3.1.10 Equilibrium constant Kp for homogeneous systems

Define partial pressure. (2 marks)

• The pressure that one gas in a mixture would exert if it alone occupied the whole container.

• The partial pressures of all the gases present add up to the total pressure.

State how a mole fraction is calculated. (2 marks)

• Mole fraction of a gas = moles of that gas ÷ total moles of all the gases.

• The mole fractions of all the gases add up to 1.

State how a partial pressure is calculated. (2 marks)

• Partial pressure = mole fraction × total pressure.

• It has the same units as the total pressure, usually Pa or kPa.

Describe how an expression for Kp is constructed. (3 marks)

• Kp = the partial pressures of the products multiplied together, divided by those of the reactants.

• Each partial pressure is raised to the power of its balancing number in the equation.

• Only gases appear in a Kp expression, so solids and liquids are left out.

Describe how the units of Kp are worked out. (3 marks)

1. Substitute the pressure unit, such as kPa, in place of each partial pressure, keeping the powers.

2. Cancel the units that appear on both the top and the bottom.

3. If all the units cancel, Kp has none, which happens when the total moles of gas are equal on each side.

State the only factor that changes the value of Kp. (1 mark)

• A change in temperature.

State the effect of increasing temperature on Kp. (2 marks)

• For an exothermic forward reaction, raising the temperature decreases Kp.

• For an endothermic forward reaction, raising the temperature increases Kp.

Explain why changing the pressure does not change the value of Kp. (3 marks)

• Raising the total pressure raises every partial pressure, so the expression is momentarily away from Kp.

• The position of equilibrium shifts towards the side with fewer moles of gas.

• The partial pressures adjust until the expression equals the original Kp again.

State the effect of a catalyst on Kp. (2 marks)

• It has no effect on the value of Kp or on the position of equilibrium.

• It only increases the rate at which equilibrium is reached, by speeding up both directions equally.

Describe how Kp is calculated from experimental data. (5 marks)

1. Write the balanced equation and the expression for Kp.

2. Work out the equilibrium moles of each gas, using the molar ratios.

3. Add them to get the total moles, and find the mole fraction of each.

4. Multiply each mole fraction by the total pressure to give its partial pressure.

5. Substitute into the expression and work out the units.

Example: at equilibrium a vessel at 200 kPa holds 0.20 mol N2, 0.60 mol H2 and 0.20 mol NH3. Calculate Kp for N2 + 3H2 ⇌ 2NH3. (5 marks)

1. Total moles = 0.20 + 0.60 + 0.20 = 1.00, so the mole fractions are 0.20, 0.60 and 0.20.

2. Partial pressures: p(N₂) = 40 kPa, p(H₂) = 120 kPa, p(NH₃) = 40 kPa.

3. Kp = p(NH₃)² ÷ (p(N₂) × p(H₂)³).

4. Kp = 40² ÷ (40 × 120³) = 1600 ÷ (40 × 1 728 000).

5. Kp = 2.3 × 10⁻⁵ kPa⁻².

Explain what can be deduced if increasing the pressure decreases the equilibrium yield of the products. (2 marks)

• The equilibrium has shifted towards the reactants, which must be the side with fewer moles of gas.

• There are therefore more moles of gas on the product side than on the reactant side.