IB Chemistry on Equilibrium Constant, Kc and Reaction Quotient, Qc.

1. Tutorial on Dynamic Equilibrium, Equilibrium
constant Kc and Reaction quotient Qc.
Prepared by
Lawrence Kok
http://lawrencekok.blogspot.com

2. Dynamic Equilibrium
Closed system
Reversible
Forward Rate, Kf
Reverse Rate, Kr
2NO2(g) N2O4(g)
combining dissociation
Conc vs time Rate vs time
Conc
Time
Conc NO2
Conc N2O4
Forward rate
brown colourless
2NO2(g) N2O4(g)
Backward rate

3. Dynamic Equilibrium
Closed system
Reversible
Forward Rate, Kf
Reverse Rate, Kr
2NO2(g) N2O4(g)
combining dissociation
Chemical system
Forward rate rxn
Rate Combining
Backward rate rxn
Rate dissociation
Rate of forward = Rate of backward
Conc of reactant and product
remain UNCHANGED/CONSTANT not EQUAL
Reversible rxn happening, same time with same rate
Conc vs time Rate vs time
Conc
Time
Conc NO2
Conc N2O4
Forward rate
brown colourless
2NO2(g) N2O4(g)
Backward rate
With time
• Conc NO2 decrease -Forward rate decrease
• Conc N2O4 increase - Backward rate increase
Forward Rate = Backward Rate
Conc NO2 and N2O4 remain UNCHANGED/CONSTANT

4. How dynamic equilibrium is achieved in closed system?
Conc of NO2 decrease ↓over time
NO2
2NO2(g) N2O4(g)
Conc of N2O4 increase ↑ over time
N2O4
1 As reaction proceeds
concentration

5. How dynamic equilibrium is achieved in closed system?
Conc of NO2 decrease ↓over time
NO2
Forward rate, Kf decrease ↓ over time
2NO2(g) N2O4(g)
Conc of N2O4 increase ↑ over time
N2O4
Reverse rate, Kr increase ↑ over time
NO2
N2O4
1
2
As reaction proceeds
concentration
As reaction proceeds
rate

6. How dynamic equilibrium is achieved in closed system?
Conc of NO2 decrease ↓over time
NO2
Forward rate, Kf decrease ↓ over time
Forward Rate = Reverse Rate
2NO2(g) N2O4(g)
Conc of N2O4 increase ↑ over time
N2O4
Reverse rate, Kr increase ↑ over time
NO2
N2O4
1
2
Conc of reactant/product remain constant
Rate
3
Time
Conc
NO2
N2O4
As reaction proceeds
concentration
As reaction proceeds
rate
At dynamic equilibrium
Time
Click here to view simulation

7. Conc vs Time
How dynamic equilibrium is achieved in a closed system?
Rate forward = ½ breakdown = ½ x 40 = 20
40 0
Rate reverse = ¼ form = ¼ x 0 = 0
Rate forward = ½ breakdown = ½ x 20 = 10
20 20
Rate reverse = ¼ form = ¼ x 20 = 5
Rate forward = ½ breakdown = ½ x 15 = 8
15 25
Rate reverse = ¼ form = ¼ x 25 = 6
Rate forward = ½ breakdown = ½ x 13 = 7
13 27
Rate reverse = ¼ form = ¼ x 27 = 7
13 27
At dynamic Equilibrium
Rate forward = Rate reverse
Breakdown (7) = Formation (7)
At dynamic Equilibrium
Conc reactant 13 /Product 27 constant
Rate vs Time
1/ 2
1/ 4
.. tan ..
  K
 
1  
rate .. cons tan t ..
reverse
 1
rate cons t forward
K
 
 
1/ 2
1   
K2
2
c 27
product
  
13
reac tan
t
1/ 4
 K
1
K
Kc or

8. Dynamic Equilibrium
Forward Rate, K1 Reverse Rate, K-1
Reversible (closed system)
At Equilibrium
Conc vs time Rate vs time
Forward rate = Backward rate
A + B
C + D
Conc
Time
Conc reactants and products remain
CONSTANT/UNCHANGE

