Part 2: Mole fractions, partial pressures and Kp

Translate a gas mixture into partial pressures and use the equilibrium expression to calculate its composition and units.

Each gas contributes to the total pressure

For an ideal gas mixture, mole fraction xi = ni/ntotal. It is dimensionless and all mole fractions sum to one. The partial pressure of a gas is the pressure it would exert alone in the same volume at the same temperature; pi = xiPtotal. All partial pressures sum to Ptotal.

An original equilibrium mixture contains 0.20 mol N₂, 0.60 mol H₂ and 0.40 mol NH₃ at 600 kPa. Total amount is 1.20 mol, giving mole fractions 1/6, 1/2 and 1/3. The corresponding pressures are 100, 300 and 200 kPa. Use equilibrium amounts, not the amounts originally admitted.

If an inert gas is present it belongs in the total-mole denominator and contributes to total pressure, but it does not appear as a reacting species in Kp. At unchanged temperature and vessel volume, adding an ideal inert gas raises total pressure without changing the reacting gases’ partial pressures.

Worked example: calculate Kp and its units

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Kp = p(NH₃)²/[p(N₂)p(H₂)³]. Using the illustrative pressures above gives 200²/(100 × 300³) = 1.48 × 10⁻⁵ kPa⁻².

The units come from kPa²/(kPa × kPa³) = kPa⁻². The powers are stoichiometric coefficients, and only gaseous species appear. Do not insert mole fractions directly unless each has first been multiplied by the total pressure or the algebra explicitly retains all Ptotal factors.

Use one pressure unit throughout. Changing from kPa to Pa changes the numerical value of a dimensional Kp. For this expression, replacing every pressure by 1000 times its value multiplies the result by 1000⁻², so the value becomes 1.48 × 10⁻¹¹ Pa⁻². It is the same physical equilibrium expressed in different units.

Calculate an unknown partial pressure systematically

For an illustrative equilibrium A(g) ⇌ 2B(g), Kp = 80.0 kPa and p(A) = 20.0 kPa. Rearranging gives p(B)² = 80.0 × 20.0 = 1600 kPa², so p(B) = 40.0 kPa. If A and B are the only gases, Ptotal = 60.0 kPa.

Where stoichiometry makes two equilibrium partial pressures equal, name that equality before substituting. It can turn a product of pressures into a square. Do not assume all product pressures are equal simply because they are products; their initial amounts and the reaction coefficients matter.

For a dissociation problem, track the mole change first. Starting with 1.00 mol N₂O₄ and dissociating 0.300 mol leaves 0.700 mol N₂O₄ and forms 0.600 mol NO₂: total 1.300 mol. Using the initial total of 1.00 in the mole-fraction denominator would make the fractions sum to 1.30, exposing the error.

Why compression can shift equilibrium without changing Kp

Compressing a gas mixture at fixed temperature initially multiplies each partial pressure by the same factor. For N₂ + 3H₂ ⇌ 2NH₃, doubling all pressures makes the pressure quotient one quarter of its equilibrium value because its numerator has power two and its denominator power four.

The reaction then forms more NH₃ to restore the quotient to Kp. This explains the familiar shift towards fewer gaseous molecules without suggesting Kp changed. If equal total gas coefficients occur on each side, those compression factors cancel and this ideal-gas argument predicts no shift. A catalyst affects how fast equilibrium is reached, not the final Kp.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.

Sources