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.
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.
