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In an ideal gas mixture, a component’s partial pressure equals its mole fraction multiplied by total pressure. A mole fraction is that component’s amount divided by total gaseous amount. Do not include liquid or solid amounts in the gas total. Adding all partial pressures should recover the total pressure; this provides a useful check before calculating the equilibrium constant. Use the pressure unit specified by the question consistently throughout the expression.
Video by Eliot Rintoul. The video’s own board label is retained; follow the specification reference and scope notes on this lesson for assessed content.
Viewing focus: Calculate partial pressures from mole fractions
After watching, attempt these questions before opening the answers. You can also use the written lesson below.
After watching 1: How is partial pressure obtained?
Mole fraction × total pressure. The fraction must refer to gaseous moles.
After watching 2: Which change alters Kp?
Temperature. Kp is temperature-dependent.
What you'll be able to do
In an ideal gas mixture, a component’s partial pressure equals its mole fraction multiplied by total pressure. A mole fraction is that component’s amount divided by total gaseous amount. Do not include liquid or solid amounts in the gas total. Adding all partial pressures should recover the total pressure; this provides a useful check before calculating the equilibrium constant. Use the pressure unit specified by the question consistently throughout the expression.
For a homogeneous gas equilibrium Kp is the product of equilibrium partial pressures of gaseous products divided by those of gaseous reactants, each raised to its equation coefficient. Initial pressure data may need an equilibrium-change calculation before substitution. In the school pressure convention, Kp units follow from the difference between product and reactant powers. If all powers cancel the numerical expression has no pressure units; otherwise a change of pressure unit changes the numerical value, so report both together.
Compressing an equilibrium mixture changes its partial pressures and may shift the composition to favour fewer gas molecules. It does not change Kp at constant temperature. If the number of gas moles is equal on each side, a uniform compression cancels within the expression. Temperature changes Kp: heating favours the endothermic direction. A catalyst changes neither Kp nor the final equilibrium mixture. These are separate from how rapidly the mixture responds after the conditions change.
Attempt these problems on paper before opening the worked reasoning. Explain your steps as well as giving a final answer.
Common mistakes to avoid
Key takeaways
Question 1: For A(g) ⇌ 2B(g), equilibrium pressures are pA = 50 kPa and pB = 100 kPa. Find Kp.
Kp = 200 kPa.
Check the assessed scope and practical or mathematical requirements.
An external explanation or interactive activity to use alongside the lesson. It may include extension material.
Close the notes and answer from memory. Revisit any explanation you cannot yet reconstruct.
Test yourself
Answer a few questions, then review anything you missed.
Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.