1. The Position of Equilibrium
Section 7.2

2. The Equilibrium Constant
The position of equilibrium for a particular reversible reaction can be defined by a constant
This constant has a numerical value found by relating the equilibrium concentrations of the products to those of the reactants
Denoted by Kc the equilibrium constant

3. Deducing the Equilibrium Constant
Consider this standard reaction:
aA + bB cC dD
Where a, b, c and d represent the stoichiometry of each substance, then:

4. Continued
The concentrations must be those that occur at equilibrium, not the starting values
Kc is constant for a given temperature
Change the temperature, you change Kc

5. Equilibrium Law
Equilibrium law: in any reversible reaction at a state of equilibrium, the rate of the forward reaction equals the rate of the reverse reaction

6. Problem #1 Consider the following reaction: N2(g) + 3H2(g) 2NH3(g)
How is the Kc calculated?

7. Problem #1 Continued
An experiment has this data at equilibrium at 500°C:
[N2] = mol dm-3
[H2] = mol dm-3
[NH3] = mol dm-3
Calculate the Kc for this experiment

8. Working Out
0.0602
A note about units: the concentrations calculated would give you varying units, Kc is unitless because you are using the “activity” which is close to concentration

9. Another Consideration
H2(g) + I2(g) HI(g)
Kc for the forward reaction:
For the reverse reaction:

10. Continued
The value of the equilibrium constant for the reverse reaction is the reciprocal of that for the forward reaction

11. How Far will a Reaction Go?
Consider the following reactions and Kc
H2(g) + I2(g) HI(g) Kc = 2 at 277°C
H2(g) + Cl2(g) HCl(g) Kc = 1018 at 277°C
The large difference illustrates how stable HCl is at this temperature compared to HI
Virtually all of the reaction between H2 and Cl2 has produced HCl

12. Continued
The magnitude of Kc is a good indicator of how far a reaction goes toward completion (under given conditions)
If Kc>>1, then the reaction has gone virtually to completion
If Kc<<1, then the reaction has hardly taken place at all

13. Continued
For Kc = 1, there are equal amounts of reactants and products
The Kc for a given reversible reaction at equilibrium only changes if the temperature changes

14. Other Considerations
The concentrations of certain substances remain constant, so they are omitted from the equilibrium constant expression
All solids have a fixed density, so these are omitted
The concentration of pure liquid water is also constant (important to consider when solutions are aqueous, water omitted)

15. Continued
If water is used as a liquid with other liquids ( not aqueous solutions), then it is included
Water must be included if it is in the gas phase

16. Le Châtelier's Principle
This principle is used to predict the qualitative effects of changes on the position of equilibrium
When a system at equilibrium is disturbed, the equilibrium position will shift in the direction which tends to minimize, or counteract, the effect of the disturbance

17. More Possible changes in conditions that we need to consider are:
Changes in the concentrations of either the reactants or the products
Changes in pressure for gas phase reactions
Changes in temperature
The presence of a catalyst

18. Changes in Concentration
Increasing the concentration of a reactant will move the position of equilibrium to the right
This favors the forward reaction and will increase the equilibrium concentrations of the products

19. More
The addition of more product to an equilibrium mixture would shift the position of equilibrium to the left
This would favor the reverse reaction
When new equilibrium concentrations are substituted into the equilibrium expression, the value of Kc remains unchanged

20. Example
Fe(H2O)63+(aq) + SCN-(aq) [Fe(H2O)5SCN]2+ + H2O(l)
yellow-brown colorless blood-red
If the concentration of either the thiocyanate ion, or the iron (III) ion is increased, then the intensity of the color increases
The shift in equilibrium to the right causes the concentration of the added reactant to fall again

21. Continued
If the concentration of iron(III) ions is decreased by adding fluoride ions, which form a stable complex with the iron (III) ions, then the intensity of the coloration decreases
The shift of the equilibrium to the left produces more aqueous iron (III) ions to counteract the reduction caused by the F-

22. Changes in Pressure
Changing the pressure only affects reactions that involve gases
Gas phase reactions will only be affected by a change of pressure if the reaction involves a change in the number of moles on the two sides of the equation

