IB Chemistry on Le Chatelier's Principle, Haber and Contact Process

1. Tutorial on Le Chatelier’s Principle, Haber and Contact
Process .
Prepared by
Lawrence Kok
http://lawrencekok.blogspot.com

2. 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
aA(aq) + bB(aq) cC(aq) + dD(aq)
coefficient
Solid/liq not included in Kc
Equilibrium Constant Kc
Conc represented by [ ]
K1
K-1
   
 a  b
c d
c
C D
A B
K 
1
K

1

K
Kc
Equilibrium Constant Kc
express in
Conc vs time Rate vs time
A + B
C + D
Conc
Time
rate .. cons tan t ..
forward
rate cons t reverse
K
K
.. tan ..
1 

1
Catalyst
Factors affecting equilibrium (closed system)
Concentration Pressure Temperature
Equilibrium constant Kc ≠ Position equilibrium

3. 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 Concentration on 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 Conc SCN- or Fe3+
Fe3+ + SCN- ↔ Fe(SCN)+2
(yellow) (red Blood)
•Equilibrium shift to right →
•Formation of complex ion Fe(SCN)2+ (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)
Click to view video

4. Factors affecting the position of Equilibrium
Effect of Concentration on 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 ↑
Decrease ConcH+
• By adding OH-
•Equilibrium shift to left ←
•Formation of CrO4
2- (yellow)
Increase Conc H+
• By adding H+
• Shift equilibrium to right →
• Formation of Cr2O7
2- (orange)
2CrO4
2- + 2H+ ↔ Cr2O7
2- + H2O
(yellow) (orange)
Click to view video
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. Factors affecting the position of Equilibrium
Effect of Concentration on 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 ↑
Decrease Conc CI-
•Adding Ag+ to form AgCI
•Equilibrium shift to right →
•Formation of Co(H2O)6
2+ (pink)
Increase Conc CI-
• Adding HCI
• Shift equilibrium to left ←
• Formation of CoCl4
2- (blue)
CoCl4
2- + 6H2O ↔ Co(H2O)6
2+ + 4CI –
(blue) (pink)
Increase Conc H2O
Click to view video
• Adding H2O
• Shift equilibrium to right →
• Formation of Co(H2O)6
2+ (pink)
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

6. 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 Pressure on 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
Reduce Vol Increase Vol
Increasing Pressure ↑
• By reducing Vol
• Equilibrium shift to left ←
• Less molecule on left side
•Pressure drop ↓
• Formation N2O4(colourless)
N2O4(g) ↔ 2NO2(g)
(colourless) (brown)
Mole ratio 1(left) ↔ 2(right)
Decreasing Pressure ↓
• By Increasing Vol
• Equilibrium shift to right →
•More molecule on right side
•Pressure increase ↑
• Formation NO2 (brown)
Click to view video
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
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 ↑

7. Factors affecting the position of Equilibrium
Effect of Pressure on 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)
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)
Mole ratio 1(left) ↔ 2(right)
Decreasing Pressure ↓
• Equilibrium shift to left ←
•More molecule on right side
•Pressure increase ↑
• Formation H2 and N2 (reactant)
Click to view video
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 4(left) ↔ 2(right)
Reduce Vol
Increase Vol

8. Factors affecting the position of Equilibrium
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 ↑
2- + 6H2O ↔ Co(H2O)6
Decrease Temp ↓
• Cooling it down
• Favour exothermic rxn
• Equilibrium shift to right →
• Increase Temp ↑ again
• Formation Co(H2O)6
2+ (pink)
Increase Temp ↑
• Heating it up
• Favour endothermic rxn
• Equilibrium shift to left ←
• Reduce Temp ↓ again
• Formation of CoCl4
2- (blue)
CoCl4
2+ + 4CI – ΔH = -ve (exothermic)
(blue) (pink)
Increase Temperature
Click to view video
• 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
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

9. 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 ↑
N2O4 (g) ↔ 2NO2(g) ΔH = + 54kJmol-1
(colourless) (brown)
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)
Click to view video
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

10. Factors affecting the position of Equilibrium
Effect of Catalyst on equilibrium constant, Kc
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
Without catalyst
Reach equilibrium slow
With catalyst
Reach equilibrium fast
N2(g) + 3H2(g) ↔ 2NH3(g) ΔH = - 92kJmol-1
Effect catalyst on Rate, Rate constant and Kc – NH3 production
Forward rate
Reverse rate
Catalyst
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

11. 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
Catalyst
• Provide alternative pathway with lower activation energy
• Fraction of molecule with energy greater than > Ea increase
• Rate increase
Without catalyst
Maxwell Boltzmann Energy distribution curve Without catalyst With catalyst
Source : http://njms2.umdnj.edu/biochweb/education/bioweb/PreK2010/EnzymeProperties.html
Maxwell Boltzmann Energy distribution curve
Fraction molecules energy > Ea
Fraction – lead to product formation

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

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

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

15. Effect of Temperature, Catalyst and Pressure on Haber Process
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)
N2(g) + 3H2(g) ↔ 2NH3(g) ΔH = - 92kJmol-1
Highest yield, HIGH Kc, HIGH Rate, Low cost
Increase yield (NH3) – Position equilibrium shift to right →
Temperature Pressure
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 →
 c K
 Rate
 Cost
Ideal conditions Practical/Industry conditions

16. Effect of Temperature, Catalyst and Pressure on Contact Process
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
Application Equilibrium constant Kc and Kinetic in Industry (H2SO4 Production)
2SO2(g) +O2(g) ↔ 2SO3(g) ΔH = - 197kJmol-1
Highest yield, HIGH Kc, HIGH Rate, Low cost
Increase yield (H2SO4) – Position equilibrium shift to right →
Temperature Pressure
High Yield Conditions
• Low temperature ↓ but rate slow
• High Pressure ↑ but too expensive
• Not economical
Industry Conditions
• Compromise Temp - 450C
• Pressure of 2atm
• Catalyst vanadium(V) oxide V2O5
• Remove SO3 produced, equilibrium
shift to right →
 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)
Low temp
Ideal conditions Practical/Industry conditions

17. IB Questions
Which of rxn not affected by change in pressure?
Mole ratio 4(left) ↔ 2(right) Mole ratio 2(left) ↔ 2(right) Mole ratio 2(left) ↔ 2(right)
4NH3(g) + 5O2(g) ↔ 4NO(g) + 6H2O(g)
Ex 1 Ex 2 Ex 3
N2(g) + 3H2(g) ↔ 2NH3(g) H2(g) + I2(g) ↔ 2HI(g)
Ex 4 Ex 5 Ex 6
2SO2(g) + O2(g) ↔ 2SO3(g)
CaCO3(s) ↔ CaO(g) + CO2(g)
Mole ratio 0(left) ↔ 2(right) Mole ratio 3(left) ↔ 2(right) Mole ratio 9(left) ↔ 10(right)
Ex 7
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
CH3COOH(l) + C2H5OH(l) ↔ CH3COOC2H5(l) + H2O(l)
CuO(s) + H2(g) ↔ Cu(s) + H2O 2CO (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 – increase temp ↑ again - CO increase ↓
d) NO change
Ex 8
Solid not included
Mole ratio 3(left) ↔ 2(right) Mole ratio 1(left) ↔ 1(right)
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 of reverse
Conc reactant/product unchange. 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
HI
1
2
2
H I
Kc 
 1.8

2
 0.7  1  0.1
1
 c K
Eq amt used instead eq conc
 46.3 c K

18. 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