9. Dynamic Equilibrium
Forward Rate, K1 Reverse Rate, K-1
Reversible (closed system)
At Equilibrium
Forward rate = Backward rate
Conc of product and reactant
at equilibrium
Conc reactants and products remain
CONSTANT/UNCHANGE
Equilibrium Constant Kc
aA(aq) + bB(aq) cC(aq) + dD(aq)
coefficient
Solid/liq not included in Kc
Conc represented by [ ]
K1
K-1
Conc vs time Rate vs time
A + B
C + D
Conc
Time

10. Dynamic Equilibrium
Forward Rate, K1 Reverse Rate, K-1
Reversible (closed system)
At Equilibrium
aA(aq) + bB(aq) cC(aq) + dD(aq)
coefficient
Solid/liq not included in Kc
C D
1
K

rate cons t forward
K
.. tan ..
1 

Kc = ratio of molar conc of product (raised to power of their respective stoichiometry coefficient)
to molar conc of reactant (raised to power of their respective stoichiometry coefficient)
Forward rate = Backward rate
Conc of product and reactant
at equilibrium
Conc reactants and products remain
CONSTANT/UNCHANGE
Equilibrium Constant Kc
Conc represented by [ ]
K1
K-1
   
 a  b
c d
c
A B
K 
1

K
Kc
Equilibrium Constant Kc
express in
Conc vs time Rate vs time
A + B
C + D
Conc
Time
Excellent Notes
Click here notes on dynamic equilibrium
rate cons t reverse
K
.. tan ..
1

11. Magnitude of Kc
Position of equilibrium
   
 a  b
c d
c
C D
A B
K 
Extend of reaction
How far rxn shift to right or left?
Not how fast
   
   a b
c d
K   
c
C D
A B
 c K  c K
92 3 10   c K
1
81  310 c K 2  8.710 c K

12. 81  310 c K
Large Kc
C D
• Position equilibrium shift to right
• More products > reactants
Magnitude of Kc
Position of equilibrium
   
 a  b
c d
c
C D
A B
K 
Extend of reaction
How far rxn shift to right or left?
Not how fast
   
   a b
c d
c
A B
K 
92 3 10   c K
Small Kc
• Position equilibrium shift to left
• More reactants > products
 
 c K  c K
2CO2(g) ↔ 2CO(g) + O2(g)
2H2(g) + O2(g) ↔ 2H2O(g)
1
2  8.710 c K
H2(g) + I2(g) ↔ 2HI(g)
Kc
• Position equilibrium lies slightly right
• Reactants and products equal amount

13. 81  310 c K
Large Kc
C D
• Position equilibrium shift to right
• More products > reactants
Magnitude of Kc
Position of equilibrium
   
 a  b
c d
c
C D
A B
K 
Extend of reaction
How far rxn shift to right or left?
Not how fast
   
   a b
c d
c
A B
K 
92 3 10   c K
Small Kc
• Position equilibrium shift to left
• More reactants > products
 
 c K  c K
2CO2(g) ↔ 2CO(g) + O2(g)
2H2(g) + O2(g) ↔ 2H2O(g)
1
2  8.710 c K
H2(g) + I2(g) ↔ 2HI(g)
Kc
• Position equilibrium lies slightly right
• Reactants and products equal amount
Reaction completion
Reactant favoured Reactant/Product equal Product favoured
c K
Temp
dependent
Extend
of rxn
Not how fast

14. Equilibrium Constant Kc
aA(aq) + bB(aq) cC(aq) + dD(aq)
   
 a  b
c d
c
C D
A B
K 
Conc of product and reactant at equilibrium
Equilibrium expression HOMOGENEOUS gaseous rxn
4NH3(g) + 5O2(g) ↔ 4NO(g) + 6H2O(g) N2(g) + 3H2(g) ↔ 2NH3(g)
NO H O
Equilibrium expression HOMOGENEOUS liquid rxn
CH3COOH(l) + C2H5OH(l) ↔ CH3COOC2H5(l) + H2O(l)
NH4CI(s) ↔ NH3(g) + HCI(g)
2SO2(g) + O2(g) ↔ 2SO3(g)
   
