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 25-1
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] Slide 25-2
Highlights of last lecture
Oxides of nitrogen
N2O, NO, NO2, NO3 + dimers: N2O2, N2O3, N2O4, N2O5
Role of nitrogen oxides in the atmosphere
NOx
HNO3 (acid rain)
PAN
Greenhouse effect
Energy balance
Radiation trapping by atmospheric gases
Relationship between increasing gases and temperature
Slide 25-3
Revision…
Is ∆S positive or negative in these chemical reactions?
dimerisation of NO2
2 NO2 (g) → N2O4 (g)
Entropy decreases (∆S < 0)
Energy confined in fewer
molecules
Ice melting then water evaporating
(all at 25°C)
H2O (s) → H2O (l) → H2O (g)
Entropy increases (∆S > 0)
Energy spread more widely
Slide 25-4
Temp dependence of ∆G
∆G = ∆H – T∆S
Notice that ∆G is temperature dependent
∆H < 0
∆H > 0
∆H < 0
∆H > 0
AND
AND
AND
AND
∆S > 0
∆S < 0
∆S < 0
∆S > 0
Spontaneous?
⇒
⇒
⇒
⇒
always
never
spontaneous at low T
spontaneous at high T
∆H < 0
∆H > 0
∆S > 0
always
at high T
∆S < 0
at low T
never
Slide 25-5
An example
A reaction that changes direction with temperature:
Na+(aq) + CH3COO-(aq) → CH3COONa(s)
What is the sign of ∆S?
negative
What is the sign of ∆H?
negative
Spontaneous?
∆H < 0
∆H > 0
∆S > 0
always
at high T
∆S < 0
at low T
never
Is the forward reaction spontaneous?
YES: At low T
At high T the reverse reaction is
spontaneous (∆G = ∆H – T∆S)
Slide 25-6
∆G = 0
When ∆G < 0
When ∆G > 0
⇒ spontaneous
⇒ non-spontaneous
What happens when ∆G = 0?
System is AT EQUILIBRIUM !
New concept ⇒ new topic
Slide 25-7
Equilibrium
Themes:
controlling equilibrium
atmospheric equilibria
commercial production of
chemicals
Key thematic concepts:
• Most chemical reactions exist in a state of equilibrium;
• The equilibrium state can be calculated and controlled;
• Industry and biology both take advantage of equilibria;
Slide 25-8
Equilibrium
Key chemical concepts:
•
Equilibrium as a dynamic process;
•
Equilibrium constants;
•
Direction of chemical change;
•
Disturbing equilibrium: Le Chatelier’s principle;
•
Connection between ∆G and equilibrium
Calculations:
•
Calculating equilibrium position;
•
Calculating direction of chemical change;
•
Calculating new equilibrium after change;
•
Calculations of equilibrium constant and free energy
Slide 25-9
Equilibrium
Many atmospheric species exist in a state of
“equilibrium”, where a reverse reaction competes with
the forward reaction.
e.g. smog reactions that proceeded both L-R and R-L
1.
2 NO2 N2O4
H
2.
H
C
H
O
C
O
O
•
O
N
O
+ •N
O
O
H
H
C
H
O
O
PAN = peroxy acetyl nitrate
C
O
Many others, e.g. in ozone depletion:
3.
2 NO + Cl2 2 NOCl
4.
Cl + O2 ClO2
Theme of this lecture: Exploring the “equilibrium
state” using atmospheric reactions as examples.
Slide 25-10
NO2 dimerization
Start with N2O4(g)…
N2O4 (colourless)
NO2 (brown)
Silberberg, p.715
N2O4 (g)
2 NO2 (g)
No further change…
Slide 25-11
NO2 dimerization
No further
change…
Slide 25-12
Dynamic Equilibrium
• Equilibrium conc’ns do not depend on starting from
reactant side or product side.
concentration
2 NO2 (g)
NO2
N2O4
time
concentration
N2O4 (g)
Equilibrium
reached
NO2
N2O4
time
Slide 25-13
Equilibrium
• Equilibrium conc’ns do not depend on starting from
reactant side or product side.
org
I2
aq
I-
I2 (org. phase) + I– (aq)
org
I3-
aq
I3– (aq)
⇒ Colour intensities of both layers are the
same, no matter how they were prepared.
Slide 25-14
Kinetics and equilibrium
N2O4 (g)
2 NO2 (g)
Rate of forward
reaction
=
Rate of reverse
reaction
[reactant] and [products] are constant…
This DOES NOT mean the reaction has stopped!
