1. Equilibrium

2. Brief Outline What is reversible reaction?
Examples of reversible reaction
Dynamic Equilibrium
Le Chatelier’s Principle
The Haber Process

3. What is reversible reaction?
A reversible reaction is a chemical reaction in which the products can be converted back to reactants under suitable conditions.
The reaction involves two reactions.
There is a "forward" reaction and a mirror image "reverse" reaction.
A reversible reaction is shown by the sign ( ) a half-arrow to the right (forward reaction) and a half-arrow to the left (reverse reaction).

4. Reversibe Reactions 2 HgO (s) 2 Hg (l) + O2 (g)
Upon heating, mercury (II) oxide decomposes to mercury (Hg) and oxygen (O2) [ Equation 1]:
Under the same conditions, mercury (Hg) and oxygen (O2) recombine form mercury (II) oxide again [Equation 2]:
Mercury and oxygen combine to form mercury oxide just as fast
as mercury oxide decomposes into mercury and oxygen
2 Hg (l) O2 (g) HgO (g)
Mercury and oxygen combine to form mercury oxide just as fast as mercury oxide decomposes into mercury and oxygen

5. 2 HgO (s) Hg (l) O2 (g)
Both reactions continue to occur, but there is no net change in the composition of the system.
The amounts of mercury(II)oxide, mercury(Hg), and oxygen (O2) remain constant. There is a state of equilibrium between two chemical reactions.

6. Dynamic Equilibrium Open System (No Equilibrium) Closed System
Evaporation
Closed System
Evaporation
Liquid Gas
Liquid Gas
Liquid Gas
(No Equilibrium)
(No Equilibrium)
(Equilibrium)

7. Dynamic Equilibrium
Dynamic equilibrium is the state in a reversible reaction
in which the forward and reverse reactions are occurring
at the same rate
Ag + Cl -
Ag Cl AgCl (s)
Cl - Ag +
Chemical
Equilibrium
AgCl (s)
Rate of Precipitation = Rate of Dissolving
HC2H3O2 (aq) H C2H3O2 -
Rate of Dissociation (ionization) = Rate of Association
HC2H3O2
H +
H + C2H3O2 -
HC2H3O2
C2H3O2 -

8. Dynamic equilibrium vs Static Equilibrium
Child ascending escalator at the same rate the escalator descends.
At the balance point (the equilibrium position), the child and escalator are moving at the same rate in opposite directions.
Static Equilibrium:
Children on see-saw.
At the balance point (the equilibrium position), there is no movement of the children or the see-saw (the opposing processes) .

9. Dynamic equilibrium vs Static Equilibrium

10. Characteristics of Dynamic Equilibrium
The concentrations of the reactants and products (macroscopic properties) remain the same but the reactions don't stop!
The reactants are still reacting to form the product and the product is still being converted back to the reactants (microscopic processes).

11. Characteristics of Dynamic Equilibrium
The rate at which the reactants change into products is exactly equal to the rate at which the products change back to the original reactants.
Thus it appears as if the reaction has stopped, but in actual fact, it is still going on.
Dynamic equilibrium can only be achieved in a closed reaction (i.e an enclosed reactor)
A closed system is one in which there is absolutely no loss/gain of material to/from the surroundings.
An open system may allow matter to escape or to enter.

12. H2 + I2 2 HI Synthesis of Hydrogen Iodide
2 HI H2 + I Dissociation of Hydrogen Iodide
These two sets represent the same chemical reaction system,
but with the reactions occurring in opposite directions.
Most importantly, note how the concentrations of all the
components are identical when the system reaches equilibrium.

13. H2 + I2 2HI The equilibrium state is independent of the direction from
which it is approached. Whether we start with an equimolar
mixture of H2 and I2 (left) or a pure sample of hydrogen iodide
(shown on the right, using twice the initial concentration of HI
to keep the number of atoms the same),
the composition after equilibrium is attained
(shaded regions on the right) will be the same.
For more information, Click Here

14. How far does the reaction go?
We have seen that when the rate forward equals the rate backwards in a chemical reaction, the system reaches a state of equilibrium. But,
How far does the reaction go?
What is the final concentration of reactants and products?

15. The adjacent graph shows the changes in the reaction rates of the forward and backward reactions:
A + B C + D
Initially (t = 0), [A] and [B] were
maximum, while [C] and [D] were zero.
The rate of the forward reaction
decreases as A and B are used up.
The rate of the reverse reaction increases as C and D are formed.
Equilibrium is attained when the two rates become equal
[A], [B], [C], and [D] remain constant at equilibrium.

