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Chemical Equilibrium
Advanced Higher Chemistry
Unit 2b
Revision
• Dynamic
• But to observe you would see no change
• Rate of forward reaction = rate of reverse reaction
• Concentrations of reactants and products are
constant, but not (necessarily) equal
2Hl(g)
H2(g) + I2(g)
To which side
does the
equilibrium lie?
How would the
graph differ if you
started with H2
and I2?
Calculating compositions
• To determine equilibrium composition of a mixture, you
need equilibrium amount of only ONE of the species
• Example:
CO(g) + 3H2(g)
CH4(g) + H2O(g)
You place 1.000 mol CO and 3.000 mol H2 in a reaction
vessel at 1200 K and allow the reaction to come to
equilibrium. The mixture is found to contain 0.387 mol
H2O. What is the molar composition of the equilibrium
mixture?
Calculation strategy
• You have starting amounts
• These will change over time
• They will remain constant at equilibrium
Amount (mol)
Starting
Change
Equilibrium
CO(g) + 3H2(g)
CH4(g) + H2O(g)
CO
3H2
CH4
H2O
1.000
-x
1.000-x
3.000
-3x
3.000-3x
0
+x
x
0
+x
x = 0.387
Therefore: 0.613 mol CO, 1.839 mol H2, 0.387 mol CH4, 0.387 mol H2O
Practice
1. You place 1.50 mol of dinitrogen trioxide into a flask,
where it decomposes at 25ºC and 1.00 atm:
N2O3(g)
NO2(g) + NO(g)
What is the composition of the reaction mixture at
equilibrium if it contains 0.45 mol of nitrogen dioxide?
2. Nitric oxide, NO, reacts with bromine to give nitrosyl
bromide, NOBr
2NO(g) + Br2(g)
2NOBr(g)
A sample of 0.0655 mol NO with 0.0328 mol Br2 gives an
equilibrium mixture containing 0.0389 mol NOBr. What is
the composition of the equilibrium mixture?
The Equilibrium Constant, Kc
• For the reaction:
aA + bB
Kc =
cC + dD
[C]c[D]d
[A]a[B]b
• The value of the equilibrium constant is
constant for a particular reaction at a given
temperature, whatever the initial starting
concentrations
Practice
Write the equilibrium constant expression
for:
1. The synthesis of ammonia by the Haber
process
2. The catalytic hydration of methane
Obtaining values for Kc
• Consider the reaction:
CO(g) + 3H2(g)
CH4(g) + H2O(g)
The equilibrium composition is:
0.613 mol CO, 1.839 mol H2, 0.387 mol CH4, 0.387 mol H2O
The volume of the reaction vessel is 10.00 l.
So the equilibrium concentration of CO is 0.613/10 = 0.0613
What is the value of the equilibrium constant?
Practice
• Suppose for the same reaction the
equilibrium compositions in a 10 l vessel
are:
1.522 mol CO, 1.566 mol H2, 0.478 mol CH4, 0.478 mol H2O
What is the vale of Kc?
Homo- / Heterogeneous Equilibria
• Consider the reaction:
3Fe(s) + 4H2O(g)
Fe3O4(s) + 4H2(g)
• Is this hetero- or homogeneous?
• For heterogeneous equilibria, pure solids and liquids are
omitted from the Kc
• Write an expression for Kc for the above reaction
• This means that equilibrium is not affected by amounts
of these substances
The Equilibrium Constant, Kp
• For the reaction:
CO(g) + 3H2(g)
CH4(g) + H2O(g)
• The equilibrium constant can be expressed in
terms of partial pressures:
Kp =
PCH4PH2O
PCOPH23
In general, for a particular reaction the value of Kp will be different from Kc
Interpretation of K
• Does a large K mean that reactants or products are
favoured?
• CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
• Kc = 1 x 10140 at 25ºC
• N2(g) + O2(g)
2NO(g)
• Kc = 4.6 x 10-31 at 25ºC
• If K is around 1, neither reactants or products are
favoured
Changing the reaction conditions
• Conditions can be changed to maximise
product:
– Concentration: adding reactants, removing
products
– Pressure: change the volume of the vessel
– Temperture
• NOT a catalyst
Concentration change
• Example:
CO(g) + 3H2(g)
CH4(g) + H2O(g)
• If cooled, water can be condensed and removed
CO
0.613
Moles
H2
CH4
1.839 0.387
(before equilibrium readjusts)
0.613
1.839
0.387
0
Equilibrium re-established
0.491
1.473
0.509
0.122
Stage of process
Original mixture
After removing water
H2O
0.387
Assuming a 1 l vessel, what is Kc for the original equilibrium and for
the re-established equilibrium?
Pressure change
• How can you change the pressure of a system?
• What is meant by partial pressure?
Pressure change
Changing the volume of a reactant container changes
the concentration of gaseous reactants and therefore
their partial pressures
Equilibrium position will therefore move
The value of Kc or Kp does NOT change
Changing pressure by adding more of an inert gas has
no effect of the equilibrium position
- No effect on partial pressures
Temperature change
• Like concentration and pressure, temperature
affects the position of equilibrium
• Unlike concentration and pressure, temperature,
temperature does affect the size of the
equilibrium constant
• Consider the following data:
Temperature (K)
Kc
298
800
1000
1200
4.9 x 1027
1.38 x 105
2.54 x 102
3.92
Is this reaction exothermic or endothermic?
Summary
• K is temperature dependant, but is
unaffected by changes in concentrations
or partial pressures
Change
Concentration
Pressure
Temperature
Catalyst
Equilibrium
position
Changes
Changes
Changes
No change
Value of K
No change
No change
Changes
No change