   5
2
4
3
6
2
4
NH O
Kc 
 
   3
2
NH
1
2
2
3
N H
Kc 
   
 0
1 1
NH HCI
3
NH CI
4
   1 1
Kc 
3 K NH HCI c 
 
   1
2
SO
2
2
2
3
SO O
Kc 
 CH COOC H   H O

   1
Equilibrium expression HETEROGENOUS rxn
CaCO3(s) ↔ CaO(g) + CO2(g)
 CaO   CO

 0
3
1
2
1
CaCO
Kc 
   1
2
1 K CaO CO c 
2 5
1
3
1
2
1
3 2 5
CH COOH C H OH
Kc 
Cu2+
(aq) + 4NH3(aq) ↔ [Cu(NH3)4]2+
   
   4
3
Cu NH
2 1
2
3 4 ( )
Cu NH
Kc



Reactant/product same phase
Reactant/product diff phase

15. Equilibrium Constant Kc Equilibrium Constant Kc
bB aA
A
2aA 2bB
B
K 2
1
aA bB
aA bB
 
b
 a
K 
c
B
A
aA bB
B
aA bB
 
 b
a
c
B
K  '
Effect on Kc
Inverse Kc
c
c K
K
' 1 
inverse
X2 coefficient
Square Kc
' 2
c c K  K
coefficient
1
2
 
b
  a
c
B
A
K
2
1
2
1
Square root c K
' 1
'  c c c K  K 2  K
 
b
 a
c
A
K 
 
b
 a
c
B
A
K 
 
b
  a
c
A
2
' 
2
1
+ 2 reactions + aA cC
aA bB bB cC
 
b
 a
ci
B
A
K 
 
 b
c
cii
C
B
K 
 
 
 
 
 
c
 a
b
K    '
a
c
b
c
C
A
B
A
C
B
Multiply both Kc
c cii ci K  K  K '
2
cii K ci K

16. N2(g) + O2(g) ↔ 2NO(g)
2NO(g) + O2(g) ↔ 2NO2(g)
19 10 3. 2    ci K
 3106 cii K
N2(g) + 2O2(g) ↔2NO2(g) K K K
2NO2(g) ↔ N2(g) + 2O2(g)
 
c ci cii
   
2.3 10 3 10
13
19 6
7 10


 
c
c
K
K
13 7 10   c K
1 1
  
7 
10
' 12
13
'
 
1.42 10
c
c
c
K
K
K
HF(ag) ↔ H+
(aq) + F -
(aq)
H2C2O4(ag) ↔ 2H+
(aq) + C2O4
2 -
(aq)
4 10 8. 6    ci K
6 3.8 10   cii K
2- ↔ 2F -
2HF(ag) ↔ 2H+
(aq) + 2F -
(aq)
2- ↔ H2C2O4(aq)
Add 2 rxn
2- ↔ H2C2O4(aq)
2- ↔ 2F -
2HF(ag) + C2O4
1 1
'' 2.6 10
  
(aq) + H2C2O4(aq)
2H+
(ag) + C2O4
' 2  4 2 7 6.8 10 4.6 10       c ci K K
5
6
3.8 10
 

cii
c K
K
' ''
K  K 
K
c c c
K
4.6 10 
7 2.6 10 5
0.12     
c
Kc for diff rxn
Adding 2 rxns
+
Inverse rxn
Adding 2 rxns
2HF(ag) + C2O4
(aq) + H2C2O4(aq)
+
HF(ag) ↔ H+
(aq) + F -
(aq)
4 6.8 10    ci K
x2 coefficient
H2C2O4(ag) ↔ 2H+
(aq) + C2O4
2 -
Inverse rxn
6 3.8 10    cii K
2HF(ag) ↔ 2H+
(aq) + 2F -
(aq)
2H+
(ag) + C2O4
' 7 4.6 10   c K
'' 5  2.610 c + K
Effect on Kc Effect on Kc
Inverse rxn Inverts expression
Doubling rxn coefficient Squares expression
Tripling rxn coefficient Cubes expression
Halving rxn coefficient Square root expression
Adding 2 reactions Multiplies 2 expression
1
c K
2
c K
3
c K
c K
ii
c
i
c K K
Square Kc
Invert Kc
Multiply Kc
1
2
3
N2(g) + 2O2(g) ↔ 2NO2(g)