23. Increase in Pressure 2SO2(g) + O2(g) 2SO3(g)
Increase the pressure and the equilibrium shifts to the side with the least moles of gas
2SO2(g) + O2(g) SO3(g)
Increase P: equilibrium → , to side with fewer moles of gas
C(s) + H2O(g) CO(g) + H2(g)
Increase P: equilibrium ←, to side with fewer moles of gas (

24. Decrease in Pressure 2SO2(g) + O2(g) 2SO3(g)
Decrease pressure and the equilibrium shifts towards the side with the most moles of gas
2SO2(g) + O2(g) SO3(g)
Decrease P: equilibrium ←, to side with more moles of gas
C(s) + H2O(g) CO(g) + H2(g)
Decrease P: equilibrium →, to side with more moles of gas

25. More About Pressure H2(g) + I2(g) 2HI(g)
Changing pressure has no effect as 2 moles of gas are converted to 2 moles of gas
The changes in concentrations that result from the changes in pressure are such that the value of Kc remains unchanged

26. Temperature Change
If the temperature of a system is increased, then the equilibrium shifts in the direction of the endothermic change, this will tend to lower the temperature
N2(g) + O2(g) NO(g) ∆H = +180 kJ mol-1
Increased T: Kc increases, equilibrium →
Decreased T: Kc decreases, equilibrium ←

27. Temperature Change
If the temperature of a system is decreased the equilibrium shifts in the direction of the exothermic change, this is to generate heat and raise the temperature
2SO2(g) + O2(g) SO3(g) ∆H = -197 kJ mol-1
Decreased T, Kc increases, equilibrium →
Increased T, Kc decreases, equilibrium ←

28. More About Temperature
Changes in temperature affect the rate constants of the forward and reverse reactions to different extents
This means that the actual value of Kc changes also

29. Presence of a Catalyst
The presence of a catalyst reduces the activation energy of both the forward and reverse reactions by the same amount
Both reactions are sped up by the same factor
Equilibrium is established more rapidly, but the position of equilibrium and the value of Kc are not affected

30. SUMMARY

31. Industrial Processes Many industrial processes involve equilibria
The goal is to produce the desired product efficiently and rapidly, but with the minimum amount of waste and the minimum input of energy
Kinetics and equilibria have to be considered

32. Haber Process Produces ammonia from nitrogen and hydrogen gases
A heated iron catalyst is used
N2(g)+ 3H2(g) NH3(g) ∆H = -92 kJ mol-1
4 moles of gas are converted to 2 moles of gas
A high pressure will favor the product

33. Continued
The forward reaction is exothermic, so a low temperature would favor the products
Issues with the process:
Low temperature causes a low reaction rate
High pressure is expensive to provide
A compromise is chosen to maximize product

34. Compromise
Typical conditions chosen for the Haber process are pressures in the range of atm and temperatures about 700K

35. Collecting Ammonia
The reaction is not allowed to reach equilibrium because the reaction rate slows as equilibrium is approached
Usually only about 20% of the reactants are converted to ammonia
The ammonia is condensed and the reactants are recycled to form more ammonia

36. Uses for Ammonia The manufacture of fertilizers
The manufacture of nitrogen-containing polymers such as nylon
Can be oxidized to produce nitric acid
Nitric acid is used to produce explosives such as TNT and dynamite
Nitric acid is also used in making dyes

37. Contact Process
The Contact process is the production of sulfuric acid by the oxidation of sulfur
First, pure sulfur is burned in air
S(s) + O2(g) SO2(g)
Then sulfur dioxide is mixed with air and passed over a vanadium (V) oxide catalyst
2SO2(g) + O2(g) SO3(g) ∆H = -196 kJ mol-1

38. Continued
High pressure would favor the formation of sulfur trioxide, but low pressure actually does a good job
Low temperature favors the product, but not too low (slows the reaction)
Compromise: K, the use of a catalyst (finely divided V2O5) and about 2 atm

39. Continued The result is over 90% conversion to sulfur trioxide
The sulfur trioxide must be reacted with water to produce sulfuric acid
SO3(g) + H2O(l) → H2SO4(l)

40. Uses of Sulfuric Acid Manufacture of: Fertilizers Polymers Detergents
Paints and pigments
Used in the petrochemical industry and in the processing of metals

41. Another use of Sulfuric Acid
Used as an electrolyte in automobile batteries (battery acid)