in
the value of the equilibrium constant.
ΔHo
reaction
-
RT
K eq = A e
ΔH o
reaction
ln K eq =- + ln A
RT

43. For example, the following data were obtained for changes in the equilibrium
Constant with changes in temperature for the reaction
1 1
N 2 (g) + O2 (g) NO (g)
2 2
T Keq ln Keq 104(1/T)
oC
25 4.5 x 10-31 -69.9 33.6
500 2.43 x 10-21 -47.5 12.9
1000 6.07 x 10-19 -41.9 7.89
1500 6.74 x 10-18 -39.5 5.64
2000 2.60 x 10-17 -38.2 4.40
2500 6.15 x 10-17 -37.3 3.61

44. 0
0 5 10 15 20 25 30 35 40
-10
-20
-30
Ln K -40
y = -1.0832x - 33.502
-50
-60
-70
-80
104(1/T)

45. ΔH reaction
slope = -
R
4 ΔH reaction
-1.0832 x 10 K = -
R
J
4
-1.0832 x 10 K x -8.314 = ΔH reaction
K-mol
J
9.00 x 10 4
= ΔH reaction
mol

46. Changes in Concentration of Reactants and Products
Increasing the concentration of the reactants
creates a situation where Q is less than K;
therefore, the system will adjust itself to
produce more product so that Q will equal K,
i.e., the equilibrium will shift toward the
product.

47. Changes in Concentration of Reactants and Products
Decreasing the concentration of the reactants
creates a situation where Q is greater than K;
therefore, the system will adjust itself to
produce more reactants so that Q will equal K,
i.e., the equilibrium will shift toward the
reactants.

48. An Example of the Effect of Concentration Change on
an Equilibrium Reaction
CH3CH2CH2CH3 (CH3 )2CHCH3
If at 25oC, the equilibrium concentrations of n-butane and
Isobutane are respectively 0.500 M and 1.25 M, what would be
the equilibrium concentrations if 1.50 mol of n-butane is added
to the equilibrium mixture?

49. K eq =
(CH3 )2CHCH3 
CH3CH 2CH 2CH3 
1.25
K eq = = 2.5
0.50
1.75 M - x 1.25 M + x
CH3CH2CH2CH3 (CH3)2CHCH3
1.25 mol + x mol
K eq = V = 2.5
1.75 mol - xmol
V
1.25 + x
= 2.5
1.75 - x
1.25 + x = 2.5 1.75 - x 
1.25 + x = 4.375 - 2.5 x
3.125 = 3.5 x
0.89 = x

50. CH3CH2CH2CH3   1.75 M -0.89 M = 0.86 M
(CH3 )2CHCH3   1.25 M + 0.89 M = 2.14 M

51. Changing the Volume
• If the number of moles of products equals the
number of moles of reactants, then a change in
volume of the reaction vessel will have no affect
on the equilibrium.
• If the number of moles of product is greater than
the number of moles of reactants, then an
increase in volume of the reaction vessel will shift
the equilibrium toward the formation of product.
A decrease in the volume will shift the
equilibrium toward the reactants.

52. Changing the Volume
• If the number of moles of reactant is greater
than the number of moles of products, then
an increase in the volume of the reaction
vessel will shift the equilibrium toward the
reactants. A decrease in volume will shift the
equilibrium toward the products.

53. Effect of Pressure Change on a Gaseous Reaction
• If the number of moles of gaseous products
equals the number of moles of gaseous reactants,
then pressure changes have no effect on the
equilibrium reaction.
• If the number of moles of gaseous products is
greater than the number of moles of gaseous
reactants, then an increase in pressure shifts the
equilibrium toward the reactants. A decrease in
pressure shifts the equilibrium toward the
products.

54. Effect of Pressure Change on a Gaseous Reaction
If the number of moles of gaseous reactants is
greater than the number of moles of gaseous
products, then an increase in pressure shifts
the equilibrium toward the products. A
decrease in pressure shifts the equilibrium
toward the reactants.