17. H2 + I2 ↔ 2HI
 HI

   1
 50 c K
2
1
2
2
H I
Kc 
2HI ↔ H2 + I2
   
 2
1
2
1
H I
' 2
HI
Kc 
0.02
' 1 1   
50
c
c K
K
2SO2 + O2 ↔ 2SO3
 
   1
2
SO
2
2
2
3
SO O
Kc 
 200 c K
1
SO2 + O2 ↔ SO3
2
1



SO
200 14.1 '    c c K K
4SO2 + 2O2 ↔ 4SO3





SO
' 2 2
   200

40000
K K
c c
K
,
 

c
Kc is 170 at 500K
Determine if rxn is at equilibrium when conc are at:
[N2] =1.50, [H2] = 1.00, [NH3] = 8.00
N2(g) + 3H2(g) ↔ 2NH3(g)
 
   3
2
NH
1
2
2
3
N H
Kc 
 NH

2
3
  1
 
3
2
2
 8.00

1.501.00


c
c
Q
N H
Q
• Rxn not at equilibrium
• Shift to right, favour product
• Qc must increase, till equal to Kc
IB Questions
Determine Kc for inversing rxn
inverse
Determine Kc for halving rxn
 
   
2
1
2
2
2
2
3






SO O
Kc
halving
Determine Kc for doubling rxn
2SO2 + O2 ↔ 2SO3
doubling
 
   1
2
SO
2
2
2
3
SO O
Kc 
 200 c K
 
   
2
1
2
2
2
2
3




SO O
Kc
1 2
4 3
 170 c  42.7 K c Q
c c Q  K

18. Kc and Qc
Initial conc of H2 , I2 and HI At equilibrium
H2(g) + I2(g) ↔ 2HI(g)
 
   1
2
HI
1
2
2
H I
Kc 
 2.52 10

2 2


c K  46.4 c K
   
46.4
2 1 2 1
1.14 10 0.12 10

 

 
Expt Initial
Conc H2
Initial
Conc I2
Initial
Conc HI
1 2.40 x 10-2 1.38 x 10-2 0
Expt Equilibrium
Conc H2
Equilibrium
Conc I2
Equilibrium
Conc HI
1 1.14 x 10-2 0.12 x 10-2 2.52 x 10-2
At equilibrium conc

19. Kc and Qc
Initial conc of H2 , I2 and HI At equilibrium
Expt Initial
Conc H2
1 2.40 x 10-2 1.38 x 10-2 0
H2(g) + I2(g) ↔ 2HI(g)
Initial
Conc HI
 
   1
2
HI
1
2
2
H I
Initial
Conc I2
Kc 
c K  46.4 c K
 4.00 c Q
 
   1
2
HI
1
2
2
H I
Initial conc of H2 , I2 and HI
Qc 
 2.52 10

2 2


   
46.4
2 1 2 1
1.14 10 0.12 10

 

 
Expt Initial
Conc H2
Initial
Conc I2
Initial
Conc HI
1 0.0500 0.0500 0.100
Expt Equilibrium
Conc H2
Equilibrium
Conc I2
Equilibrium
Conc HI
1 1.14 x 10-2 0.12 x 10-2 2.52 x 10-2
At equilibrium conc
Not at equilibrium
H2(g) + I2(g) ↔ 2HI(g)
 
  
4.00
0.100 2
  c Q
0.050 0.050

20. Kc and Qc
Expt Initial
Conc H2
1 2.40 x 10-2 1.38 x 10-2 0
H2(g) + I2(g) ↔ 2HI(g)
2 2


c K  46.4 c K
2 1 2 1
  c Q
c K
2.52 10
0.100 2
Constant at
fixed Temp
Initial
Conc HI
 
   1
2
HI
1
2
2
H I
Initial
Conc I2
Kc 
At equilibrium
Independent of
initial conc
Initial conc of H2 , I2 and HI
 4.00 c Q
 
   1
2
HI
1
2
2
H I
Initial conc of H2 , I2 and HI
Qc 
 
   
46.4
1.14 10 0.12 10

 

 
Expt Initial
Conc H2
Initial
Conc I2
Initial
Conc HI
1 0.0500 0.0500 0.100
Expt Equilibrium
Conc H2
Equilibrium
Conc I2
Equilibrium
Conc HI
1 1.14 x 10-2 0.12 x 10-2 2.52 x 10-2
At equilibrium conc
Not at equilibrium
H2(g) + I2(g) ↔ 2HI(g)
 
