1. Chemical Equilibrium 1. 2. 3. 4. A B + A + C + B D A + B C + D 1. 4.
Reaction begins.
No products yet formed.
High rate of collisions between A & B.
Rate of forward reaction HIGH.
2 & 3
Products formed
Collisions between reactants decrease.
Rate of forward reaction DECREASES
Reverse reaction begins.
Rate of forward reaction EQUAL to rate of reverse reaction.
Dynamic equilibrium established.
Concentrations constant.

2. EQUILIBRIM REACTIONS e.g. N2(g) + 3H2(g) 2NH3(g)
Most reactions DO NOT go to completion.
Reactions that do not go to completion are REVERSIBLE.
Reversible reactions exist in a state of EQUILIBRIUM.
Equilibrium is reached when the rate of the forward reaction equals the rate of the reverse reaction.
The reaction proceeding from L R as the equation is written is known as the forward reaction.
e.g. N2(g) + 3H2(g) NH3(g)

3. Chemical Equilibrium A + B A + C + B D A + B C + D …………… ……………… ……………
…………..
……………………………………………
………………
……………
……………….

4. Chemical Equilibrium A + B A + C + B D A + C + B D
DYNAMIC CHEMICAL EQUILIBRIUM
Reactants
Concentration
Products
time
Forward
Reaction Rate
Reverse

5. Equilibrium Position - Left
The equilibrium reaction does not mean the amounts of products and reactants are equal.
If the products react more easily than the reactants but BOTH RATES are very slow.
We say the Equilibrium is shifted to the LEFT.
There will be higher concentration of reactants. A + C + B D
time
Reactants
Products
Concentration
Ratio of: Products
Reactants
Would be LESS than one.
< 1

6. Equilibrium Position Middle
If the reactants and the products both have similar reaction rates.
The Equilibrium position will lie in the middle.
There will be the same concentration of reactants & Products. A + C + B D
Reactants
Concentration
Ratio of:
Products
Reactants
Would be EQUAL TO one.
Products
time
= 1
Products
Reactants
= 1

7. Equilibrium Position - Right
If the reactants react ……… ………… than the products.
The Equilibrium is ………….. to the ……………….
There will be higher concentration of …………. A + B C + D
……………
Ratio of:
Products
Reactants
Would be ………………. than ONE.
Conc
……………. 1
time

8. Equilibrium Position - Right
If the reactants react more easily than the products.
The Equilibrium is shifted to the RIGHT.
There will be higher concentration of Products. A + B C + D
Products
Ratio of:
Products
Reactants
Would be GREATER than one.
Conc
Reactants
> 1
time

9. Equilibrium Position SHIFT
To ………… A + C + B D
To ………
Concentration
time
……………
……………….
Reactants
Concentration
Concentration
……………….
Products
time
time
Products
Reactants
> 1
Products
Reactants
< 1
Products
Reactants
= 1

10. Equilibrium Position SHIFTA + C +
SHIFTED To Left B D
To right
Concentration
time
Reactants
Products
Reactants
Reactants
Concentration
Concentration
Products
Products
time
time
Products
Reactants
> 1
Products
Reactants
< 1
Products
Reactants
= 1

11. Adding Reactants Conc reactants products A + C + B D time
More …………………….added
Reactant concentration ……………………
Forward reaction rate …………………..
Reactant concentration ………………/Product concentration ………….
Reverse reaction rate ……………………….
New equilibrium established - ………….. ratio of PRODUCTS/REACTANTS - equilibrium shifted to …………………

12. Adding Reactants reactants Conc. products A + C + B D time
Reactant added
reactants
Equilibrium Shifts
New equilibrium
established
Conc.
products A + C + B D
time
More Reactant added
Reactant concentration increases & Forward reaction rate increases
Product concentration decreases & Reverse reaction rate decreases
Reactant concentration decreases as reactants used up.
Product concentration increases as new products formed.
New equilibrium established - higher ratio of PRODUCTS/REACTANTS - equilibrium shifted to RIGHT

13. Adding Products Eqm. Shifts New Equilibrium Reactants Conc Products A
Product added A + C + B D
time
More PRODUCT added
PRODUCT concentration increases (Instantly - Reactant decreases )
3. REVERSE reaction rate increases (fast then slower)
4. PRODUCT concentration decreases, REACTANT concentration increases (fast then slower)
5. Forward reaction rate increases
6. New equilibrium established - lower ratio of PRODUCTS/REACTANTS - equilibrium shifted to LEFT.

14. Conditions of an equilibrium
At equilibrium both reactions continue to occur - the system is DYNAMIC.
The system is CLOSED – nothing added or removed.
The concentrations of reactants and of products remain constant.
Rate of the forward reaction equals rate of the back reaction.
Equilibrium can be obtained from either side.