16. Chemical Equilibrium Consider this reaction:
If NO2 is reddish brown and N2O4 is colorless:
What is happening here?
What properties are changing?
What is happening over time? After a long time?

17. The final state depends on:
Get time progression
Check silberberg
N2O4  2NO2
Low T High T
The final state depends on:
The chemical nature of the reactants and products
The conditions of the system (temperature, pressure, volume).

18. The Equilibrium Constant
For a reaction of the type
the following is a CONSTANT (at a given T)
aA + bB
cC + dD
The equilibrium expression for this reaction would be
Kc =
[C]c[D]d
[A]a[B]b
Kc : equilibrium constant
[C] and [D] : concentration of products
[A] and [B] : concentration of reactants
If Kc is known, then we can PREDICT concentrations of products or reactants

19. What is the equilibrium expression for the reaction?
Your Turn
Equilibrium mix
What is the equilibrium expression for the reaction? .
Tier 1

20. Your Turn
Keq ~ 92/(1 x 1) = 81

21. What is the equilibrium expression for the reaction?
Your Turn
Equilibrium mix
What is the equilibrium expression for the reaction? . .
Tier 1

22. Your Turn
Equilibrium mix .
Keq ~ 22/(6 x 4) = 0.17

23. What Does the Value of K Mean?
If K >> 1, the reaction is product-favored; product predominates at equilibrium.

24. What Does the Value of K Mean?
If K >> 1, the reaction is product-favored; product predominates at equilibrium.
If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium.

25. The Concentrations of Solids and Liquids Are Essentially Constant
Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression
PbCl2 (s)
Pb2+ (aq) + 2 Cl−(aq)
The Concentrations of Solids and Liquids Are Essentially Constant
Kc = [Pb2+] [Cl−]2

26. CaCO3 (s)
CO2 (g) + CaO(s)
As long as some CaCO3 or CaO remain in the system, the amount of CO2 above the solid will remain the same.

27. What Are the Equilibrium Expressions for these equilibria?
2SO2(g) + O2(g) SO3(g)
Fe3+(aq) + SCN-(aq) [Fe(SCN)]2+(aq)
4NH3(g) + 5O2(g) NO(g) + 6H2O(g)

28. Example Give the expression of the equilibrium constant
N2 (g) H2 (g) NH3 (g)
Given [N2]=0.1M,[H2]=0.125M, [NH3]=0.11M, find the equilibrium constant for the reaction.

29. Example
Find the relationship between K1 and K2, the equilibrium constants of these two reactions:
2A + 2B C K1
C A + B K2
What can you deduce about the equilibrium constant for the forward and backward reaction?
K2 = 1/K1

30. Example
The following reaction is an esterification reaction producing ethyl ethanoate. CH3CO2H(l) + C2H5OH(l) CH3CO2C2H5(l) + H2O(l) The value of Kc at 250C is 4.0. This equilibrium can be approached experimentally from the opposite direction. What is the value for Kc for the reaction, the hydrolysis of ethyl ethanoate. CH3CO2C2H5(l) + H2O(l) CH3CO2H(l) + C2H5OH(l)

31. Example
The following reaction takes place at 460ºC, where the equilibrium constant K has a value of 85.
SO2(g) + NO2(g) NO (g) + SO3 (g)
At a certain moment, the concentrations of the reactant and products were measured to be: [SO2] = 0.04, [NO2] = 0.5M, [NO] = 0.3M, [SO3] = 0.02M
(a) Is this system at equilibrium?
(b) If not, in which direction must the reaction go to reach
equilibrium?
54.47

32. Le Chatelier’s Principle
states that when a system in dynamic equilibrium is subjected to a change, the position of equilibrium will shift to minimise the change.
The possible changes in conditions include
Changes in concentration o f either the reactants or products.
Changes in pressure for gas phase reactions
Changes in temperature
Presence of a catayst
Has industrial significance as it allows chemists to alter the reaction conditions to produce an increased amount of product, hence increase the profitability of the chemical process.