  
4.00
0.050 0.050
Difference between
Predict the
direction of rxn
c Q
Conc of
product/reactant
at equilibruimconc
Reaction quotient
at particular time
Not at equilibrium
conc
Varies NOT constant

21. Kc and Qc
H2(g) + I2(g) ↔ 2HI(g)
 
   1
2
HI
1
2
2
H I
Kc 
 2.52 10

2 2


c K  46.4 c K
   
46.4
2 1 2 1
1.14 10 0.12 10

 

 
At equilibrium conc
c c Q  K c c Q  K
c c Q  K
Reaction at
Initial conc of H2 , I2 and HI
Expt Initial equilibrium
Conc H2
Initial
Conc I2
Initial
Conc HI
1 0.0500 0.0500 0.100
Initial conc of H2 , I2 and HI
Expt Initial
Conc H2
Initial
Conc I2
Initial
Conc HI
1 0.0250 0.0350 0.300

22. Kc and Qc
H2(g) + I2(g) ↔ 2HI(g)
 
   1
2
HI
1
2
2
H I
Kc 
 2.52 10

2 2


c K  46.4 c K
   
46.4
2 1 2 1
1.14 10 0.12 10

 

 
At equilibrium conc
c c Q  K c c Q  K
c c Q  K
Reaction at
equilibrium
Initial conc of H2 , I2 and HI
Qc   
  c Q
More product > reactant
→
HI
0.300 2
Rxn shift left more reactant
c c Q  K
 c Q
Bring Qc down
Initial conc of H2 , I2 and HI
0.100 2
  c Q
More reactant > product
HI
Rxn shift right → more product
Bring Qc up  c Q
c c Q  K
 c Q
Expt Initial
Conc H2
Initial
Conc I2
Initial
Conc HI
1 0.0500 0.0500 0.100
 
   1
2
1
2
2
H I
Qc 
 
  
4.00
0.050 0.050
 c Q
Expt Initial
Conc H2
Initial
Conc I2
Initial
Conc HI
1 0.0250 0.0350 0.300
 
   1
2
1
2
2
H I
  
103
0.0250 0.0350
Click here to view notes

23. Kc from reaction stoichiometry
4 diff initial conc of H2 , I2 and HI At equilibrium Kc = 46.4 ( 730K)
H2(g) + I2(g) ↔ 2HI(g)
2.52 10
2 2


2 1 2 1
K  same  46.4 c
 
   1
2
HI
1
2
2
H I
Kc 
 
   
46.4
1.14 10 0.12 10

 

 
c Rxn 1 K
same
Rxn 2, 3, 4
diff initial conc H2(g) + I2(g) ↔ 2HI(g)
 
   1
2
HI
1
2
2
H I
Kc 

24. Kc from reaction stoichiometry
4 diff initial conc of H2 , I2 and HI At equilibrium Kc = 46.4 ( 730K)
H2(g) + I2(g) ↔ 2HI(g)
2.52 10
2 2


2 1 2 1
K  same  46.4 c
 
   1
2
HI
1
2
2
H I
Kc 
 
   
46.4
1.14 10 0.12 10

 

 
c Rxn 1 K
same
Rxn 2, 3, 4
diff initial conc
more products
H2(g) + I2(g) ↔ 2HI(g)
 c Q
HI
Rxn shift to right
more reactants Rxn shift to left
 product

reac t
Qc
tan

 product

reac t
Qc
tan

 c Q
 
   1
2
1
2
2
H I
Kc 

25. Kc from reaction stoichiometry
4 diff initial conc of H2 , I2 and HI At equilibrium Kc = 46.4 ( 730K)
H2(g) + I2(g) ↔ 2HI(g)
2.52 10
2 2


