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Equilibrium Constants

Key Concepts

Equilibrium Constants (K):

are derived from experimental data
are temperature dependent
provide a measure of the equilibrium position

if K is large (> 10²) products of the reaction are favoured
if K is small, reactants are favoured.

remain the same value for a constant temperature even if the equilibrium concentrations of the reactants or products are altered
do NOT change in the presence of a catalyst (a catalyst changes the time taken to reach equilibrium but does not alter the equilibrium position or equilibrium constant)
For the reaction: aA + bB cC + dD
where A, B, C and D are the chemical species and
a, b, c and d are the stoichiometric coefficients,

K₁ = [C]^c[D]^d

[A]^a[B]^b
For the same reaction written in reverse: cC + dD aA + bB
where A, B, C and D are the chemical species and
a, b, c and d are the stoichiometric coefficients,

K₂ = [A]^a[B]^b

[C]^c[D]^d

K₂ = 1/K₁ and K₁ = 1/K₂
If the concentration of species in the reaction is used to calculate the equilibrium constant then it can be given the symbol K_c. If the equilibrium constant has no subscript, that is it is given as K, this refers to K_c and the concentration of each species in the reaction is used to calculate the equilibrium expression.
If a solvent takes part in a reaction, its concentration is said to remain constant and is incorporated into the value of K.
If a solid is present in a reaction, its concentration is said to remain constant and is incorporated into the value of K.

Writing Equilibrium Expressions

Reactions Involving Aqueous Species

All species are present in aqueous solution eg:
Ag⁺_(aq) +2NH₃_(aq) Ag(NH₃)₂⁺_(aq)

K = [Ag(NH₃)₂⁺]

[Ag⁺][NH₃]²
Solvent takes part in reaction, eg:
NH_3(aq) + H₂O_(l) NH₄⁺_(aq) + OH^-_(aq)

K = [NH₄⁺][OH^-]

[NH₃]

The concentration of the water as solvent is said to be constant and is incorporated into the value of K.
A solid is present in the reaction, eg
Pb(NO₃)_2(aq) + 2KI_(aq) PbI_2(s)

K = 1

[Pb(NO₃)₂][KI]²

The concentration of the solid is said to be constant and is incorporated into the value of K.

Reactions Involving Gases

All species are present as gases, eg:
CO_(g) + 2H_2(g) CH₃OH_(g)

K = [CH₃OH]

[CO][H₂]²
A solid is present in the reaction, eg:
CaCO_3(s) CaO_(s) + CO_2(g)
K = [CO₂]
The concentration of solids are said to be constant and are incorporated into the value of K.

Some Special Special Cases

Calculating Equilibrium Constant : Concentration of All Species Known

For the reaction COCl_2(g)

CO_(g) + Cl_2(g) at 900^oC at equilibrium,
the concentration of COCl_2(g) is 3.0 x 10^-6M while that of CO_(g) and Cl_2(g) is 4.97 x 10^-4M.
Calculate the value of K.

Write the equilibrium expression using the balanced chemical equation:

K = [CO][Cl₂]

[COCl₂]
Extract all the relevant data from the question:
[CO] = 4.97 x 10^-4M
[Cl₂] = 4.97 x 10^-4M
[COCl₂] = 3.0 x 10^-6M
K = ?
Substitute the values into the equilibrium expression and solve:

K = [4.97 x 10^-4][4.97 x 10^-4]

[3.0 x 10^-6]

K = 0.082

Calculating Equilibrium Constant : Initial Concentrations and Equilibrium Concentration of One Species Known

A mixture of 5.0 mol H_2(g) and 10.0 mol I_2(g) are placed in a 5L container at 450^oC and allowed to come to equilibrium. At equilibrium the concentration of HI_(g) is 1.87M.
Calculate the value for K for this reaction.

Write the balanced chemical equation for the reaction: H_2(g) + I_2(g) 2HI_(g)
Write the equilibrium expression for the reaction:

K = [HI]²

[H₂][I₂]

Calculate the equilibrium concentrations for all reactants and products:

	[H₂]	[I₂]	[HI]
initial concentration M = moles/Volume	5.0mol/5L = 1.0M	10.0/5L = 2.0M	0M
equilibrium concentration expression Let x=amount used in reaction	1.0 - x	2.0 - x	1.87M
calculate x using mole ratio from equation	x = ½ x 1.87 x = 0.935	x = ½ x 1.87 x = 0.935	1.87M
calculate equilibrium concentrations	1.0 - 0.935 = 0.065M	2.0 - 0.935 = 1.065M	1.87M

Substitute the equilibrium concentrations into the equation and solve for K:

K = [1.87]²

[0.065][1.065]

K = 50.5

Calculating Equilibrium Constant : Initial Reactant Concentration and %Dissociation Known

4.0 moles of NOCl_(g) were placed in a 2L vessel at 460^oC. The following dissociation reaction occurred:
2NOCl_(g)

2NO_(g) + Cl_2(g)
At equilibrium the NOCl_(g) was 33% dissociated. Calculate the equilibrium constant for this reaction.

Write the equilibrium expression for the reaction:

K = [NO]²[Cl₂]

[NOCl]²

Calculate the equilibrium concentrations for all reactants and products:

	[NOCl]	[NO]	[Cl₂]
initial concentration M = moles/Volume	4.0mol/2L = 2.0M	0M	0M
equilibrium concentration expression Let x=amount used in reaction	2.0 - 2x	2x	x
calculate x using 33% dissociation of NOCl	2x = 33/100 x 2.0 2x = 0.66M
and using mole ratio from equation		2x = 0.66M	x = ½ x 0.66 x = 0.33M
calculate equilibrium concentrations	2.0 - 0.66 = 1.34M	0.66M	0.33M