This document discusses dynamic chemical equilibrium. It explains that at equilibrium, the rates of the forward and reverse reactions are equal and the concentrations of reactants and products remain constant. It describes how changing concentration, pressure, temperature affect the position of equilibrium according to Le Chatelier's principle - the system shifts to counteract the applied stress. Increasing concentration favors the side with fewer moles of gas. Increasing pressure favors the side with fewer total gas molecules. Increasing temperature favors the endothermic direction for endothermic reactions and exothermic direction for exothermic reactions. Catalysts increase the rates of both forward and reverse reactions but do not change the position of equilibrium or equilibrium constant.

1. Dynamic Equilibrium
Reversible (closed system)
Forward Rate, K1 Reverse Rate, K-1
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
rate cons t reverse
rate cons t forward
K
K
.. tan ..
.. tan ..
1
1 

Catalyst
Factors affecting equilibrium (closed system)
Concentration Pressure Temperature
Equilibrium constant Kc ≠ Position equilibrium

2. Factors affecting the position of Equilibrium
Effect of Concentration on the position of equilibrium
Increase Conc SCN- or Fe3+
•Equilibrium shift to right →
•Formation of complex ion Fe(SCN)2+ (red blood)
Fe3+ + SCN- ↔ Fe(SCN)+2 (yellow) (red Blood)
Increase Concentration
• Rate of rxn increase ↑
• Position of equilibrium shift to a side to decrease conc again ↓
•Kc, equilibrium constant - no change
•Rate constant, forward/backward - no change
Decrease Conc Fe3+
• By adding OH- will shift equilibrium to left ←
•Fe(SCN)2+ breakdown to form more Fe3+ (yellow) Decrease Conc SCN-
• By adding Ag+ will shift equilibrium to left
• Fe(SCN)2+ breakdown to form more SCN- (yellow)
• Increase Conc ↑ - position of equilibrium shift to right/left - Conc is Reduced ↓
• Decrease Conc ↓ – position of equilibrium shift to right/left - Conc is Increased ↑
Click to view video
Le Chatelier’s Principle
• System in dynamic equilibrium is disturbed, the position of equilibrium will shift so as to cancel out the effect of change and a new equilibrium can be established again

3. Effect of Concentration on the position of equilibrium
Decrease Conc H+
• By adding OH-
•Equilibrium shift to left ←
•Formation of CrO42- (yellow)
Increase Conc H+
• By adding H+
• Shift equilibrium to right →
• Formation of Cr2O72- (orange)
2CrO42- + 2H+ ↔ Cr2O72- + H2O (yellow) (orange)
Click to view video
Factors affecting the position of Equilibrium
• Increase Conc ↑ - position of equilibrium shift to right/left - Conc is Reduced ↓
• Decrease Conc ↓ – position of equilibrium shift to right/left - Conc is Increased ↑
Increase Concentration
• Rate of rxn increase ↑
• Position of equilibrium shift to a side to decrease conc again ↓
•Kc, equilibrium constant - no change
•Rate constant, forward/backward - no change
Le Chatelier’s Principle
• System in dynamic equilibrium is disturbed, the position of equilibrium will shift so as to cancel out the effect of change and a new equilibrium can be established again

4. Effect of Concentration on the position of equilibrium
Decrease Conc CI-
•Adding Ag+ to form AgCI
•Equilibrium shift to right →
•Formation of Co(H2O)62+ (pink)
Increase Conc CI-
• Adding HCI
• Shift equilibrium to left ←
• Formation of CoCl42- (blue)
CoCl42- + 6H2O ↔ Co(H2O)62+ + 4CI – (blue) (pink)
Increase Conc H2O
• Adding H2O
• Shift equilibrium to right →
• Formation of Co(H2O)62+ (pink)
Click to view video
Factors affecting the position of Equilibrium
• Increase Conc ↑ - position of equilibrium shift to right/left - Conc is Reduced ↓
• Decrease Conc ↓ – position of equilibrium shift to right/left - Conc is Increased ↑
Increase Concentration
• Rate of rxn increase ↑
• Position of equilibrium shift to a side to decrease conc again ↓
•Kc, equilibrium constant - no change
•Rate constant, forward/backward - no change
Le Chatelier’s Principle
• System in dynamic equilibrium is disturbed, the position of equilibrium will shift so as to cancel out the effect of change and a new equilibrium can be established again

