Chemistry
Topic 4: Equilibrium
Dynamic equilibrium: A reversible reaction in which the rate of the forward reaction is equal to
the rate of the backward reaction and the amount of the reactants and products remains constant.
The condition for a dynamic equilibrium was that the reaction must take place in a closed system
and the result was that the macroscopic properties remain constant.
N2 + 3H2 ⇌ 2NH3
ΔH = -98 kJmol-1
When the temperature is increased, the reverse endothermic reaction is favoured (equilibrium
position shifts to the left) therefore using up the added heat. Therefore, when the temperature is
increase [N2] and [H2] increase while [NH3] decreases.
A low temperature favours the forward exothermic reaction, but since this would mean a slow
reaction, a moderate temperature of about 450 oC is used along with a catalyst at 250atm. Even
under these conditions the proportion of NH 3 present in the dynamic equilibrium is approximately
10%.
KC Expression
2
[ NH 3 ]
K C= 3
[ N2] [ H2]
Kc is the equilibrium constant
[NH3] is the concentration of NH3
[N2] is the concentration of N2
[H2] is the concentration of H2
For the Haber process the units of Kc are
Kc = (moldm-3)2/(moldm-3) (moldm-3)3
= mol-2dm6
The numerical size of the equilibrium gives an indication of the position of the equilibrium. If the
equilibrium constant is greater than the original constant, then there is a greater number of
products as the equilibrium moves right.
Equilibrium Calculations
Example 1
0.1 moles of ethanol and 0.2 moles of ethanoic acid are mixed in a total volume of 1dm 3. At
equilibrium there were 0.117 moles of ethanoic acid present. Calculate the value of K c and its units.
CH3CO2H + C2H5OH ⇌ CH3CO2C2H5 + H2O
Initial Moles 0.2 0.1 0 0
Reacting Moles 0.2 – 0.117 = -0.083 +0.083 +0.083
0.083
Equilibrium Moles 0.117 0.017 0.083 0.083
Topic 4: Equilibrium
Dynamic equilibrium: A reversible reaction in which the rate of the forward reaction is equal to
the rate of the backward reaction and the amount of the reactants and products remains constant.
The condition for a dynamic equilibrium was that the reaction must take place in a closed system
and the result was that the macroscopic properties remain constant.
N2 + 3H2 ⇌ 2NH3
ΔH = -98 kJmol-1
When the temperature is increased, the reverse endothermic reaction is favoured (equilibrium
position shifts to the left) therefore using up the added heat. Therefore, when the temperature is
increase [N2] and [H2] increase while [NH3] decreases.
A low temperature favours the forward exothermic reaction, but since this would mean a slow
reaction, a moderate temperature of about 450 oC is used along with a catalyst at 250atm. Even
under these conditions the proportion of NH 3 present in the dynamic equilibrium is approximately
10%.
KC Expression
2
[ NH 3 ]
K C= 3
[ N2] [ H2]
Kc is the equilibrium constant
[NH3] is the concentration of NH3
[N2] is the concentration of N2
[H2] is the concentration of H2
For the Haber process the units of Kc are
Kc = (moldm-3)2/(moldm-3) (moldm-3)3
= mol-2dm6
The numerical size of the equilibrium gives an indication of the position of the equilibrium. If the
equilibrium constant is greater than the original constant, then there is a greater number of
products as the equilibrium moves right.
Equilibrium Calculations
Example 1
0.1 moles of ethanol and 0.2 moles of ethanoic acid are mixed in a total volume of 1dm 3. At
equilibrium there were 0.117 moles of ethanoic acid present. Calculate the value of K c and its units.
CH3CO2H + C2H5OH ⇌ CH3CO2C2H5 + H2O
Initial Moles 0.2 0.1 0 0
Reacting Moles 0.2 – 0.117 = -0.083 +0.083 +0.083
0.083
Equilibrium Moles 0.117 0.017 0.083 0.083