15. Reminder of dynamic equilibrium
Dynamic equilibrium animation

16. Chemical Equilibrium Equilibrium is reached when the rate of the
forward reaction equals the rate of the
reverse reaction.
H2(g) + I2(g)
2HI(g)
(H = -13kJ/mol)
The H value always refers to the forward reaction. 13 kJ of energy is liberated for every mole of HI formed.
For the whole reaction:
H2(g) + I2(g)
2HI(g)
(H = -26kJ)

17. Chemical Equilibrium Reaction Rate H2 + I2 2 HI 2 HI H2 + I2 H2 + I2
Explanation P142
2 HI
H2 + I2
Time

18. Chemical Equilibrium Reaction Rate H2 + I2 2 HI 2 HI H2 + I2 H2 + I2
Explanation P142
2 HI
H2 + I2
Time

19. At equilibrium the concentration of all sub-
Equilibrium Constant
At equilibrium the concentration of all sub-
stances are constant at a fixed temperature.
Each equilibrium has a constant e.g. for
H2(g) + I2(g)
2HI(g)
Kc =
[H2] means conc. of ...
N.B
Kc is only determined by the concentrations of solutions and gases. Pure liquids & solids are not included in the equation – their concentrations are constant.

20. Equilibrium Constant In General: aA + bB cC + dD [C]c [D]d Kc =
When Kc has a high value, there will be more PRODUCTS – (on the RIGHT) - we say the equilibrium lies to The RIGHT (vice versa for a low value).

21. Equilibrium Constant Calculations
1. The balanced equation must be known.
The concentration of a solid or liquid remains constant - these are not included in the equation.
3. The concentration of the solvent is constant and not included in the equation.
examples: PP
4. The value of Kc is given without units.

22. Kc Calculation Examples
Write expressions for Kc for each of the following reactions:
C3H8(g) O2(g)  3CO2(g) H2O(g)
Ca(s) + 2H2O(ℓ)  Ca(OH)2(aq) H2(g)
AgNO3(aq) NaCℓ(s)  AgCℓ(s)+ NaNO3(aq)
D. Na2CO3(s) HCℓ (aq) 2NaCℓ(aq) + H2O(ℓ) CO2(g)

23. Kc Calculation Examples
Write expressions for Kc for each of the following reactions:
C3H8(g) O2(g)  3CO2(g) H2O(g)
Ca(s) + 2H2O(ℓ)  Ca(OH)2(aq) H2(g)
AgNO3(aq) NaCℓ(s)  AgCℓ(s)+ NaNO3(aq)
D. Na2CO3(s) HCℓ (aq) 2NaCℓ(aq) + H2O(ℓ) CO2(g)
[CO2]3[H2O]4
[C3H8] [O2 ]5
Kc =
Kc =
[Ca(OH)2][H2]
[NaNO3]
[AgNO3]
Kc =
[NaCl]2[CO2]
[HCl]2
Kc =

24. Kc Calculation Examples7
[NH3]2
[H2]3 [N2]
(4/2)2
(1/2)3 (6/2)
Kc = = =
3H N2  2NH3
Initial: 7mol 8mol 0mol
Reacts: 6mol 2mol
Eqm: 1mol 6mol 4mol
[ ]: /2 6/2 4/2

25. Calculating K
Given: N2(g) + 3H2(g) ↔ 2NH3(g) At equilibrium: [N2] = 0.30 M [H2] = 0.10 M [NH3] = 0.20 M

26. Calculating Concentrations at Equilibrium
Methane gas reacts with water vapour to produce carbon monoxide gas and hydrogen gas according to this equation:
CH4(g) + H2O(g) ↔ CO(g) H2(g)
At equilibrium: [CO] = M [H2] = M [CH4] = M. If K is 5.67, calculate the concentration of water vapour
Homework: complete the worksheet: Equilibrium Exercise 1

27. Carbon monoxide is a primary starting material in the synthesis of many organic compounds, including methanol, CH3OH(l). At C, K is 6.4 x 10 −7 for the decomposition of carbon dioxide into carbon monoxide and oxygen. Calculate the concentrations of all entities at equilibrium if mol of CO2(g) is placed in a closed container heated at C.

28. If 0. 50 mol N2O4(g) is placed in a 1
If 0.50 mol N2O4(g) is placed in a 1.0 L closed container at 150 0C, what will be the concentrations of N2O4(g) and NO2(g) at equilibrium? (K=4.50)
When hydrogen and iodine are placed in a closed container at 440 0C, they react to form hydrogen iodide. At this temperature, the equilibrium constant, K is Determine the concentrations of all entities at equilibrium if 4.00 mol of hydrogen and 2.00 mol of iodine are placed in a 2.00 –L reaction vessel.
H2(g) I2(g) ↔ HI(g)

29. The following reaction has an equilibrium constant of 25.0 1100 K.
H2(g) I2(g) ↔ HI(g)
If 2.00 mol of hydrogen gas, H2(g) and 3.00 mol of iodine gas, I2(g) are placed in a 1.00 L reaction vessel at 1100K, what is the equilibrium concentration of each gas?

30. Changing Equilibrium Conditions
Predicting the effect
Le Chatelier’s Principle
If the conditions of an equilibrium system are changed, a process takes place which tends to oppose or cancel the effect of the change.

