Chemical Equilibrium

What is Kc and Kp?

Kc = Equilibrium constant in terms of concentration

Kp = Equilibrium constant in terms of partial pressure

Write the expression of Kc

Liquid D is omitted as its concentration does not change.

Write the expression for Kp

Liquid D is omitted as it has no partial pressure

Fill in this table for the equation:

aA_(g) + bB_(g) ⇌ cC_(g) + dD_(l)

ΔH –ve

(a + b) > (c + d)

Sulfur dioxide reacts with oxygen to form sulfur trioxide in a reversible reaction.

2SO_2(g) + O_2(g) ⇌ 2SO_3(g)

Write an expression for the equilibrium constant, Kp, of this reaction.

2SO_2(g) + O_2(g) ⇌ 2SO_3(g)

Industrially, the conversion of sulfur dioxide and oxygen into sulfur trioxide is carried out at slightly above atmospheric pressure.

Suggest why this condition is chosen.

An increase in pressure favours the forward reaction (because the RHS has fewer gaseous moles) and speeds up the rate of reaction

A pressure only slightly above atmospheric pressure does not require too much energy to maintain and is not too hazardous.

A chemist investigated methods to improve the synthesis of sulfur trioxide from sulfur dioxide.

2SO_2(g) + O_2(g) ⇌ 2SO_3(g)

The chemist:

• mixed together 1.00mol SO₂ and 0.500mol O₂ with a catalyst at room temperature

• compressed the gas mixture to a volume of 250cm³

• allowed the mixture to reach equilibrium at constant temperature without changing the total gas volume.

At equilibrium, 82.0% of the SO₂ had been converted into SO₃

Determine the concentrations of SO₂, O₂ and SO₃ present at equilibrium and calculate K_c for this reaction, including its units

K_c = [SO₃]²/[SO₂]²[O₂]

K_c = [3.28]²/[0.70]²[0.36]

57.6 mol^-1dm³

Must have 3sf

Explain what would happen to the pressure as the system was allowed to reach equilibrium.

2SO_2(g) + O_2(g) ⇌ 2SO_3(g)

Pressure decreases as there are fewer moles of gaseous products

The chemist repeated the experiment at the same temperature with 1.00mol SO₂ and an excess of O₂.

The gas mixture was still compressed to a volume of 250cm³

State and explain, in terms of K_c, how the equilibrium yield of SO₃ would be different from the yield in the first experiment above.

K_c does not change (with pressure / concentration).

If [O₂] increases, the denominator of the K_c expression is greater, the yield of SO₃ (the numerator) must increase to keep K_c constant.

The chemist repeats the experiment three further times. In each experiment, the chemist makes one change, but uses the same initial amounts of SO₂ (1.00mol) and O₂ (0.500mol).

Complete the table to show the predicted effect of each change compared with the original experiment.

The reaction is exothermic

Increasing concentration effect on equilibrium

• If the reactant concentration is increased, the POE will shift to the right (towards the product) to use up more reactant

• If the production concentration is increased, the POE will shift to the left (towards the reactant)

Increasing the concentration of OH^- ions

The PO will shift to the right, giving a higher yield of I^- and IO^-

The colour would change from brown to colourless

Ethene to Ethanol

• Low temp gives good yield but slow rate

• High pressure gives good yield and high rate, but too high pressure leads to high energy costs for pumps to produce the pressure

• High pressure also leads to unwanted polymerisation of ethene to polyethene

Contact process

• Low temp gives good yield but slow rate

• High pressure gives slightly better yield and high rate, but too high pressure leads to high energy costs for pumps to produce the pressure

Haber process

• Low temp gives good yield but slow rate

• High pressure gives slightly better yield and high rate, but too high pressure leads to high energy costs for pumps to produce the pressure

Production of methanol from CO

• Low temp gives good yield but slow rate

• High pressure gives slightly better yield and high rate, but too high pressure leads to high energy costs for pumps to produce the pressure

Effects of catalysts on equilibrium

A catalyst has no effect on the position of equilibrium, but it will speed up the rate at which the equilibrium is achieved. It does not effect the position of equilibrium because it speeds up the rates of the forward and backward reactions by the same amount.

What does K_c only change with?

K_c only changes with temperature.

It does not change if pressure or concentration is altered. A catalyst has no effect on K_c.

Value of K_c changing

The larger the K_c, the greater the amount of products.

If K_c is smaller, the equilibrium favours the reactants

In a container of volume 600cm³ there were initially 0.500mol of H₂ and 0.600 mol of Cl₂ .

At equilibrium there were 0.200 mol of HCl.

Calculate Kc

• ICEE Table (Initial moles, Change in moles, Equilibrium moles, Equilibrium concentration)

Initially there were 1.5 moles of N₂ and 4 mole of H₂, in a 1.5 dm³ container.

At equilibrium 30% of the Nitrogen had reacted.

Calculate K_c

30% of the nitrogen had reacted

30% x 1.5 = 0.45 moles reacted.

3 x 0.45 moles of H₂ must have reacted and 2 x 0.45 moles of NH₃ must have formed

Moles of reactant at equilibrium = Initial moles - moles reacted

Moles of N₂ at equilibrium = 1.5 - 0.45 = 1.05

Moles of H₂ at equilibrium = 4.0 - (0.45 × 3) = 2.65

Moles of NH₃ at equilibrium = 0 + (0.45 × 2) = 0.9

What is the partial pressure of a gas?

The pressure that the gas would have, if it alone occupied the volume occupied by the whole mixture.

If a mixture of gases contains 3 different gases then the total pressure will equal the 3 partial pressure added together

Mole fraction

Mole fraction for a 3 part mixture

How to calculate the partial pressure of a gas

Calculate mole fraction of the gas (moles/total moles)

Then multiply the mole fraction by the total pressure provided (Mole fraction of gas 1 x total pressure) to calculate the partial pressure of that gas

A mixture contains 0.2 moles N₂, 0.5 moles O₂ and 1.2 moles of CO₂.

If the total pressure is 3atm, calculate the partial pressures of the 3 gases

• Calculate molar fraction

• Mole fraction x pressure

Units of K_p

atm

What’s the units of

1 mole of N₂ and 3 moles of H₂ are added together and the mixture is allowed to reach equilibrium.

At equilibrium 20% of the N₂ has reacted.

If the total pressure is 2atm what is the value of K_p?

• ICEMP Table

  Initial moles

  Change in moles

  Equilibrium moles

  Molar fraction

  Partial pressure

Answer = 0.0469 atm²

Rules for writing K_p expression

K_p expressions only contain gaseous substances. If the equation has a substance with another state, it is not included in the expression.