Summary Complete Equilibrium II Revision Notes (A Level Edexcel)

EQUILIBRIUM II
1. Be able to deduce an expression for Kp, for homogenous and
heterogeneous systems, in terms of equilibrium partial pressures in
atm.
Equilibria:
- Dynamic equilibrium: the rate of the forward reaction is equal to the rate of
the backward reaction. The concentrations of the reactants and products
stays constant.
- According to Le Chatelier’s Principle, a system in dynamic equilibrium reacts
to oppose changes.
o The equilibrium position will shift.
- The equilibrium can be reached from either the reactants or the products.
Equilibrium expressions for Kc:

- For a reaction aA + bB ⇌ cC + dD, we can write the equilibrium expression
as:
a b
[A] [B]
o Kc =
[C]c [D] d
o This equilibrium expression comes directly from the equation, not from
experimental data (as with kinetics).
- A high Kc (Kc > 1) shows that the equilibrium lies to the right hand side.
- A low Kc (Kc < 1) shows that the equilibrium lies to the left hand side.
- When Kc = 1, the equilibrium is in the middle and equal concentrations of
reactants and products are being made.
Homogenous vs heterogeneous equilibria:
- We exclude solids and separate liquid phases from the expression.
o Solids have a constant concentration.
o The solvent (normally water) is usually in a separate liquid phase.
Substances in a separate liquid phase are in such large excess that
their concentrations can usually be approximated as constant.
- Homogeneous – all the reactants and products are in the same phase.
- Heterogeneous – the reactants and productions are not all in the same phase.
- For the reaction Ag+ (aq) + Fe2+ (aq) ⇌ Ag (s) + Fe3+ (aq), we write the
equilibrium expression as:
3+
[ Fe ]
o Kc = =
[Ag+ ][Fe 2+ ]
o We exclude Ag (s) as it is solid.

Equilibrium expressions for Kp:
- It is often easier to measure the pressure of a gas rather than its
concentration.
- The equilibrium constant expressed in terms of pressure is given the symbol
Kp.
- In Kp expressions, we only ever include gases (not solids, liquids or solutions).
- For the reaction CH4 (g) + H2O (g) ⇌ CO (g) + 3 H2 (g), we write the
equilibrium expression as:

, 3
(p CO ) ( pH )
o Kp = 2

(p CH )( p H O )
4 2

- When calculating Kp values, we use the partial pressures of each gaseous
substance (in atm) rather than their concentrations (as with K c).
- Partial pressure = the pressure a gas wold exert if it were alone in the volume
occupied.
o Each gas exerts a partial pressure that is proportional to the number of
moles of the gas in the mixture (pV = nRT means that, assuming
constant temperature and volume, p α n).
o Partial pressure = molar fraction x total pressure
o PA = X A x P
o The total pressure is the sum of the partial pressures of all the gases in
the mixture.
- Molar fraction = the number of moles of a gaseous substance relative to the
total number of moles of all the gaseous substances.
number of moles of the gas
o Molar fraction =
total number of moles of gas in the mixture


2. Be able to calculate a value, with units where appropriate, for the
equilibrium constant (Kc and Kp) for homogeneous and
heterogeneous reactions, from experimental data.
Calculating Kc values:
- There are three types of possible exam calculation.
o 1. You are given the concentrations of all reactants and products at
equilibrium.
o 2. You are given the moles of all reactants and products at equilibrium.
o 3. You are given the initial moles of all reactants and products, and the
concentration of one reactant or product at equilibrium.
- You must give both the numerical value and the units of K c.
o Kc will be in factors of mol dm-3 and can be worked out based on the
expression.
1. You are given the concentrations of all reactants and products at equilibrium.
- Method:
o Plug in the concentrations into the equilibrium expression to get K c.
- Example:
o CH3COOH (l) + C2H5OH (l) ⇌ H2O (l) + CH3COOC2H5 (l)
o In a system at 298 K, at equilibrium, we have 0.1 mol dm -3 CH3COOH,
0.1 mol dm-3 C2H5OH, 0.5 mol dm-3 H2O and 0.5 mol dm-3 CH3COOC2H5.
[ CH 3 COOC 2 H5 ] [ H2 O]
o Kc =
[ CH3 COOH ] [ C2 H5 OH]
o We have to include H2O in the equilibrium expression here, as it is not
in a separate liquid phase. The reactants and products are all pure
liquids, so H2O is not acting as a solvent.