Unless otherwise stated, all images in this file have been reproduced from:
Blackman, Bottle, Schmid, Mocerino and Wille,
Chemistry, 2007 (John Wiley)
ISBN: 9 78047081 0866
Slide 29-1
Chem 1101
A/Prof Sébastien Perrier
Room: 351
Phone: 9351-3366
Email: [email protected]
Prof Scott Kable
Room: 311
Phone: 9351-2756
Email: [email protected]
A/Prof Adam Bridgeman
Room: 222
Phone: 9351-2731
Email: [email protected] 29-2
Highlights of last lecture
Equilibrium constants
Various types of equilibrium constants introduced
e.g. Ka, Kb, Ksp, Kp
Equilibrium constants involving liquids and solids
Concentration of liq. and solids not included
Equilibrium calculations
Solving equilibrium problems
The “small x” approximation
ANSWERS from last lecture:
e.g. (3): 1.21x10-3 mol/L PbI2 will dissolve
e.g. (4): [NOCl] = 0.48 M; [NO] = 2.0 x 10-2 M; [Cl2] = 1.0 x 10-2 M
Slide 29-3
Revision: Kp and Kc
Pressure and concentration are linked by the ideal gas law:
n
P
=
or… conc. =
V
RT
PV=nRT
Using the NOCl example:
Kc=1.6x10-5 mol/L
2NOCl(g) 2NO(g) + Cl2(g)
2
 PNO   PCl 

