1. IB CHEMISTRY
Topic 7 Equilibrium
Higher level

2. 7.1 Equilibrium
OBJECTIVES
• A state of equilibrium is reached in a closed system when the rates of the forward and reverse reactions are equal. • The equilibrium law describes how the equilibrium constant (Kc) can be determined for a particular chemical reaction. • The magnitude of the equilibrium constant indicates the extent of a reaction at equilibrium and is temperature dependent. • The reaction quotient (Q) measures the relative amount of products and reactants present during a reaction at a particular point in time. Q is the equilibrium expression with non-equilibrium concentrations. The position of the equilibrium changes with changes in concentration, pressure, and temperature. • A catalyst has no effect on the position of equilibrium or the equilibrium constant. • The characteristics of chemical and physical systems in a state of equilibrium. • Deduction of the equilibrium constant expression (Kc) from an equation for a homogeneous reaction. • Determination of the relationship between different equilibrium constants (Kc) for the same reaction at the same temperature. • Application of Le Châtelier’s principle to predict the qualitative effects of changes of temperature, pressure and concentration on the position of equilibrium and on the value of the equilibrium constant.

3. Concentration change in a reaction
As the rate of reaction is dependant on the concentration of reactants... the forward reaction starts off fast but slows as the reactants get less concentrated
FASTEST AT
THE START
SLOWS DOWN AS REACTANTS ARE USED UP
THE STEEPER THE GRADIENT, THE FASTER THE REACTION
TOTAL CONVERSION TO PRODUCTS
In an ordinary reaction; all reactants end up as products; there is 100% conversion

4. Haber Process example

5. Equilibrium reactions
Initially, there is no backward reaction but, as products form, it speeds up and provided the temperature remains constant there will come a time when the backward and forward reactions are equal and opposite; the reaction has reached equilibrium.
FASTEST AT THE START
NO BACKWARD REACTION
FORWARD REACTION SLOWS DOWN AS REACTANTS ARE USED UP
BACKWARD REACTION
STARTS TO INCREASE
In an equilibrium reaction, not all the reactants end up as products; there is not a 100% conversion.
BUT IT DOESN’T MEAN THE REACTION
IS STUCK IN THE MIDDLE
AT EQUILIBRIUM THE BACKWARD AND FORWARD REACTIONS ARE EQUAL AND OPPOSITE

6. IMPORTANT REMINDERS
• a reversible chemical reaction is a dynamic process
• everything may appear stationary but the reactions are moving both ways
• the position of equilibrium can be varied by changing certain conditions
Trying to get up a “down” escalator gives an excellent idea of a non- chemical situation involving dynamic equilibrium.
Summary When a chemical equilibrium is established ...
• both the reactants and the products are present at all times
• the equilibrium can be approached from either side
• the reaction is dynamic - it is moving forwards and backwards
• the concentrations of reactants and products remain constant

7. The Equilibrium Constant Kc
What is heterogeneous Equilibria?
When all reactants and products are in one phase, the equilibrium is homogeneous.
If one or more reactants or products are in a different phase, the equilibrium is heterogeneous.

8. The equilibrium law Simply states
“If the concentrations of all the substances present at equilibrium are raised to the power of the number of moles they appear in the equation, the product of the concentrations of the products divided by the product of the concentrations of the reactants is a constant, provided the temperature remains constant”
There are several forms of the constant; all vary with temperature.
Kc the equilibrium values are expressed as concentrations of mol dm-3

9. The Equilibrium constant (Kc)
for an equilibrium reaction of the form...
aA bB cC dD
then (at constant temperature)
[C]c . [D]d = a constant, (Kc)
[A]a . [B]b
where [ ] denotes the equilibrium concentration in mol.dm-3
Kc is known as the Equilibrium Constant

10. Magnitude of Kc
If Kc is very large, then the reaction goes to completion (mostly products). Kc >>1 If Kc is very small, then the reaction would be considered to not occur (mostly reactants). Kc <<1

11. ”When a change is applied to a system in dynamic equilibrium, the
Le Chatelier’s Principle
”When a change is applied to a system in dynamic equilibrium, the
system reacts in such a way as to oppose the effect of the change.”
A(s) + heat  B(g)
Increase heat, increase B (decrease A)
Decrease pressure, increase B (decrease A)
Increase A, increase B (decrease A)

13. Factors affecting the position of equilibrium
TEMPERATURE
• temperature is the only thing that can change the value of the equilibrium constant.
• altering the temperature affects the rate of both backward and forward reactions
• it alters the rates to different extents
• the equilibrium thus moves producing a new equilibrium constant.
• the direction of movement depends on the sign of the enthalpy change.
REACTION TYPE DH
INCREASE TEMP
DECREASE TEMP
EXOTHERMIC -
TO THE LEFT
TO THE RIGHT
ENDOTHERMIC +
Predict the effect of a temperature increase on the equilibrium position of...
H2(g) + CO2(g) CO(g) + H2O(g) DH = kJ mol-1 moves to the RHS
2SO2(g) + O2(g) SO3(g) DH = - ve moves to the LHS

