1. Chemical Equilibrium

2. Equilibrium – A state of balance
Static Equilibrium – When a meter stick is balanced in the middle neither end is moving this is an example of static equilibrium

3. Dynamic Equilibrium – When a person is running on a treadmill they are moving forward while the belt is moving forward but over all neither is progressing in either direction as the two movements balance each other this is known as Dynamic equilibrium

4. In chemical reactions a state of dynamic equilibrium exists
When magnesium ribbon is burned in air, a bright light is given off and a white substance called magnesium oxide is formed
2Mg + O MgO
It is very difficult to turn the magnesium oxide back into magnesium and oxygen chemists often say the reaction has “gone to completion” meaning the reactants are almost completely turned to products.

5. Many reactions do not go to completion
German chemist Fritz Haber discovered that nitrogen gas and hydrogen gas react to form ammonia
Ammonia is used in the manufacture of fertilizers, explosive sand nylon

6. Ammonia production Haber+
THE NITROGEN AND HYDROGEN ARE CONTINUALLY FORMING AMMONIA WHILST THE AMMONIA IS CONTINUALLY DECOMPOSING TO FOR NITROGEN AND HYDROGEN

7. The above reaction is reversible ie
The above reaction is reversible ie. It takes place in the forward and backward direction
Such reactions are said to be equilibrium reactions and are indicated using the equilibrium arrow sign

8. When a reaction is in dynamic equilibrium the rate of forward reaction is the same as the rate of backwards reaction
This means that for the above reaction there is no change in the overall amounts of either nitrogen or hydrogen or ammonia present in the reaction system

9. Definition
Chemical equilibrium is a state of dynamic balance where the rate of the forward reaction equals the rate of the reverse reaction
NB examine figure 17.2 p233

10. Le Chatelier’s Principle
Many industrial chemical processes involve equilibrium reactions
It is important for chemists to understand these so they can try to shift the balance to the right ( ie. To maximise the amount of product formed)
French Chemist Henri Le Chetalier studied equilibrium reactions and put forward a rule known as Le Chetalier’s Principle

15. Changes in Temperature
Increase in temp an endothermic reaction is favoured
Decrease in temp an exothermic reaction is favoured
H I2 → 2HI ΔH = kJ/mol
2HI → H2 + I ΔH = kJ/mol

16. Changes in concentration
H2 or I2 (increased) SHIFTS TO THE RIGHT to make more 2HI
HI (increased) SHIFTS TO THE LEFT to make more H2 or I2
A decrease in the concentration of HI has the opposite effect , hydrogen and iodine react forming more HI SHIFTS TO THE RIGHT
A decrease in H2 or I 2 SHIFTS TO THE LEFT

17. Adding a catalyst It has no effect on the on the equilibrium
A catalyst increases the rate of the reaction by lowering the activation energy and the activation energy for the reverse reaction is lowered to the same extent.

18. Changes in pressure Pressure changes only effect gases.
Increase in pressure the reaction that produces less molecules is favoured
N2 + 3H2 produces 2NH3
A decrease in pressure the reaction that produces more molecules is favoured

19. Changes in the vol of the container
An increase in volume decreases the pressure.
A decrease in volume increases the pressure

20. Le Chatelier’s Principle
To demonstrate the effect of temp and concentration changes
CoCl H2O Co(H2O)62+ 4Cl-
Blue Red
Cr2O H2O CrO H+
Orange Yellow
FeCl CNS Fe(CNS) Cl-
Yellow Red
Adding acid shifts to left and adding base shifts to right

21. Industrial applications of Le Chatelier’s Principle
Manufacture of Ammonia by the Haber Process - Ammonia is one of the most important chemicals made by the chemical industry
80% of ammonia is used to make fertilisers
It is made by reacting nitrogen gas (from air) with hydrogen gas (from natural gas) in the presence of an iron catalyst