5. Effect of Pressure on the position of equilibrium
Increasing Pressure ↑
• By reducing Vol
• Equilibrium shift to left ←
• Less molecule on left side
•Pressure drop ↓
• Formation N2O4(colourless)
Increase pressure ↑ - favour rxn with a decrease ↓in pressure/number of molecule Decrease pressure ↓ - favour rxn with a increase ↑ in pressure/number of molecule
Decreasing Pressure ↓
• By Increasing Vol
• Equilibrium shift to right →
• More molecule on right side
•Pressure increase ↑
• Formation NO2 (brown)
Increase pressure ↑ – collision more frequent - shift equilibrium to left - reduce number of molecule - pressure decrease again ↓ Decrease pressure ↓ – collision less frequent – shift equilibrium to right – increase number of molecule – pressure increase again ↑
Click to view video
Factors affecting the position of Equilibrium
Le Chatelier’s Principle
• System in dynamic equilibrium is disturbed, the position of equilibrium will shift so as to cancel out the effect of change and a new equilibrium can be established again
N2O4(g) ↔ 2NO2(g) (colourless) (brown)
Increase Pressure
• Rate of rxn increase ↑
• Position of equilibrium shift to a side to decrease pressure again ↓
•Kc, equilibrium constant - no change
•Rate constant, forward/backward - no change
Reduce Vol
Increase Vol
Mole ratio 1(left) ↔ 2(right)

6. Effect of Pressure on the position of equilibrium
N2(g) + 3H2(g) ↔ 2NH3(g) ( 4 vol/mole ) (2 vol/mole)
Increasing Pressure ↑
• Equilibrium shift to right →
• Less molecule on left side
•Pressure drops ↓
• Formation of NH3 (product)
Decreasing Pressure ↓
• Equilibrium shift to left ←
• More molecule on right side
•Pressure increase ↑
• Formation H2 and N2 (reactant)
Click to view video
Factors affecting the position of Equilibrium
Increase pressure ↑ - favour rxn with a decrease ↓in pressure/number of molecule Decrease pressure ↓ - favour rxn with a increase ↑ in pressure/number of molecule
N2O4(g) ↔ 2NO2(g) (colourless) (brown)
Increasing Pressure ↑
• By reducing Vol
• Equilibrium shift to left ←
• Less molecule on left side
•Pressure drop ↓
• Formation N2O4(colourless)
Decreasing Pressure ↓
• By Increasing Vol
• Equilibrium shift to right →
• More molecule on right side
•Pressure increase ↑
• Formation NO2 (brown)
Mole ratio 1(left) ↔ 2(right)
Mole ratio 4(left) ↔ 2(right)
Reduce Vol
Increase Vol

7. Effect of Temperature on position of equilibrium
Decrease Temp ↓
• Cooling it down
• Favour exothermic rxn
• Equilibrium shift to right →
• Increase Temp ↑ again
• Formation Co(H2O)62+ (pink)
Increase Temp ↑
• Heating it up
• Favour endothermic rxn
• Equilibrium shift to left ←
• Reduce Temp ↓ again
• Formation of CoCl42- (blue)
CoCl42- + 6H2O ↔ Co(H2O)62+ + 4CI – ΔH = -ve (exothermic) (blue) (pink)
Increase Temp ↑ – Favour endothermic rxn – Absorb heat to reduce Temp again ↓ Decrease Temp ↓ – Favour exothermic rxn – Release heat to increase Temp again ↑
Increase Temperature
• Rate of rxn increase
• Rate constant also change
• Rate of forward/reverse increase but to diff extend
• Position equilibrium shift to endo to decrease Temp
• Kc, equilibrium constant change
Click to view video
Factors affecting the position of Equilibrium
Le Chatelier’s Principle
• System in dynamic equilibrium is disturbed, the position of equilibrium will shift so as to cancel out the effect of change and a new equilibrium can be established again

