When a reaction is at equilibrium, the concentrations of the reactants and products do not change because the rates of forwards and reverse reactions are equal. This is known as dynamic equilibrium.
Homogeneous: products and reactants are in the same phase
Heterogeneous: products and reactants not in the same phase.
At equilibrium, the ratio of concenrations between products and reactants stays constant and can be calulated.
The equilibrium constant, Kc
Kc is caluclated by the concentration of products divided by the concentration of reactants.
Kc = [products] / [reactants] note: this is for an equilibrium reaction
e.g. for the reaction C2H5OH + CH3CO2H ⇌ CH3CO2C2H5 + H2O
ethanol + ethanoic acid ⇌ ethyl ethanoate + water
Kc = [CH3CO2C2H5] [H2O] / [C2H5OH] [CH3CO2H]
Kc should be quoted to a temperature. The units of Kc also vary. For this reaction, Kc has no units. moldm-3 moldm-3 / moldm-3 mold-3 The units cancel out.
Units of Kc
2A + B ⇌ 2C Kc = [C]² / [A]² [B] Units: M M / M M M note: M stands for mold-3
Units: 1 / moldm-3 or mold-1dm3
D ⇌ 2E Kc = [E]² / [D] Units: M M / M or moldm-3
F ⇌ G + H + I Kc = [G] [H] [I] / [F] Units: M M M / M or mol2dm-6
2X + Y ⇌ Z Kc = [Z] / [X]² [Y] Units: M / M M M or 1 / moldm-3 moldm-3
Take care when calcaluting Kc's units. Remember to cancel out your moldm-3 's as the units could be anything. ANYTHING!
Calculating Kc with mol at start
C2H5OH + CH3CO2H ⇌ CH3CO2C2H5 + H2O
mol at start: 0.1 0.1 0 0
mol at eqm: 0.033 ? ? ? to calculate the mols at eqm you must look at the change
change: -0.067 -0.067 +0.067 +0.067
as the stoichometery is 1:1, if ethanol decreases by 0.067, so shall ethanoic acid and so shall the products be increased by 0.067.
mol at eqm: 0.033 0.033 0.067 0.067
We can now calculate concentration using M = n x 1000 / v as you will be given a volume for a mixture to use. If the volumes all cancel out you can calulate Kc using mol, like you can here.
Kc = (0.067 / v) x (0.067 / v) / (0.033 / v) x (0.033 / v) = 4.1 no units
When stoichometry is not 1:1
A + 2B ⇌ C + 2D
mol at start: 1 1 0 0
mol at eqm: 0.6 ? ? ?
as the stoichometry is not 1:1, calculating mol at eqm is a bit moe confusing.
change: -0.4 - 2 x 0.4 +0.4 + 2 x 0.4
mol at eqm: 0.6 0.2 0.4 0.8
As there are twice as many moles of B than A, that means B changes by twice as much as A.
As there is the same amount of moles of A and C, that means C increases by the same amount A decreases.
As there are twice as many moles of D than A, that means D increases twice as much as A decreases.
Changing temperature on a eqm
For an equilibrium with a forwards endothermic reaction (takes in heat energy in order to react), if the temperature is decreased this causes the equilibrium to shift to the left in order to reverse the change (as the reverse reaction is exothermic and will release heat energy).
Kc will decrease because if a reaction moves to the left, more reactants are made than products so [reactants] will be greater than [products]. Remember, Kc = [products] / [reactants]
Kc will increase for a equilibrium shifting to the right as more products are made than reactants.
Type of reaction Temp change Effect on Kc Products Reactants Direction of eqm
endothermic decrease decrease decrease increase to the left
endothermic increase increase increase decrease to the right
exothermic decrease increase increase decrease to the left
exothermic increase decrease decrease increase to the right
Changing concentration on a eqm
Kc does not change unless temperature changes. Why Kc is quoted to a temperature,
Le Chatelier's principle: When a change occurs to a system at equilibrium, the equilibrium will move in the direction that opposes the change to minimise it.
A + B ⇌ C + D Kc = [C] [D] / [A] [B]
If more C is added to the equilbrium you would expect Kc to increase as [products] would increase. HOWEVER, the equilibrium will shift to the left to produce more reactants to reverse the change. So [reactants] will also increase canceling out the increase in [products] so Kc remains unchanged.
Kc and the position of eqm
The size of Kc can tell us in which direction the eqm is over.
Kc = [products] / [reactants]
If Kc is over 1, that means [products] predominate over [reactants] and the equilibrium is over to the right.
If Kc is under 1, that means [reactants] predominates over [products] so the equilbirum is over to the left.
When the equilibrium constant is over 10 to the power of 10, that means the reaction is close to completion.
When Kc us under 10 to the power of -10, thiss means the reaction is next to not taking place at all.
Catalysts and Kc
Remember! Only changing temperature at eqm has an effect on Kc.
Catalysts do NOT effect Kc or the position of equilibrium as catalysts effect the rates of the forward and negative reactions equally. Finding an alternative pathway for both reactions with a lower activation energy, increasing both rates of reactions equally.
Catalysts do increase the rate that equilibrium is attained. Which is why catalysts are used in industry when working with equilbirums.