22. Reaction
iron catalyst Nitrogen + hydrogen ammonia N2 + 3H2 2NH3 H = -92.4kj/mol The amount of ammonia made depends on the temperature and pressure inside the vessel

23. Using Le Chatelier’s principle Fritz Haber selected the right conditions for the manufacture of ammonia
There are 4 moles on the left hand side but only 2 moles on the right hand side using le Chatelier’s principle this tells us a good yield of ammonia will be obtained at high pressure
NB: A high pressure plant is expensive to build and maintain a compromise pressure of 200 atmospheres is used in practice

24. The reaction is exothermic going from left to right, therefore decreasing the temperature will push the reaction in this direction
However if temperature is too low the rate of reaction becomes too slow and even though more ammonia will be formed it will be formed much more slowly
A compromise temperature of 500⁰C is used in practice

25. Le Chatelier’s principle tells us that manufacture of ammonia by the Haber process is best achieved at high pressure and low temperature
The iron catalyst does not change the equilibrium but is very important as it helps the system reach equilibrium much more quickly

26. Manufacture of Sulfuric Acid by the Contact Process
Sulfuric acid is used in the manufacture of many materials such as paints, detergents, fertilizers, plastics, fibres, car batteries etc.
It is manufactured in the Contact process which involves passing sulfur dioxide gas and oxygen gas over a catalyst
There needs to be very close contact between the catalyst and the two gases for the reaction to proceed satisfactorily

27. Reaction
vanadium pentoxide Sulfur dioxide + Oxygen sulfur trioxide 2SO2 + O2 2SO3 H = -196 kj/mol The sulfur trioxide formed is reacted with water to make sulfuric acid NB: Platinum could also be used as a catalyst but is found to be easily poisoned by impurities in the reactants

28. Le Chatelier’s Principle indicates high pressure would favour the formation of product
This is true but due to high costs a pressure just above atmospheric pressure is used in practice
Another reason is that high pressure tend to liquefy sulfur dioxide

29. As the reaction is exothermic from left to right low temperature will drive the reaction in this reaction
However too low a temperature will greatly slow the rate of the reaction so a compromise temperature of 450⁰C is used as it is found the catalyst works best at this temperature

30. The equilibrium constant (Kc)
When a system reaches a state of equilibrium chemists and chemical engineers need to know the concentrations of reactants and products that exist in the equilibrium mixture
This can be found either by experiment or by performing calculations

31. Law of Chemical Equilibrium
Provided the temperature remains constant there is a simple mathematical relationship between the concentrations of reactants and products in an equilibrium mixture
Consider the reaction where a moles of reactant A react with b moles of reactant B to form c moles of reactant C and d moles of reactant D

32. The equilibrium constant

33. For the reaction between nitrogen and hydrogen to form ammonia the Equilibrium constant expression is
N H NH3

34. When writing Equilibrium constant expressions .............
Products go above the line and reactants below the line
The concentrations of the products are multiplied together, the concentration of the reactants are multiplied together (multiplication sign is left out)
The concentration of all species is raised to the power of their coefficients in the balanced equation

35. Every Equilibrium reaction has its own value for Kc at the particular temperature at which the reaction takes place
At 458⁰C the value of Kc for the reaction
H2 + I HI is 46
This constant value is always obtained at this temperature no matter what concentrations of hydrogen + iodine are reacted

36. NB: In the above example the value of Kc at 46 is quite large it tells us that the numerator is much greater than the denominator (ie the reactants “went into” the product 46 times!)
This indicates how far the reaction is pushed to the right, the larger the value of Kc the more the reaction is pushed towards products
A low value of Kc indicates only a small fraction of reactants have formed products

37. 2 SO2 + O SO3
At equilibrium the conc of SO2 , O2 and SO3 were 0.07 mol/l 0.035mol/l and 0.03mol/l respectively.
Calculate the value of the equilibrium constant KC for this reaction
Kc = [SO3] (0.03)2
[SO2]2[O2] (0.07)2(0.035)
KC = 5.25mol/l