14. Factors affecting the position of equilibrium
CATALYSTS
Catalysts work by providing an alternative reaction pathway involving a lower activation energy. Ea
MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY
EXTRA MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER
MOLECULAR ENERGY
NUMBER OF MOLECUES WITH
A PARTICULAR ENERGY

15. Factors affecting the position of equilibrium

16. EXTRA: So why does pressure shift the equilibrium but not the equilibrium constant?
N2 + 3H2 ⇌ 2NH 𝐾 𝑐 = [NH3] 2 N2 [H2] 3
The concentration of the gas is determined by the pressure as all the particles act as a unit.
Partial pressure of A, PA = mole fraction of A x total pressure
PA = xA x P
𝐾 𝑐 = [NH3] 2 N2 [H2] 3 = 𝑃NH32 𝑃N2 ×𝑃H2 3 = (𝑥 NH3 ×𝑃)2 ( 𝑥 N2 ×𝑃) ( 𝑥 H2 ×𝑃) 3 = 𝑥 NH3 2𝑃2 𝑥 N2 𝑃 𝑥 H2 3𝑃 3 = 𝑥 NH3 2 𝑥 N2 𝑥 H2 3𝑃 2

17. N2 H2
NH3
Ratio 1 3 2
10atm mole fraction 500K
0.15
0.45
0.4
300atm mole fraction 500K
0.0375
0.1125
0.85

18. For 10atm: For 300atm: 𝐾 𝑐 = 𝑥 NH3 2 𝑥 N2 𝑥 H2 3𝑃 2
= (0.45) 310 2 =
= 0.12
For 300atm:
𝐾 𝑐 = 𝑥 NH3 2 𝑥 N2 𝑥 H2 3𝑃 2
= (0.1125) =
= 0.15
Therefore despite an increase in pressure the Kc is the same (graph approximation errors) even though the equilibrium shifts dramatically to the right (doubles NH3).

19. Equilibrium example: The Haber Process
N2(g) H2(g) NH3(g) DH = - 92 kJ mol-1
Conditions Pressure 200 atmospheres
Temperature 450°C
Catalyst iron
Equilibrium theory favours
low temperature exothermic reaction - higher yield at lower 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
Which is better? A low yield in a shorter time or
a high yield over a longer period.
The conditions used are a compromise with the catalyst
enabling the rate to be kept up, even at a lower temperature.

20. N2(g) + 3H2(g) 2NH3(g) DH = - 92 kJ mol-1

21. IMPORTANT USES OF AMMONIA AND ITS COMPOUNDS
MAKING
FERTILISERS 80% of the ammonia produced goes to make fertilisers such as
ammonium nitrate (NITRAM) and ammonium sulphate
NH HNO3 ——> NH4NO3
2NH H2SO4 ——> (NH4)2SO4
NITRIC ACID ammonia can be oxidised to nitric acid
nitric acid is used to manufacture... fertilisers (ammonium nitrate)
explosives (TNT)
polyamide polymers (NYLON)

22. Combining Equilibrium constants
Effect on equilibrium expression
Effect on Kc
Reversing the reaction
Inverts the expression
1/Kc or Kc-1
Doubling the reaction coefficients
Squares the expression
Kc2
Halving the reaction coefficients
Square roots the expression
√Kc
Adding together two reactions
Multiplies the two expressions
Kc.Kc’

23. The Reaction Quotent (Q)
for reaction of the form...
aA bB cC dD
then (at constant temperature)
[C]c . [D]d = a constant, (Q)
[A]a . [B]b
This is the same as the equilibrium constant, but the system is not necessarily in equilibrium with Q.

24. Using Q Chemically: Can be used to determine if a reaction will occur
Physically: Can be used to determine if a salt will dissolve
Q > K
Products are greater
Reverse reaction is favoured
Q = K
Both reactions occurring in equilibrium
Q < K
Reactants are greater
Forward reaction is favoured

25. 17.1 The equilibrium law
Higher level
OBJECTIVES
• Le Châtelier’s principle for changes in concentration can be explained by the equilibrium law. • The position of equilibrium corresponds to a maximum value of entropy and a minimum in the value of the Gibbs free energy. • The Gibbs free energy change of a reaction and the equilibrium constant can both be used to measure the position of an equilibrium reaction and are related by the equation, ∆G = -RT lnK. • Solution of homogeneous equilibrium problems using the expression for Kc. • Relationship between ∆G and the equilibrium constant. • Calculations using the equation ∆G = -RT lnK

26. Higher level
EXTRA:
Q problems

27. Problem 1: (chemical reaction) In the following reaction:
Higher level
Problem 1: (chemical reaction)
In the following reaction:
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
[CO] = 1M, [H2O] =1M, [CO2] = 2M, [H2] = 2M
The Kc for this reaction is 1.0. In which direction will the reaction shift?
Q = [CO2][H2] [CO][H2O] = 2×2 1×1 = 4
Hence Q>K, so the products are in excess and the equilibrium will shift to the left.

