OxfordAQA A-Level Chemistry: Equilibrium constant Kp for homogeneous systems (9620)

This study guide teaches section 3.1.12 Equilibrium constant Kp for homogeneous systems of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards. Its single subsection is 3.1.12.1. Because the specification lists Kp under International A2 physical chemistry, every outcome here is International A-level only. Unit 4 (Organic 2 and Physical 2) examines it, and Unit 5 (Practical and synoptic) may test it too, since that unit covers the whole specification.

Use it with the Kp revision notes and the Kp practice questions. The rest of the course is on the 9620 chemistry hub; to tick off outcomes, print the course checklist; and for a quick gap check, use the free 10-minute diagnostics.

Kp extends the International AS work on Kc and Le Chatelier’s principle (section 3.1.7). Revise that first on the Kc and Le Chatelier page. Mole calculations come from the amount of substance guide, and the rate side of catalysts is in the kinetics guide.

What this topic covers

Section	Outcome from the specification	Stage
3.1.12.1	Derive partial pressure from mole fraction and total pressure	International A-level only
3.1.12.1	Construct an expression for Kp for a homogeneous system in equilibrium	International A-level only
3.1.12.1	Perform calculations involving Kp	International A-level only
3.1.12.1	Predict the qualitative effects of changes in temperature and pressure on the position of equilibrium	International A-level only
3.1.12.1	Predict the qualitative effects of changes in temperature on the value of Kp	International A-level only
3.1.12.1	Understand that a catalyst can affect the rate of attainment of an equilibrium but not the value of the equilibrium constant	International A-level only

The specification lists no required practical for 3.1.12.

What Kp is

The specification gives two facts to start from. Kp is deduced from the equation for a reversible reaction occurring in the gas phase. And Kp is the equilibrium constant calculated from partial pressures for a system at constant temperature.

“Homogeneous” means every reactant and product is in the same phase. In this section, that phase is the gas phase, so every species in the equation appears in the Kp expression.

The section introduction links Kp to industry: it lets you calculate how an equilibrium yield depends on the partial pressures of reactants and products.

Mole fraction and partial pressure

The mole fraction of a gas is its share of the total amount of gas:

mole fraction of A = amount of A (mol) / total amount of gas (mol)

The partial pressure of a gas is the pressure that gas would exert if it were alone in the container. You derive it from the mole fraction:

partial pressure of A = mole fraction of A x total pressure

Mole fraction has no units. Partial pressure takes the same unit as the total pressure. Two checks catch most slips: the mole fractions sum to 1, and the partial pressures sum to the total pressure.

Worked example 1: from moles to partial pressures

A gas-phase equilibrium is set up at constant temperature:

4HCl(g) + O₂(g) ⇌ 2Cl₂(g) + 2H₂O(g)

The equilibrium mixture contains 0.480 mol HCl, 0.300 mol O₂, 0.360 mol Cl₂ and 0.360 mol H₂O. The total pressure is 300 kPa. Calculate the partial pressure of each gas.

total amount = 0.480 + 0.300 + 0.360 + 0.360 = 1.500 mol

HCl : 0.480 / 1.500 = 0.320    p = 0.320 x 300 = 96.0 kPa
O2  : 0.300 / 1.500 = 0.200    p = 0.200 x 300 = 60.0 kPa
Cl2 : 0.360 / 1.500 = 0.240    p = 0.240 x 300 = 72.0 kPa
H2O : 0.360 / 1.500 = 0.240    p = 0.240 x 300 = 72.0 kPa

checks: 0.320 + 0.200 + 0.240 + 0.240 = 1.000
        96.0 + 60.0 + 72.0 + 72.0 = 300 kPa

Constructing a Kp expression

Build Kp exactly as you built Kc, but replace each concentration [X] with a partial pressure p(X). Products go on top, reactants underneath, and each term carries its stoichiometric coefficient as an index. Only equilibrium partial pressures belong in the expression.

For the reaction in Worked example 1:

              p(Cl2)^2 x p(H2O)^2
   Kp  =  ---------------------------
              p(HCl)^4 x p(O2)

Units. Write the pressure unit for every term and cancel. The power that remains is (gas moles among the products) minus (gas moles among the reactants), which this guide calls Δn. Here Δn = 4 − 5 = −1, so with kPa the units are kPa⁻¹. If Δn = 0, Kp has no units.

Worked example 2: the value of Kp

Use the partial pressures from Worked example 1 to calculate Kp.

Kp = 72.0^2 x 72.0^2 / (96.0^4 x 60.0)
   = 72.0^4 / (96.0^4 x 60.0)
   = (0.750)^4 / 60.0          since 72.0 / 96.0 = 0.750
   = 0.3164 / 60.0
   = 5.27 x 10^-3 kPa^-1

Keep unrounded values until the final step.

Calculations involving Kp

Many questions give starting amounts and one equilibrium amount. Work through a table:

1. Write the balanced equation.
2. Set up rows for initial, change and equilibrium amounts in mol.
3. Use the mole ratio from the equation to fill the change row.
4. Total the equilibrium amounts of every gas, including any that started at zero.
5. Convert to mole fractions, then partial pressures.
6. Substitute into Kp and deduce the units.

Worked example 3: Kp from starting amounts

Carbon dioxide reacts with hydrogen over a nickel catalyst:

CO₂(g) + 4H₂(g) ⇌ CH₄(g) + 2H₂O(g)

A mixture of 1.00 mol CO₂ and 4.00 mol H₂ reaches equilibrium. It then contains 0.600 mol CH₄, and the total pressure is 380 kPa. Calculate Kp.

