Topic 13 · page 2 of 12

Partial pressures and Kp

For gases it is often easier to use pressures than concentrations. Each gas contributes a partial pressure, found from its mole fraction, and Kp has the same shape as Kc with partial pressures in atm.

13 min read13.2

Read first: Writing Kc

Gases share a pressure 13.2

In a mixture of gases, every gas fills the whole container and contributes part of the total pressure. That part is its partial pressure. The total pressure is the sum of the partial pressures.

To find a partial pressure you need the share of the molecules that belong to that gas. That share is the mole fraction.

The mole fractions of all the gases add up to 1, and the partial pressures add up to PtotalP_\text{total}. Use both as checks.

Worked examplePartial pressures of a mixture

1.00 mol of N₂ and 3.00 mol of H₂ are mixed at a total pressure of 4.00 atm. Find each partial pressure.

  1. Total moles of gas = 1.00 + 3.00 = 4.00 mol
  2. xNX2x_{\ce{N2}} = 1.00 ÷ 4.00 = 0.250, and xHX2x_{\ce{H2}} = 3.00 ÷ 4.00 = 0.750
  3. pNX2p_{\ce{N2}} = 0.250 × 4.00 = 1.00 atm, and pHX2p_{\ce{H2}} = 0.750 × 4.00 = 3.00 atm
  4. Check: 1.00 + 3.00 = 4.00 atm, the total pressure.

For gases, the percentage by volume is the percentage by moles, so the mole fraction is the volume percentage ÷ 100. Air at 1.00 atm is about 78% N₂ and 21% O₂ by volume, so pNX2p_{\ce{N2}} ≈ 0.78 atm and pOX2p_{\ce{O2}} ≈ 0.21 atm.

Partial pressures at equilibrium 13.2

At equilibrium you often know the amount of each gas and the total pressure. Work out the total moles of gas first, then each mole fraction.

Worked exampleAn equilibrium mixture of N₂O₄ and NO₂

An equilibrium mixture at a total pressure of 2.00 atm contains 0.80 mol of N₂O₄ and 0.40 mol of NO₂ (illustrative amounts).

  1. Total moles = 0.80 + 0.40 = 1.20 mol
  2. xNX2OX4x_{\ce{N2O4}} = 0.80 ÷ 1.20 = 0.667, and xNOX2x_{\ce{NO2}} = 0.40 ÷ 1.20 = 0.333
  3. pNX2OX4p_{\ce{N2O4}} = 0.667 × 2.00 = 1.33 atm, and pNOX2p_{\ce{NO2}} = 0.333 × 2.00 = 0.667 atm
Two bars of equal length. The top bar shows 1.20 mol of gas: 0.80 mol of N2O4 and 0.40 mol of NO2. The bottom bar shows a total pressure of 2.00 atm split in the same proportions: 1.33 atm for N2O4 and 0.67 atm for NO2.
The same split, twice. The mole fractions divide the total pressure in the same proportions as they divide the moles.

The expression for Kp 13.2

KpK_\text{p} has exactly the same shape as KcK_\text{c}. Replace each concentration with the equilibrium partial pressure of the gas, and write it as pp rather than with square brackets.

Worked exampleThree Kp expressions

NX2(g)+3 HX2(g)⇌2 NHX3(g)Kp=pNHX3 2pNX2  pHX2 3 \ce{N2(g) + 3H2(g) <=> 2NH3(g)} \qquad K_\text{p} = \frac{p_{\ce{NH3}}^{\,2}}{p_{\ce{N2}}\;p_{\ce{H2}}^{\,3}}
2 SOX2(g)+OX2(g)⇌2 SOX3(g)Kp=pSOX3 2pSOX2 2  pOX2 \ce{2SO2(g) + O2(g) <=> 2SO3(g)} \qquad K_\text{p} = \frac{p_{\ce{SO3}}^{\,2}}{p_{\ce{SO2}}^{\,2}\;p_{\ce{O2}}}
NX2OX4(g)⇌2 NOX2(g)Kp=pNOX2 2pNX2OX4 \ce{N2O4(g) <=> 2NO2(g)} \qquad K_\text{p} = \frac{p_{\ce{NO2}}^{\,2}}{p_{\ce{N2O4}}}

Check yourself

  1. A mixture of 0.30 mol CO and 0.70 mol H₂ is at a total pressure of 5.0 atm. Calculate the mole fraction and partial pressure of each gas.
  2. Dry air is 78.0% N₂, 21.0% O₂ and 1.0% Ar by volume. Give the partial pressure of each gas at a total pressure of 1.00 atm and at 2.00 atm.
  3. Write the expression for KpK_\text{p} for 2 NO(g)+OX2(g)⇌2 NOX2(g)\ce{2NO(g) + O2(g) <=> 2NO2(g)}.
  4. Write the expression for KpK_\text{p} for HX2(g)+IX2(g)⇌2 HI(g)\ce{H2(g) + I2(g) <=> 2HI(g)}.
  5. At equilibrium, a mixture of 1.20 mol SO₂, 0.60 mol O₂ and 8.20 mol SO₃ is at a total pressure of 2.0 atm. Calculate the partial pressure of each gas.
  6. A student calculates the partial pressure of a gas as p=x×Pp = x \times P with xx = 0.25 and PP = 8.0 atm. She gets 32 atm. Where did the error come from, and what is the correct value?

Answers cover these until you have tried the questions

  1. Total = 1.00 mol. xCOx_{\ce{CO}} = 0.30 ÷ 1.00 = 0.30 and xHX2x_{\ce{H2}} = 0.70. pCOp_{\ce{CO}} = 0.30 × 5.0 = 1.5 atm and pHX2p_{\ce{H2}} = 0.70 × 5.0 = 3.5 atm (sum 5.0 atm).
  2. At 1.00 atm: N₂ 0.780 atm, O₂ 0.210 atm, Ar 0.010 atm. At 2.00 atm every partial pressure doubles: N₂ 1.56 atm, O₂ 0.420 atm, Ar 0.020 atm.
  3. Kp=pNOX2 2pNO 2  pOX2K_\text{p} = \dfrac{p_{\ce{NO2}}^{\,2}}{p_{\ce{NO}}^{\,2}\;p_{\ce{O2}}}
  4. Kp=pHI 2pHX2  pIX2K_\text{p} = \dfrac{p_{\ce{HI}}^{\,2}}{p_{\ce{H2}}\;p_{\ce{I2}}}
  5. Total = 1.20 + 0.60 + 8.20 = 10.00 mol. pSOX2p_{\ce{SO2}} = (1.20 ÷ 10.00) × 2.0 = 0.24 atm; pOX2p_{\ce{O2}} = (0.60 ÷ 10.00) × 2.0 = 0.12 atm; pSOX3p_{\ce{SO3}} = (8.20 ÷ 10.00) × 2.0 = 1.64 atm. Check: 0.24 + 0.12 + 1.64 = 2.00 atm.
  6. She divided by xx in place of multiplying (8.0 ÷ 0.25 = 32). The correct value is 0.25 × 8.0 = 2.0 atm. A partial pressure can never be bigger than the total pressure.