2 1 2 1
K  same  46.4 c
 
   1
2
HI
1
2
2
H I
Kc 
 
   
46.4
1.14 10 0.12 10

 

 
c Rxn 1 K
same
 
2
   1
2
HI
1
H I
2
Kc 
Qc = Kc - rxn at equilibrium, no side/shift occur
Qc < Kc – rxn shift right, favour product
Qc > Kc – rxn shift left, favour reactant
Rxn 2, 3, 4
diff initial conc
more products
H2(g) + I2(g) ↔ 2HI(g)
 c Q
Rxn shift to right
more reactants Rxn shift to left
 product

reac t
Qc
tan

 product

reac t
Qc
tan

 c Q
c c Q  K
c c Q  K c c Q  K

26. Kc and Qc
H2(g) + I2(g) ↔ 2HI(g)
 
   1
2
HI
1
2
2
H I
Kc 
2.52 10
2 2


c K  46.4 c K
2 1 2 1
 4.00 c Q  
   1
2
HI
1
2
2
H I
Initial conc of H2 , I2 and HI
Qc 
 
   
46.4
1.14 10 0.12 10

 

 
Expt Initial
Conc H2
Initial
Conc I2
Initial
Conc HI
1 0.0500 0.0500 0.100
At equilibrium conc
Not at equilibrium
H2(g) + I2(g) ↔ 2HI(g)
 
  
4.00
0.100 2
  c Q
0.050 0.050
c c Q  K c c Q  K
Reaction at
equilibrium
 4.00 c Q  46.4 c < K
More reactant > product
Rxn shift right → more product
Bring Qc up  c Q
 c Q
c c Q  K

27. Kc and Qc
H2(g) + I2(g) ↔ 2HI(g)
 
   1
2
HI
1
2
2
H I
Kc 
c K  46.4 c K
 103 c Q
 
   1
2
HI
1
2
2
H I
Qc 
 2.52 10

2 2


   
46.4
2 1 2 1
1.14 10 0.12 10

 

 
Initial conc of H2 , I2 and HI
At equilibrium conc
Not at equilibrium
H2(g) + I2(g) ↔ 2HI(g)
  c Q
c c Q  K
c c Q  K
Reaction at
equilibrium
0.300 2
More product > reactant
→
Rxn shift left more reactant
c c Q  K
 c Q
Bring Qc down  c Q
Expt Initial
Conc H2
Initial
Conc I2
Initial
Conc HI
1 0.0250 0.0350 0.300
 
  
103
0.0250 0.0350
 103 c Q  46.4 c > K

28. How dynamic equilibrium is shifted when H2 is added ?
N2(g) + 3H2(g) ↔ 2NH3(g) 07. 4  c K
At equilibrium
Conc reactant/product no change
Equilibrium Conc H2 = 0.82M
Equilibrium Conc N2 = 0.20M
Equilibrium Conc NH3 = 0.67M
 
   3
2
NH
1
2
2
3
N H
Kc 
 
2
0.67
 0.20  1  0.82
3
 c K
 4.07 c K

29. How dynamic equilibrium is shifted when H2 is added ?
N2(g) + 3H2(g) ↔ 2NH3(g) 07. 4  c K
Equilibrium disturbed
H2 added. More reactant
At equilibrium
Conc reactant/product no change
NH
0.67
 2.24 c Q
Equilibrium Conc H2 = 0.82M
Equilibrium Conc N2 = 0.20M
Equilibrium Conc NH3 = 0.67M
 
   3
2
NH
1
2
2
3
N H
Kc 
 
2
0.67
 0.20  1  0.82
3
 c K
New Conc H2 = 1.00M
Conc N2 = 0.20M
Conc NH3 = 0.67M
 
   3
2
1
2
2
3
N H
Qc 
 
2
 0.20  1  1.00
3
 c Q
 4.07 c K

30. How dynamic equilibrium is shifted when H2 is added ?
N2(g) + 3H2(g) ↔ 2NH3(g) 07. 4  c K
Equilibrium disturbed
H2 added. More reactant
At equilibrium
Conc reactant/product no change
At new equilibrium
Conc reactant/product no change
NH
0.67
 2.24 c Q
Equilibrium Conc H2 = 0.82M
Equilibrium Conc N2 = 0.20M
Equilibrium Conc NH3 = 0.67M
 
   3
2
NH
1
2
2
3
N H
Kc 
 
2
0.67
 0.20  1  0.82
3
 c K
New Conc H2 = 1.00M
Conc N2 = 0.20M
Conc NH3 = 0.67M
 
   3
2
1
2
2
3
N H
Qc 
 
2
 0.20  1  1.00
3
 c Q
 4.07 c K
New Equilibrium Conc H2 = 0.90M
New Equilibrium Conc N2 = 0.19M
New Equilibrium Conc NH3 = 0.75M
NH
 