8. Decrease Temp ↓
• Cooling it down ↓
• Favour exothermic rxn
• Equilibrium shift to left ←
• Increase Temp ↑
• Formation N2O4 (colourless)
Increase Temp ↑
• Heating it up ↑
• Favour endothermic rxn
• Equilibrium shift to right →
• Reduce Temp ↓
• Formation NO2 (brown)
N2O4 (g) ↔ 2NO2(g) ΔH = + 54kJmol-1 (colourless) (brown)
Click to view video
Factors affecting the position of Equilibrium
Le Chatelier’s Principle
• System in dynamic equilibrium is disturbed, the position of equilibrium will shift so as to cancel out the effect of change and a new equilibrium can be established again
Effect of Temperature on position of equilibrium
Increase Temp ↑ – Favour endothermic rxn – Absorb heat to reduce Temp again ↓ Decrease Temp ↓ – Favour exothermic rxn – Release heat to increase Temp again ↑
Increase Temperature
• Rate of rxn increase
• Rate constant also change
• Rate of forward/reverse increase but to diff extend
• Position equilibrium shift to endo to decrease Temp
• Kc, equilibrium constant change

9. Catalyst
• Provide an alternative pathway with lower activation energy
• Increase forward and reverse rate to the same extent/factor
• Position of equilibrium and Kc UNCHANGED
• Catalyst shorten time to reach equilibrium
Effect of Catalyst on equilibrium constant, Kc
Without catalyst Reach equilibrium slow
With catalyst Reach equilibrium fast
Effect catalyst on Rate, Rate constant and Kc – NH3 production
N2(g) + 3H2(g) ↔ 2NH3(g) ΔH = - 92kJmol-1
Factors affecting the position of Equilibrium
Forward rate
Reverse rate
Catalyst
• Rate of rxn increase
•Forward/reverse rate increase to SAME extend
• Kc equilibrium constant NO change
•Position equilibrium NO change
•Product/reactant yield NO change
Catalyst

10. Effect of catalyst on Rate of Reaction
Catalyst
• Provide alternative pathway with lower activation energy
• Greater proportion of colliding molecule with energy greater than > Ea
• Rate increase
Source : http://njms2.umdnj.edu/biochweb/education/bioweb/PreK2010/EnzymeProperties.html
Catalyst
• Provide alternative pathway with lower activation energy
• Fraction of molecule with energy greater than > Ea increase
• Rate increase
Maxwell Boltzmann Energy distribution curve
Without catalyst
With catalyst
Without catalyst
Maxwell Boltzmann Energy distribution curve
Fraction molecules energy > Ea
Fraction – lead to product formation

11. How position equilibrium shift when H2 is added ?
N2(g) + 3H2(g) ↔ 2NH3(g)
 4.07 c K
Qualitatively (prediction) Quatitatively
Le Chatelier’s Principle
At equilibrium
Conc reactant/product
no change
Equilibrium disturbed
H2 added. More reactant
N2(g) + 3H2(g) ↔ 2NH3(g)
Position equilibrium shift to right
- Reduce conc H2
- More product form
Shift to right
Qc and Kc
At equilibrium
Conc reactant/product
no change
Equilibrium disturb
H2 added.
New equilibrium
Conc reactant/product
no change
Eq Conc H2 = 0.82
Eq Conc N2 = 0.20
Eq Conc NH3= 0.67
New Conc H2 = 1.00
Conc N2 = 0.20
Conc NH3 = 0.67
New Eq Conc H2 = 0.90
New Eq Conc N2 = 0.19
New Eq Conc NH3 = 0.75
 
   3
2
1
2
2
3
N H
NH
Kc 
 
   1 3
2
0.20 0.82
0.67
 c K
 4.07 c K
 
   3
2
1
2
2
3
N H
NH
Qc 
 
   1 3
2
0.20 1.00
0.67
 c Q
 2.24 c Q
 
   3
2
1
2
2
3
N H
NH
Kc 
 
   1 3
2
0.19 0.90
0.75
 c K
 4.07 c K
Shift to the right
- Increase product
- Qc = Kc again

