1) A reversible reaction between NO2 and N2O4 reaches dynamic equilibrium in a closed system when the forward and backward reaction rates become equal. 2) As the reaction proceeds, the concentration of reactants and products remain constant but the individual reaction rates change over time. 3) At equilibrium, the forward and backward reaction rates are equal and the concentrations of NO2 and N2O4 stop changing and remain constant.

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

2. How dynamic equilibrium is achieved in closed system?
Conc of NO2 decrease ↓over time
Forward rate, Kf decrease ↓ over time
Forward Rate = Reverse Rate
NO2
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
At dynamic equilibrium
As reaction proceeds Concentration
As reaction proceeds Rate
Time
Click here to view simulation

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

4. Dynamic Equilibrium
Reversible (closed system)
Forward Rate, K1 Reverse Rate, K-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)
Conc of product and reactant
at equilibrium
At Equilibrium
Forward rate = Backward rate
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
   
   a b
c d
c
A B
C D
K 
1
1


K
K
Kc
Equilibrium Constant Kc
express in
Conc vs time Rate vs time
A + B
C + D
Conc
Time
Click here notes on dynamic equilibrium
Excellent Notes
rate cons t reverse
rate cons t forward
K
K
.. tan ..
.. tan ..
1
1 


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

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

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

8. N2(g) + O2(g) ↔ 2NO(g)
2NO(g) + O2(g) ↔ 2NO2(g)
19 2.3 10   ci K
6  310 cii K
2NO2(g) ↔ N2(g) + 2O2(g)
13
19 6
7 10
2.3 10 3 10


 
   
 
c
c
c ci cii
K
K
N2(g) + 2O2(g) ↔ 2NO2(g) K K K
13 7 10   c K
' 12
13
'
1.42 10
7 10
1 1
 

  
c
c
c
K
K
K
HF(ag) ↔ H+
(aq) + F -
(aq)
H2C2O4(ag) ↔ 2H+
(aq) + C2O4
2 -
(aq)
4 6.8 10   ci K
6 3.8 10   cii K
2HF(ag) + C2O4
2- ↔ 2F -
(aq) + H2C2O4(aq)
2HF(ag) ↔ 2H+
(aq) + 2F -
(aq)
2H+
(ag) + C2O4
2- ↔ H2C2O4(aq)
' 2  4 2 7 6.8 10 4.6 10       c ci K K
5
6
'' 2.6 10
3.8 10
1 1
 

  
cii
c K
K
4.6 10 2.6 10 0.12 7 5
' ''
    
 

c
c c c
K
K K K
Kc for diff rxn
Adding 2 rxns
+
Inverse rxn
Adding 2 rxns
2HF(ag) + C2O4
2- ↔ 2F -
(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
2- ↔ H2C2O4(aq)
Add 2 rxn
' 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
c K
1
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)

9. H2 + I2 ↔ 2HI
 50 c K
 
   1
2
1
2
2
H I
HI
Kc 
2HI ↔ H2 + I2
   
 2
1
2
1
' 2
HI
H I
Kc 
0.02
50
' 1 1   
c
c K
K
2SO2 + O2 ↔ 2SO3
 
   1
2
2
2
2
3
SO O
SO
Kc 
 200 c K
SO2 + O2 ↔ SO3
200 14.1 '    c c K K
2
1
4SO2 + 2O2 ↔ 4SO3
   
40000
200
,
' 2 2

 
c
c c
K
K K
N2(g) + 3H2(g) ↔ 2NH3(g)
 
   3
2
1
2
2
3
N H
NH
Kc 
Kc is 170 at 500K
Determine if rxn is at equilibrium when conc are at:
[N2] =1.50, [H2] = 1.00, [NH3] = 8.00
 
   
 
1.501.00
8.00
3
2
1
2
2
3


c
c
Q
N H
NH
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
1
2
2
2
2
3
 


 



SO O
SO
Kc
halving
Determine Kc for doubling rxn
2SO2 + O2 ↔ 2SO3
doubling
 
   1
2
2
2
2
3
SO O
SO
Kc 
 200 c K
 
   
2
1
2
2
2
2
3
 


 



SO O
SO
Kc
1 2
4 3
 170 c  42.7 K c Q
c c Q  K

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

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

12. Kc from reaction stoichiometry
H2(g) + I2(g) ↔ 2HI(g)
K  same  46.4 c
 
   1
2
1
2
2
H I
HI
Kc 
4 diff initial conc of H2 , I2 and HI At equilibrium Kc = 46.4 ( 730K)
 
   
46.4
1.14 10 0.12 10
2.52 10
2 1 2 1
2 2

 


 

c Rxn 1 K
same
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
 
reac t
product
Qc
tan

 
reac t
product
Qc
tan

 c Q
c c Q  K
c c Q  K c c Q  K
 
   1
2
1
2
2
H I
HI
Kc 

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

14. Kc and Qc
H2(g) + I2(g) ↔ 2HI(g)
 
   1
2
1
2
2
H I
HI
Kc 
 103 c Q
 
   1
2
1
2
2
H I
HI
Qc 
 
   
46.4
1.14 10 0.12 10
2.52 10
2 1 2 1
2 2

 


 

c K  46.4 c K
Initial conc of H2 , I2 and HI
At equilibrium conc
Not at equilibrium
H2(g) + I2(g) ↔ 2HI(g)
c c Q  K
c c Q  K
Reaction at
equilibrium
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
0.300 2
  c Q
103 c Q  46.4 c > K

15. 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)  4.07 c K
Equilibrium disturbed
H2 added. More reactant
At equilibrium
Conc reactant/product no change
At new equilibrium
Conc reactant/product no change
 2.24 c Q
Equilibrium Conc H2 = 0.82M
Equilibrium Conc N2 = 0.20M
Equilibrium Conc NH3 = 0.67M
 
   3
2
1
2
2
3
N H
NH
Kc 
 
   1 3
2
0.20 0.82
0.67
 c K
New Conc H2 = 1.00M
Conc N2 = 0.20M
Conc NH3 = 0.67M
 
   3
2
1
2
2
3
N H
NH
Qc 
 
   1 3
2
0.20 1.00
0.67
 c Q
 4.07 c K
New Equilibrium Conc H2 = 0.90M
New Equilibrium Conc N2 = 0.19M
New Equilibrium Conc NH3 = 0.75M
 
   1 3
2
0.19 0.90
0.75
 c K
 
   3
2
1
2
2
3
N H
NH
Kc 
 4.07 c K
Shift to the right
- Increase product
- New Conc achieve
- Qc = Kc again
c c Q  K

16. 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)  4.07 c K
  4.07 c c Q K
Equilibrium disturbed
H2 added. More reactant
c c Q  K
Equilibrium shift to right
Rate forward Kf > Rate reverse Kr
 c Q
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

17. Click here to view simulation
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Simulation on Dynamic equilibrium
Click here on equilibrium constant