                 CO2      H2       CH4      H2O
initial / mol    1.00     4.00     0        0
change / mol    -0.600   -2.40    +0.600   +1.20
equilibrium      0.400    1.60     0.600    1.20     total = 3.80 mol

p / kPa          40.0     160      60.0     120      (fraction x 380)

Kp = p(CH4) x p(H2O)^2 / (p(CO2) x p(H2)^4)
   = 60.0 x 120^2 / (40.0 x 160^4)
   = 864 000 / (2.62 x 10^10)
   = 3.30 x 10^-5 kPa^-2        (Δn = 3 − 5 = −2)

The change row follows the 1 : 4 : 1 : 2 ratio. A common slip is to subtract 0.600 mol from the hydrogen instead of 4 × 0.600 = 2.40 mol.

Worked example 4: when the total pressure cancels

For H₂(g) + Br₂(g) ⇌ 2HBr(g) at a fixed temperature, Kp = 36.0. A vessel starts with 0.500 mol H₂ and 0.500 mol Br₂. Calculate the amount of HBr at equilibrium.

Let y mol of H₂ react. At equilibrium: H₂ = (0.500 − y), Br₂ = (0.500 − y), HBr = 2y. The total is still 1.000 mol, because Δn = 0.

Kp = p(HBr)^2 / (p(H2) x p(Br2))

Every p is (amount / 1.000) x P, so P and the total amount cancel:

36.0 = (2y)^2 / (0.500 − y)^2
 6.0 = 2y / (0.500 − y)          (square root of both sides)
3.00 − 6y = 2y
       y = 0.375

HBr at equilibrium = 2 x 0.375 = 0.750 mol

Kp has no units here, and you never needed the total pressure. That is true for any reaction with the same number of moles of gas on each side.

Changing the total pressure

Le Chatelier’s principle works here just as it did for Kc. A higher total pressure favours whichever side of the equation has fewer gas molecules; a lower total pressure favours the side with more. At constant temperature, the value of Kp does not change.

Writing each partial pressure as (mole fraction × P) shows why. A factor of P raised to the power Δn appears in the expression. If P changes, the mole fractions must adjust to keep Kp fixed, and that adjustment is the shift in position.

Worked example 5: lower pressure, higher yield

Ethane can be dehydrogenated: C₂H₆(g) ⇌ C₂H₄(g) + H₂(g). At a fixed temperature, take Kp = 25.0 kPa. Starting from pure ethane, let α be the fraction that reacts. Then 1 mol of ethane gives (1 − α) mol ethane, α mol ethene and α mol hydrogen, with a total of (1 + α) mol.

Kp = p(C2H4) x p(H2) / p(C2H6) = α^2 x P / (1 − α^2)

rearranged:  α = √( Kp / (Kp + P) )

P = 100 kPa:  α = √(25.0 / 125) = 0.447   so 44.7% converted
P = 400 kPa:  α = √(25.0 / 425) = 0.243   so 24.3% converted

Kp is the same at both pressures. The higher pressure gives a smaller conversion, because the equilibrium moves towards the side with fewer moles of gas (1 mol on the left, 2 on the right). This is how Kp lets you calculate the effect of pressure on an equilibrium yield.

Changing temperature: the only change that alters Kp

A temperature change moves the position of equilibrium and changes the value of Kp.

Forward reaction	Temperature raised	Position	Kp
Exothermic (ΔH negative)	the reverse reaction, which is endothermic, is favoured	left	falls
Endothermic (ΔH positive)	the forward reaction, which is endothermic, is favoured	right	rises

Cooling has the opposite effect on both the position and Kp.

Applying it. The CO₂ and H₂ reaction in Worked example 3 is exothermic. Raising the temperature moves the position of equilibrium to the left, so the partial pressures of CH₄ and H₂O fall relative to those of CO₂ and H₂, and Kp gets smaller. The ethane dehydrogenation in Worked example 5 is endothermic, so a higher temperature gives a larger Kp and a greater conversion of ethane.

Catalysts

A catalyst raises the rates of the forward and the reverse reaction by the same factor. The specification’s point is that a catalyst can affect the rate of attainment of an equilibrium but does not affect the value of the equilibrium constant. The equilibrium partial pressures, and so the yield, are the same with or without it; they are simply reached sooner.

Summary of effects

What you change	Where the equilibrium goes	Kp
Higher total pressure, same T	towards fewer gas molecules	same
Lower total pressure, same T	towards more gas molecules	same
Any pressure change with Δn = 0	nowhere	same
Higher temperature	in the endothermic direction	different (table above)
Catalyst added	nowhere; reached in less time	same

Common errors

• Totalling the initial amounts instead of the equilibrium amounts before working out mole fractions.
• Leaving out a gas that started at zero when you add up the equilibrium total.
• Using the wrong mole ratio in the change row, for example subtracting 0.600 mol of H₂ when the equation needs 4 × 0.600 mol.
• Writing [X] in a Kp expression. Kp uses p(X).
• Dropping a power, such as p(HCl)⁴.
• Giving units when Δn = 0, or getting the sign of the power wrong when Δn is negative.
• Claiming that a higher pressure increases Kp, or that a catalyst raises the yield.
• Rounding partial pressures early and drifting in the final significant figure.

Next steps

Shrink this page down with the Kp revision notes, and then work through the Kp practice questions. Older Kc material sits in the AS equilibria question set and the AS equilibria summary.

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.1.12 Equilibrium constant Kp for homogeneous systems (International A2), subsection 3.1.12.1.