12. Factors affecting the position of Equilibrium
Effect of Temperature on equilibrium constant, Kc
N2O4 (g) ↔ 2NO2(g) ΔH = + 54kJmol-1
Temp increase ↑ – Kc increase ↑
A B ΔH = +ve Rate reverse = k r
Rate forward = kf
Kc
 
A
B
Kc 
r
f
c K
K
K  rate cons t reverse
rate cons t forward
K
K
r
f
.. tan ..
.. tan ..

Temp affect rate constant
Temp changes
c K
Increase Temp ↑
Position equilibrium shift to right Endo side – Absorb heat Temp decrease ↓
More product , less reactant  
reac t
product
Kc tan

 c K
Forward rate constant, kf > reverse rate, kr
r
f
c K
K
K 
Decrease Temp ↓
Position equilibrium shift to left Exo side – Release heat Temp increase ↑
More reactant , less product
 
reac t
product
Kc tan

Forward rate constant, kf < reverse rate, kr
r
f
c K
K
K 
 c K
Conclusion :
Endo rxn – Temp ↑ – Kc ↑ – Product ↑

13. A B ΔH = -ve
Factors affecting the position of Equilibrium
Effect of Temperature on equilibrium constant, Kc
Temp increase ↑ – Kc decrease ↓
Rate reverse = k r
Rate forward = kf
Kc
 
A
B
Kc 
r
f
c K
K
K  rate cons t reverse
rate cons t forward
K
K
r
f
.. tan ..
.. tan ..

Temp affect rate constant
Temp changes
c K
Increase Temp ↑
Position equilibrium shift to left Endo side – Absorb heat Temp decrease ↓
More Reactant, less product  
reac t
product
Kc tan

 c K
Forward rate constant, kf < Reverse rate, kr
r
f
c K
K
K 
Decrease Temp ↓
Position equilibrium shift to right Exo side – Release heat Temp increase ↑
More Product , less reactant
 
reac t
product
Kc tan

Forward rate constant, kf > Reverse rate, kr
r
f
c K
K
K 
 c K
Conclusion :
Exo rxn – Temp ↑ – Kc ↓ – Product ↓
H2(g) + I2(g) ↔ 2HI(g) ΔH = -9.6kJmol-1

14. N2(g) + 3H2(g) ↔ 2NH3(g) ΔH = - 92kJmol-1
Haber process
• Production ammonia making fertiliser
• Reversible process N2(g) + 3H2(g) ↔ 2NH3(g)
• Optimum yield conditions are :
Pressure – 400 atm, Temp – 400C, Catalyst - Iron
Application Equilibrium constant Kc and Kinetic in Industry (NH3 Production)
Highest yield, HIGH Kc, HIGH Rate, Low cost
Increase yield (NH3) – Position equilibrium shift to right →
Low Temp ↓
•Position shift right (exo) - Release heat – Temp ↑
•Low ↓ Temp – Yield NH3 high ↑ BUT Rate slow
High Pressure ↑
- Position shift right - less mole of gas – Pressure ↓
- High ↑ Pressure – Yield NH3 high – BUT cost high
(Not economical)
High Yield Conditions
• Low temperature ↓ but rate slow
• High Pressure ↑ but too expensive
• Not economical
Industry Conditions
• Compromise Temp -400C
• Pressure - 400atm
• Catalyst iron – Increase Rate
• Remove NH3 produced, equilibrium
shift to right →
Effect of Temperature, Catalyst and Pressure on Haber Process
Temperature Pressure
 c K
Rate 
Cost 
Ideal conditions Practical/Industry conditions