 
2
RT

  RT 
[NO ] [Cl2 ]
Kc =
=
2
2
[NOCl ]
 PNOCl 


 RT 
2
Check :
Kp=4.1x10-4 atm
 1
2
(PNO ) PCl  RT
=
×
2
(PNOCl )  1

 RT
( )
2
3



2



4.1 × 10 − 4
=
= 1.6 × 10 −5 = Kc
RT 0.0821 × (273 + 35)
= Kp ×
1
RT
Kp
Slide 29-4
Revision: Kp and Kc
The preceding derivation is just one example of a
general relationship:
K p = Kc (RT )
or
Kc = K p (RT )
∆n gas
− ∆n gas
To avoid confusion, just remember ONE of these
Slide 29-5
Le Chatelier’s Principle
“If a change is imposed on a system at equilibrium, the
position of the equilibrium will shift in a direction that
tends to reduce that change.”
Although Le Chatelier’s Principle over-simplifies the
situation, it works remarkably well.
Slide 29-6
This lecture…
Disturbing equilibrium.
Predicting direction of chemical change after
disturbance (Le Chatelier’s Principle).
The effect of
The effect of
The effect of
The effect of
changing concentration
changing volume
changing pressure
changing temperature
The effect of catalysts
The connection between equilibrium and ∆G
Slide 29-7
Change in concentration
Consider the following reversible reaction:
PCl3(g) + Cl2(g) PCl5(g)
Kc(523 K) = 24.0
[PCl5 ]
At eq' m, Q =
= 24.0
[PCl3 ][Cl2 ]
What happens if we inject more Cl2(g)?
ANS: Q gets smaller, therefore the eq’m shifts to produce more product,
i.e. acts to remove Cl2
• If the concentration increases, the system acts to consume some of it
• If the concentration decreases, the system acts to produce some of it
Slide 29-8
Experiment
Effect of a change in concentration
[Fe(OH2)6]3+(aq) + SCN-(aq) [Fe(OH2)5(SCN)]2+(aq) + H2O(l)
brown
light yellow
dark red
Predict what will happen when I:
1. Add [Fe(OH2)6]3+(aq)
2. Add SCN-(aq)
3. Add F-(aq)
F- is a VERY strong
binding ligand and forms
FeF6 (colourless)
Slide 29-9
Effect of a change in pressure
There are 3 ways in which the pressure of a system can be
changed:
1. Add or remove a gaseous reactant or product;
This is the same effect as changing concentration above.
2. Add an inert gas (ie not involved in the reaction);
Although this changes total pressure, it does not change the
PARTIAL pressures of the species. Therefore adding an inert
gas (at constant volume) HAS NO EFFECT
3. Change the volume of the container.
Slide 29-10
Changing volume
Consider our old friend:
2 NO2(g) N2O4(g)
If the volume decreases, the partial pressures of
everything increase. By Le Chatelier’s Principle, the
system will move to minimise the change, ie. move to the
side with least moles of gas.
Predict the outcome of the following
experiment: Compressing a syringe
containing an equilibrium mixture of
NO2/N2O4.
Slide 29-11
Changing volume
Volume reduced
Note:
NO2
N2O4
final colour
will be
DARKER
2 NO2(g) N2O4(g)
Slide 29-12
Changing temperature
Same reaction:
2 NO2(g) N2O4(g)
∆H = -58 kJ/mol
Is the reaction endothermic or exothermic?
What happens to the equilibrium if the
temperature is raised or lowered?
ANS: Treat the heat as a reactant:
2 NO2(g) N2O4(g) + 58 kJ/mol
Slide 29-13
Experiment
2 NO2(g) N2O4(g) + 58 kJ/mol
By Le Chatelier’s Principle, the reaction will move to minimise the change, i.e.
for an exothermic reaction such as this, when the temperature is raised the
reaction will move to the left to remove the heat.
Slide 29-14
Molecular visualisation:
N2O4 (g) + heat → 2 NO2 (g)
Slide 29-15
Changing temperature
Consider the following reaction in which ammonia is
synthesised from N2 and H2 at 500 K:
N2(g) + 3H2(g) 2 NH3(g)
∆H = -92 kJ/mol
Is the reaction endothermic or exothermic?
If we increase the temperature to 600K, what do
we expect to happen to the equilibrium?
What is happening to K when the temperature is
increased?
Slide 29-16
Changing temperature
ANS:
Write the reaction with the heat included, e.g.
N2(g) + 3H2(g) 2 NH3(g) + 92 kJ
By Le Chatelier’s Principle, the reaction will move to minimise
the change, i.e. for an exothermic reaction such as this,
when the temperature is raised the reaction will move to the
left to remove the heat.
Slide 29-17
Changing temperature
ANS:
Write the reaction with the heat included, e.g.
N2(g) + 3H2(g) 2 NH3(g) + 92 kJ
As temperature is increased, eq’m moves to left
As reaction moves to left (less
products), K must be getting smaller.
This is the only time where we have
made a change to eq’m that has
changed K.
Temp (K)
K (L2/mol2)
500
90
600
3
700
.3
800
.04
Slide 29-18
Summary
N2O4 (g)
2NO2 (g)
Change
Addition of N2O4(g)
Addition of NO2(g)
Removal of N2O4(g)
Removal of NO2(g)
Addition of He(g) at const. V
Decrease container volume
Increase container volume
Increase temperature
Decrease temperature
∆H = +58 kJ/mol
Shift
Right
Left
Left
Right
None
Left
Right
Right
Left
Slide 29-19
Plotting the change in concentrations
Let’s examine the change in concentration of all species involved when the eq’m
is perturbed. Again consider the ammonia synthesis reaction:
N2(g) + 3H2(g) 2 NH3(g) + 92 kJ
Add some H2(g) and follow the change in concentrations:
Note that the change
cannot be entirely
removed.
Slide 29-20
Adding a catalyst
A catalyst can greatly accelerate a reaction
without being consumed (e.g. Cl atoms in catalytic
ozone depletion). Catalysts do not appear in the
final overall reaction. They therefore CANNOT
affect the equilibrium position of a reaction…
only the rate.
(see Kinetics, next semester)
Slide 29-21
Catalysis
Uncatalyzed Reaction
energy
Ea
Catalyzed Reaction
reactant
Ea(cat)
∆ H°
product
Reaction Coordinate
Slide 29-22
Catalysis
Chemistry 4th Edition, Silberberg, Fig 16.22, p.706
Uncatalysed:
A + B → products (slow)
Catalysed:
A + catalyst → C
(fast)
C + B → products (faster)
+ catalyst
• Overall, the catalyst is not consumed.
Slide 29-23
Catalysis
Catalysts: increase rate coefficient by providing alternative
reaction path (or mechanism) with lower activation energy.
Catalyst does not change equilibrium, only rate of attaining
equilibrium.
Enormous practical importance (~ all modern chemical
products).
Examples:
- nitrogeneous fertilizers (N2 + 3H2 → 2NH3 ) – Pt catalyst
- removal of NO in vehicle exhaust – Pd oxide catalyst
- hydrogenating natural oils for margarine
- cracking petroleum → petrol, nylon, etc. – zeolite catalyst
Slide 29-24
Catalysis
Slide 29-25
Catalysis
Adapted from Chemistry 4th Edition, Silberberg, Fig B16.5, p.710:
Slide 29-26
Catalysis Demonstration
2H2O2 (aq) → 2H2O (l) + O2 (g)
Slide 29-27
Heterogeneous Catalysis
Chemistry 4th Edition, Silberberg, Fig 16.24, p.708:
Slide 29-28
The connection between K and ∆G
You must have noticed that we can predict the
direction of spontaneous chemical change using both Q
vs K and using ∆G
Same three examples as in L27:
N2(g) + O2(g) 2 NO(g)
Keq = 1x10-30
∆G0 = 170 kJ/mol
2CO(g) + O2(g) 2 CO2(g)
Keq = 2.2x1022 L/mol
∆G0 = -128 kJ/mol
N2O4(g) 2NO2(g)
Keq = 0.211 mol/L
∆G0 = -4 kJ/mol
Which side does equilibrium lie?
Slide 29-29
The connection between K and ∆G
What is the relationship between K and ∆G 0?
Same three examples as in L27:
N2(g) + O2(g) 2 NO(g)
Keq = 1x10-30
∆G0 = 170 kJ/mol
2CO(g) + O2(g) 2 CO2(g)
Keq = 2.2x1022 L/mol
∆G0 = -128 kJ/mol
N2O4(g) 2NO2(g)
Keq = 0.211 mol/L
∆G0 = -4 kJ/mol
Firstly notice the range of values!
This suggests a log relationship.
∆G0 = -RT ln Keq
Slide 29-30
Summary: Le Chatelier’s Principle
“If a change is imposed on a system at equilibrium, the
position of the equilibrium will shift in a direction that
tends to reduce that change.”
Increase reactant concentrations
⇒ more products produced
Increase pressure / decrease volume
Q⇒ K
⇒ less gas produced
Increase temperature
⇒ endothermic reaction favoured
Add catalyst
⇒ no change
K changes,
then Q ⇒ K
Q= K
Slide 29-31
Summary from Chemistry 4th Edition, Silberberg, Table
17.4, p.752:
Slide 29-32
Example questions
CONCEPTS
Concept of using Le Chatelier’s Principle
Predicting direction of chemical change
Predicting change in K with temperature.
CALCULATIONS
Work out final eq’m concentrations after
disturbing equilibrium
Slide 29-33