31. Changing Equilibrium Conditions
2. The equilibrium constant
N2(g) + 3H2(g) NH3(g) (H < 0)
Kc =
When the concentration of the N2 is increased, the ratio (Kc) will be smaller.
To restore the value of Kc and the equilibrium,more N2 will have to react with H2
This will diminish the concentrations of reactants.

32. N2(g) + 3H2(g) 2NH3(g) (H < 0)
Adding Reactants
N2(g) + 3H2(g) NH3(g) (H < 0)
When the concentration of the N2 is increased:
the concentration of reactants increases instantaneously
The ratio Kc (momentarily) decreases
Forward reaction rate increases and the concentrations of REACTANTS is LOWERED whilst the concentration of NH3 INCREASES.
The reverse reaction rate INCREASES as more NH3 forms.
A new equilibrium is established which has shifted to the RIGHT.
The value of the ratio Kc is restored to its original value
Kc =
Overall Kc UNCHANGED! Kc
time

33. N2(g) + 3H2(g) 2NH3(g) (H < 0)
Effect of Temperature
N2(g) + 3H2(g) NH3(g) (H < 0)
Kc =
Increasing the temperature
will favour the ENDOTHERMIC reaction.
The REVERSE reaction will be favoured
More REACTANTS produced
the reaction shift to the LEFT. Kc
time
The ratio (Kc) will therefore DECREASE.

34. Effect of Pressure Kc = N2(g) + 3H2(g) 2NH3(g) (H < 0)
Increasing the PRESSURE
Equilibrium will shift to REDUCE pressure.
Reaction will favour the side with the LEAST number of MOLES of GAS – RHS NH3.
More NH3 produced.
Kc goes go up (this eg)!
Kc = Kc
time
The ratio (Kc) will therefore INCREASE.

35. Effect of a Catalyst Ea Ea
E react
E prod
Potential Energy
Potential Energy
E react
E prod
A CATALYST lowers the ACTIVATION ENERGY of the reaction by providing a different reaction pathway.
Activation Energy is lowered for BOTH FORWARD AND REVERSE REACTIONS.
BOTH rates are therefore increased by the same amount and so the EQUILIBRIUM DOES NOT SHIFT!!!!

36. Industrial Preparation
N2(g) + 3H2(g) NH3(g) (H < 0)
Favorable conditions:
High
LOW
In practice a ……………………. Temperature is used. Too ………………… will slow the reaction down.
Kc = Kc
time

37. Industrial Preparation
N2(g) + 3H2(g) NH3(g) (H < 0)
Favourable conditions:
High [N2] & [H2]
High Pressure
LOW Temperature
LOW [NH3]
In practice a compromise temperature is used. Too low will slow the reaction down.
Kc = Kc
time

38. Changing Equilibrium Conditions
Change
Shift

39. Equilibrium Disturbance
2SO2(g) + O2(g) --> 2SO3(g)
Conc of O2 is increased
Temperature is increased
Pressure is increased

40. Equilibrium in solutions
Heterogeneous equilibrium
Depending on circumstances, a solution may be:
Under saturated – forward rate greater than reverse (salt dissolving)
Over saturated – reverse greater than forward – salt precipitates
Saturated – forward & reverse the same rate (equilibrium)

41. Equilibrium in solutions
Heterogeneous equilibrium
Depending on circumstances, a solution may be:
Under saturated – forward rate greater than reverse (salt dissolving)
Over saturated – reverse greater than forward – salt precipitates
Saturated – forward & reverse the same rate (equilibrium)

42. Solubility of salts All nitrates are soluble
All alkali metal & ammonium salts are soluble
Chlorides, bromides & iodides are soluble - except Ag, Hg, Cu & Pb
Sulphates are soluble – except Pb, Ca, Ag &Hg
Carbonates, phosphates & sulphates of alkali metals & ammonium are soluble
Hydroxides of alkali metals, ammonium & barium are soluble
Sulphides of alkali metals, alkaline earth metals and ammonium are soluble

43. Equilibrium in solutions
The equation for the equilibrium reaction
of a saturated salt solution can be represented
as follows:
AB (s) A+ (aq) + B- (aq)

44. Equilibrium in solutions
The equilibrium constant is called the
solubility product and is calculated as
follows:
Ca(OH)2(s) Ca2+(aq) OH-(aq)
Ksp = [Ca2+][OH-] 2

45. Equilibrium in solutions
Temperature Change:
Solubility curves show us that solubility's of most salts increase with increase in temperature.

46. Equilibrium in solutions
Change in concentration :
NaCl (s) Na+(aq) + Cl-(aq)
Adding HCl to the above equilibrium, causes
the equilibrium to shift to the left. NaCl is
therefore precipitated until the equilibrium
is restored.
Disturbance of the equilibrium by increas-
ing the concentration of one kind of ion is
called the common ion effect.

47. Equilibrium in solutions
Take note that the common ion effect is
not restricted to solubility equilibria
only.
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
Adding a few drops of indicator and NH4Cl
will show that the equilibrium will shift to the
left - Explain.