15. Highest yield, HIGH Kc, HIGH Rate, Low cost
Increase yield (H2SO4) – Position equilibrium shift to right →
High Yield Conditions
• Low temperature ↓ but rate slow
• High Pressure ↑ but too expensive
• Not economical
Temperature Pressure
Contact process
• Production sulphuric acid
• Process involve 3 stages
Stage 1 – S + O2 (g) → SO2(g) Stage 2 - 2SO2(g) + O2(g) ↔ 2SO3(g) Stage 3 – SO3(g) + H2O → H2SO4
2SO2(g) + O2(g) ↔ 2SO3(g) ΔH = - 197kJmol-1
Industry Conditions
• Compromise Temp - 450C
• Pressure of 2atm
• Catalyst vanadium(V) oxide V2O5
• Remove SO3 produced, equilibrium
shift to right →
Effect of Temperature, Catalyst and Pressure on Contact Process
Application Equilibrium constant Kc and Kinetic in Industry (H2SO4 Production)
 c K
Rate 
Cost 
Low Temp ↓
•Position shift right (exo) - Release heat – Temp ↑
•Low ↓ Temp – Yield NH3 high BUT Rate slow
High Pressure ↑
- Position shift right - less mole of gas – Pressure ↓
- High ↑ Pressure – Yield NH3 high – BUT cost high
(Not economical)
Ideal conditions Practical/Industry conditions
Low temp

16. IB Questions
Which of rxn not affected by change in pressure?
4NH3(g) + 5O2(g) ↔ 4NO(g) + 6H2O(g)
N2(g) + 3H2(g) ↔ 2NH3(g) H2(g) + I2(g) ↔ 2HI(g)
2SO2(g) + O2(g) ↔ 2SO3(g)
CO is toxic. Rxn take place in catalytic converter.
At equilibrium, will CO increase, decrease or unchanged
a) Pressure increase/by decreasing vol
b) Pressure increase by adding O2
c) Temp increase
d) Platinum catalyst added
CaCO3(s) ↔ CaO(g) + CO2(g)
CH3COOH(l) + C2H5OH(l) ↔ CH3COOC2H5(l) + H2O(l)
2CO CuO(s) + H2(g) ↔ Cu(s) + H2O(g) (g) + O2(g) ↔ 2CO2(g)
a) Shift to right – decrease number molecule ↓ -CO decrease ↓
b) Shift to right – decrease conc O2 ↓ - CO decrease ↓
c) Shift to left –endo rxn – decrease ↓ temp again -CO increase ↓
d) NO change
Ex 1 Ex 2 Ex 3
Mole ratio 4(left) ↔ 2(right) Mole ratio 2(left) ↔ 2(right) Mole ratio 2(left) ↔ 2(right)
Ex 4 Ex 5 Ex 6
Mole ratio 0(left) ↔ 2(right) Mole ratio 3(left) ↔ 2(right) Mole ratio 9(left) ↔ 10(right)
Ex 7
Ex 8
Mole ratio 3(left) ↔ 2(right) Mole ratio 1(left) ↔ 1(right)
Solid not included
Solid not included
Ex 9 2CO(g) + O2(g) ↔ 2CO2(g) ΔH = -566kJmol-1 Ex 10 Reversible rxn bet hydrogen and iodine shown below
H2 + I2 ↔ 2HI
a) Outline characteristic of homogenous sys in equilibrium
b) Predict the position eq when pressure increase from 1 to 2 atm
c) Kc at 500k - 160. Kc at 700K - 54. Deduce the enthalpy of forward rxn.
d) 1.60 mol H2 and 1 mol I2 allowed to reach equilibrium in 4 dm3
vessel. Amt HI formed at eq is 1.8 mol. Find Kc
a) Reactant/product on same phase, Rate forward = Rate reverse
Conc reactant/product unchanged. Macroscopic property (same)
b) No change in position equilibrium (molecules both sides same)
c) Rxn exo/heat given out. H = -ve
d) Moles- H2 = 1.6 – 0.9 = 0.7, I2 = 1 – 0.9 = 0.1, HI = 1.8
 
   1
2
1
2
2
H I
HI
Kc 
 
   1 1
2
0.7 0.1
1.8
 c K
Eq amt used instead eq conc
 46.3 c K