Slide 25-15
NO2 dimerization
N2O4 (g)
2 NO2 (g)
Three people did an experiment on the equilibrium of this
reaction…
Expt
[NO2]
(init.)
[N2O4]
(init.)
[NO2]
(final)
[N2O4]
(final)
1
0
0.1000
0.1018
0.0491
2
0.1000
0
0.0627
0.0185
3
0.0500
0.0500
0.0837
0.0332
Who is right?
Or… what’s the point?
Slide 25-16
Equilibrium constant
The fundamental observation of chemical
equilibrium was first stated in 1864 by two
Norwegian chemists, Guldberg and Waage:
“At a given temperature, a chemical
system reaches a state in which a
particular ratio of reactant and product
concentrations has a constant value.”
A+B C+D
Keq =
[C] x [D] Π [products]
=
[A] x [B] Π [reactants]
Π = “product of ”, like Σ = “sum of ”
Slide 25-17
Equilibrium constant
N2O4 2NO2
N2O4 NO2 + NO2
Keq =
[NO2] x [NO2]
[N2O4]
In general,
Keq =
=
[NO2]2
[N2O4]
aA + bB cC + dD;
[D]d x [C]c
[A]a x [B]b
Slide 25-18
NO2 dimerization
N2O4 (g)
2 NO2 (g)
Three people did an experiment on the equilibrium of this
reaction…
Expt
[NO2]
(init.)
[N2O4]
(init.)
[NO2]
(final)
[N2O4]
(final)
Keq
1
0
0.1000
0.1018
0.0491
0.211 mol/L
2
0.1000
0
0.0627
0.0185
0.212 mol/L
3
0.0500
0.0500
0.0837
0.0332
0.211 mol/L
So… they were all correct, and the
point is that the eq’m position in ALL
cases is given by:
[NO]2
[N2O4]
= 0.211 mol/L
Slide 25-19
Position of eq’m and K
Three examples:
N2(g) + O2(g) 2 NO(g)
Keq = 1x10-30
2CO(g) + O2(g) 2 CO2(g)
Keq = 2.2x1022 L/mol
N2O4(g) 2NO2(g)
Keq = 0.211 mol/L
Q: Which compounds are favoured at equilibrium?
Hint: remember that K = Π[prod.] / Π[react.]
Small Keq: Equilibrium favours reactants
Large Keq: Equilibrium favours products
Medium Keq: reactants and products are
present at equilibrium
Slide 25-20
Approaching Equilibrium
The reaction quotient, Q, is calculated in exactly the same
way as K, except using current concentrations rather than
eq’m concentrations.
Keq = 4.34 L/mol
e.g. 2SO2(g) + O2(g) 2SO3(g)
I put the following concentrations into a flask:
Compound
Concentration (M)
Expt 1
Concentration (M)
Expt 2
SO2
2.00
0.500
O2
1.50
0.100
SO3
3.00
0.350
Is the reaction at eq’m? If not, which way will it move to
achieve eq’m?
Slide 25-21
Approaching Equilibrium
Keq = 4.34 L/mol
e.g. 2SO2(g) + O2(g) 2SO3(g)
Compound
Concentration (M)
Expt 1
Concentration (M)
Expt 2
SO2
2.00
0.500
O2
1.50
0.100
SO3
3.00
0.350
Expt 1:
Q=
[SO3]2
[SO2]2 x [O2]
=
(3.00)2
(2.00)2 x (1.50)
= 1.5 L/mol
Therefore the system is NOT at eq’m
Q < K therefore not enough products and reaction proceeds
to right
Slide 25-22
The Reaction Quotient
Keq = 4.34 L/mol
e.g. 2SO2(g) + O2(g) 2SO3(g)
Compound
Concentration (M)
Expt 1
Concentration (M)
Expt 2
SO2
2.00
0.500
O2
1.50
0.100
SO3
3.00
0.350
Expt 2:
class problem…
Slide 25-23
Example questions
CONCEPTS
The equilibrium state
The dynamic nature of eq’m (kinetic picture)
The relationship between ∆H and eq’m
How to write Keq for any reaction
The reaction quotient
CALCULATIONS
Work out Keq from eq’m concentrations
Work out Q from any set of concentrations
Work out whether a system is at eq’m
Work out the direction of chemical change.
Slide 25-24