2
0.75
 0.19  1  0.90
3
 c K
 
   3
2
1
2
2
3
N H
Kc 
 4.07 c K

31. How dynamic equilibrium is shifted when H2 is added ?
• Add H2 , Qc decrease
• Position equilibrium shift to right
• Rate forward and reverse increase
• New equilibrium conc achieved when
Rate forward Kf = Rate reverse Kr
• More product NH3 ,but Kc unchanged
N2(g) + 3H2(g) ↔ 2NH3(g) 07. 4  c K
Equilibrium disturbed
H2 added. More reactant
At equilibrium
Conc reactant/product no change
At new equilibrium
Conc reactant/product no change
NH
0.67
 2.24 c Q
Equilibrium Conc H2 = 0.82M
Equilibrium Conc N2 = 0.20M
Equilibrium Conc NH3 = 0.67M
 
   3
2
NH
1
2
2
3
N H
Kc 
 
2
0.67
 0.20  1  0.82
3
 c K
New Conc H2 = 1.00M
Conc N2 = 0.20M
Conc NH3 = 0.67M
 
   3
2
1
2
2
3
N H
Qc 
 
2
 0.20  1  1.00
3
 c Q
 4.07 c K
New Equilibrium Conc H2 = 0.90M
New Equilibrium Conc N2 = 0.19M
New Equilibrium Conc NH3 = 0.75M
NH
 
2
0.75
 0.19  1  0.90
3
 c K
 
   3
2
1
2
2
3
N H
Kc 
 4.07 c K
c c Q  K
Shift to the right
- Increase product
- New Conc achieve
- Qc = Kc again

32. How dynamic equilibrium is shifted when H2 is added ?
• Add H2 , Qc decrease
• Position equilibrium shift to right
• Rate forward and reverse increase
• New equilibrium conc achieved when
Rate forward Kf = Rate reverse Kr
• More product NH3 ,but Kc unchanged
At equilibrium
Conc reactant/product no change
c c Q  K
Rate forward Kf = Rate reverse Kr
N2(g) + 3H2(g) ↔ 2NH3(g) 07. 4  c K
  4.07 c c Q K
c c Q  K

33. How dynamic equilibrium is shifted when H2 is added ?
• Add H2 , Qc decrease
• Position equilibrium shift to right
• Rate forward and reverse increase
• New equilibrium conc achieved when
Rate forward Kf = Rate reverse Kr
• More product NH3 ,but Kc unchanged
Rate forward Kf = Rate reverse Kr
N2(g) + 3H2(g) ↔ 2NH3(g) 07. 4  c K
  4.07 c c Q K
Equilibrium disturbed
H2 added. More reactant
c c Q  K
Equilibrium shift to right
 c Q
Rate forward Kf > Rate reverse Kr
At equilibrium
Conc reactant/product no change
c c Q  K c c Q  K

34. How dynamic equilibrium is shifted when H2 is added ?
• Add H2 , Qc decrease
• Position equilibrium shift to right
• Rate forward and reverse increase
• New equilibrium conc achieved when
Rate forward Kf = Rate reverse Kr
• More product NH3 ,but Kc unchanged
Rate forward Kf = Rate reverse Kr
N2(g) + 3H2(g) ↔ 2NH3(g) 07. 4  c K
  4.07 c c Q K
Equilibrium disturbed
H2 added. More reactant
c c Q  K
Equilibrium shift to right
 c Q
Rate forward Kf > Rate reverse Kr
At equilibrium
Conc reactant/product no change
At new equilibrium
Conc reactant/product no change
Qc increase until Qc = Kc
 c Q
Rate forward Kf = Rate reverse Kr
c c Q  K c c Q  K c c Q  K

35. Click here simulation using paper clips Click here to view simulation Click here simulation on reversible rxn
Click here on reversible rxn
Simulation on Dynamic equilibrium
Click here on equilibrium constant

36. Acknowledgements
Thanks to source of pictures and video used in this presentation
Thanks to Creative Commons for excellent contribution on licenses
http://creativecommons.org/licenses/
Prepared by Lawrence Kok
Check out more video tutorials from my site and hope you enjoy this tutorial
http://lawrencekok.blogspot.com