28. EXTRA: Solubility product constant (Ksp)
Higher level
EXTRA: Solubility product constant (Ksp)
For an insoluble salt: MX(s) ⇌ M+(aq) + X- (aq) As most the salt is in the form of MX, the Keq will not change significantly. A more common expression, solubility product constant is used: Ksp = [M][X] (similarly as before, if 2 ions are given off then that ion in the equation is squared)

29. Problem 2: (salt solubility)
Higher level
Problem 2: (salt solubility)
0.5dm3 of M NaOH is mixed with 0.5dm3 of 0.002M MgSO4. Determine if a precipitate will form. 
Ksp(MgOH) = 1.8 x 10-11 
2NaOH + MgSO Na2SO4 + Mg(OH)2 
n(OH-) = C x V = M x 0.5dm3 = mol
n(Mg2+) = C x V = 0.002M x 0.5dm3 = 0.001mol
C(OH-) = n/V = mol/1dm3 = M 
C(Mg2+) = n/V = 0.001mol/1dm3 = 0.001M
Ksp = 1.8 x 10-11 
[Mg2+][OH-] 2 = (0.001) ( )2 = 2.5x10-12 
The ion-product is less than the Ksp so there should NOT be a precipitate.

30. Higher level
K problems

31. Ratio, Initial, Change, Equilibrium
1. Kc from concentrations:
Higher level
Hydrogen reacts with iodine to form HI. If mole of iodine was in the reaction initially and only 0.02 detected at equilibrium, what is the equilibrium constant. H2 + I2  2HI R I C E H2 I2 HI 1 1 2
0.100
0.100
0.02
Ratio, Initial, Change, Equilibrium

32. Higher level R I C E H2 I2 HI 1 1 2
0.100
0.100
-0.08
-0.08
+0.16
0.02
0.02
0.16
[H2] [I2]
Kc=
[HI]2 =
[.020] [.020]
[.160]2
= 64

33. R I C E H2 I2 HI 1 1 2 0.200 0.200 2. Concentrations from Kc:
Higher level
Hydrogen reacts with iodine to form HI. If 0.200M of iodine was in the reaction initially and the equilibrium constant is 49.5, what is the concentration of HI at equilibrium? H2 + I2  2HI R I C E H2 I2 HI 1 1 2
0.200
0.200

34. When writing x, use whole number ratios.
Higher level R I C E H2 I2 HI 1 1 2
0.200
0.200 -x -x
+2x
0.2 - x
0.2 - x 2x
[H2][I2]
Kc =
[HI]2 =
[0.2 -x]2
[2x]2
= 49.5
2x/[0.2-x] = 7.04, 2x = x, x=0.156
H2 (I2 also): = M
HI: 2(0.156) = M

35. 3. When Kc is very small < 10-3:
Higher level
The Kc for this reaction N2 + O2  2NO is 4.8 x Initially there were mol N2 and moles of O2. How many moles of NO are there at equilibrium? R I C E N2 O2 NO 1 1 2
0.033

36. Higher level R I C E N2 O2 NO 1 1 2
0.033 -x -x
+2x
0.033-x x 2x
[N2][O2]
Kc =
[NO]2 =
[0.033-x][ x]
[2x]2
= 4.8 x 10-31
[0.033-x][ x]
[2x]2
= 4.8 x 10-31

37. This is the equilibrium [NO2]
Higher level
[0.033-x][ x]
[2x]2
= 4.8 x 10-31
Small Kc: Thus numerator is small and x must be small: x is negligible when adding or subtracting
[0.033][0.0081]
[2x]2
= 4.8 x 10-31
[2x]2
= 1.28 x 10-34 2x
= 1.13 x 10-17
This is the equilibrium [NO2]

38. Higher level
G and K problems

39. Gibbs free energy and equilibrium
Higher level
Gibbs free energy and equilibrium
K gives and indication of G:
K = 1
Equilibrium
∆G=0
K>1
Products favoured
∆G<1
K<1
Reactants favoured
∆G>1

40. Gibbs free energy and equilibrium
Higher level
Gibbs free energy and equilibrium
The mathematical relationship is found in your data booklet: ∆G° = -RTlnK Rearranging for K: lnK = - ∆G° RT

41. ∆G = ∆H - T∆S = -824.2 – (300x-0.2705) = =743.1 kJ/mol
Higher level
Problem 1: Determine K for the following reaction at 300K, given that ∆H = kJ/mol and ∆S = J/Kmol.
2Fe(s) + 3/2O2(g) ⇌ Fe2O3(s)
∆G = ∆H - T∆S = – (300x ) = =743.1 kJ/mol
lnK = - ∆G° RT = × J/mol J/Kmol x300K = 298
K = e298 = 2.6 x10129
Therefore the reaction is highly favourable