33. Concentration
A change in the concentration of any component in the equilibrium mixture will result in the shift in position of the equiibrium.
Example
If one of the products is removed from the mixture, the equilibrium will shift to the right to replace the substance removed.
In the reaction below, what happens when
water is removed / added ?
ethanol , C2H5OH is added / removed ?
CH3CO2H(l) + C2H5OH(l) CH3CO2C2H5(l) + H2O(l)
More carboxylic acid and alcohol would react to replace it producing more ester.
This can be done by adding more conc sulfuric acid that provides H+ which acts as the catalyst for the reaction but also as a dehydrating agent removing water from the mixture.
The addition of acid favours the prouction of ester since by removing water, it shifts the equilibrium to the right.

34. Pressure
The equilibrium positions of certain gas phase reactions are affected by changes in external pressure.
These are reactions where there is a change in the amount (number of moles) of substance during the course of the reaction.
Example
An increase in pressure will shift the position of the equilibrium towards the side that involves fewer molecules.
In the reaction below, what happens when pressure is
increased ?
decreased ?
N2O4(g) NO2(g)
colourless brown
Increase in pressure cause the colour is initially darker as the gas is squeezed into a smaller vol – the conc is increased . Then the equilibrium shifts to the left , the side with fewer molecules and the colour lightens to almost the original level

35. Temperature
A increase in temperature will always favour the endothermic process and will shift the equilibrium towards that direction. Lowering the temperature will favour the exothermic process.
A change in temperature will alter the Kc.
Example
In the reaction below, what happens when the temperature is
increased ?
decreased ?
N2O4(g) NO2(g) ΔHƟ = +57kJmol-1
colourless brown
Increase in temp (putting in a hot water bath) cause the colour to be darker as the equilibrium shifts to the right that is endothermic in order to reduce the effect of the higher temperature.
Decrease in temp (putting in ice bath) will cause the colour to be lighter.

36. Example
Draw a table showing how the position of equilibrium in the reaction A,B and C would be affected by the followng changes:
Increased temperature
Increased pressure
Reaction A : interconversion of oxygen and ozone
3O2(g) O3(g) ΔHƟ = +284 kJmol-1
Reaction B : the reaction between sulfur dioxide and oxygen in the presence of a platinium/rhodium catalyst.
SO2(g) + 2O3(g) SO3(g) ΔHƟ = -197 kJmol-1
Reaction C : the reaction between hydrogen and carbon dioxide.
CO2(g) + 2H2(g) CO(g) + H2O(g) ΔHƟ = +41 kJmol-1

37. Catalyst
A catalyst has no effect on the position of an equilibrium at a particular temperature as increases the rate of the forward and the reverse reaction equaly.
It decreases the time it takes the system to reach equilibrium.
"A catalyst provides an alternative route for the reaction with a lower activation energy."
It does not "lower the activation energy of the reaction". There is a subtle difference between the two statements that is easily illustrated with a simple analogy.
Suppose you have a mountain between two valleys so that the only way for people to get from one valley to the other is over the mountain. Only the most active people will manage to get from one valley to the other.
Now suppose a tunnel is cut through the mountain. Many more people will now manage to get from one valley to the other by this easier route. You could say that the tunnel route has a lower activation energy than going over the mountain.
But you haven't lowered the mountain! The tunnel has provided an alternative route but hasn't lowered the original one. The original mountain is still there, and some people will still choose to climb it.

38. Conditions affecting the position of Equilibrium
Increase in temperature:
The equilibrium moves in the endothermic direction, taking in the extra energy
Kc changes
A catalyst does not alter the position of equilibrium but speeds up the forward and reverse reactions equally
Kc : no change
Increase pressure:
The equilibrium moves to the side with fewer gaseous moles, reducing pressure
Increase the concentration of a species
The equilibrium moves to the side without the species, reducing its concentration
Concentration
Pressure
Temperature
Catalyst
An equilibrium system opposes changes

39. Ammonia and the Haber Process
N₂ (g) + 3H₂ (g) ↔ 2NH₃ (g)
Typical Conditions
Pressure: 20000kPa (200 atm)
Temperature: °C
Catalyst: Iron
Equilibrium Theory Favours
Low temperature: exothermic reaction – higher yield at low temperature
High pressure: decrease in number of gaseous molecules
Kinetic Theory Favours
High temperature: greater average energy + more frequent collisions
High pressure: more frequent collisions for gaseous molecules
Catalyst: lower activation energy
Compromise Conditions
The conditions used are a compromise with the catalyst enabling the rate to be kept up, even at a lower temperature