AQA · GCSE · Combined Science
Course version: 8464 · Version from Sept 2018
The effect of pressure changes on equilibrium (HT only): a decrease in pressure causes the equilibrium position to shift towards the side with the larger number of molecules as shown by the symbol equation for that reaction.
As a higher-tier concept, when the pressure on a system at dynamic equilibrium is decreased, the equilibrium position shifts towards the side with the larger number of gas molecules.
- 2 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
As a higher-tier concept, when the pressure on a system at dynamic equilibrium is decreased, the equilibrium position shifts towards the side with the larger number of gas molecules. Producing more gas molecules increases the total volume occupied, which opposes the reduction in pressure. To apply the rule, count the gas molecules on each side of the symbol equation using the balancing numbers, ignoring solids and liquids. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the left side has 1 + 3 = 4 molecules and the right side has 2, so lowering pressure shifts the equilibrium position to the left, reducing the yield of ammonia. If both sides have equal gas molecule counts, pressure changes have no effect on the equilibrium position.
What earns the marks
- State that decreasing pressure shifts the equilibrium position towards the side with more gas molecules.
- Count gas molecules on each side of the symbol equation using the balancing numbers, ignoring solids and liquids.
- Apply the rule to a named equation, for example N₂(g) + 3H₂(g) ⇌ 2NH₃(g), where 2 molecules on the right shift to 4 on the left.
- Explain that the shift opposes the decrease in pressure by producing more gas molecules.
- Recognise that if the number of gas molecules is equal on both sides, changing pressure does not change the equilibrium position.
Common mistakes and how to put them right
Counting atoms rather than molecules.
Put it right: Count the balancing number of each gaseous species, so 3H₂ contributes three molecules.
Including solids or liquids in the molecule count.
Put it right: Only gases affect pressure, so ignore species such as CaCO₃(s) or H₂O(l).
Assuming decreased pressure always shifts equilibrium to the left.
Put it right: Identify which side has more gas molecules first; the shift may be to the right.
Exam tips
- Write the gas molecule totals above each side of the equation before deciding the direction of the shift.
- Use the phrase 'shifts to the side with more gas molecules' and then name that side, for example 'to the left, towards nitrogen and hydrogen'.
- If the equation is unfamiliar, count only species labelled (g) and check whether the totals differ.
Try the lesson
The start of the The effect of pressure changes on equilibrium (HT only): a decrease in pressure causes the equilibrium position to shift towards the side with the larger number of molecules as shown by the symbol equation for that reaction. lesson, made from checked teaching notes for this specification point.
Have a go first
In the equilibrium 2NO₂(g) ⇌ N₂O₄(g), a chemist decreases the pressure. In which direction does the equilibrium position shift, and why?
Listen
As a higher-tier concept, when the pressure on a system at dynamic equilibrium is decreased, the equilibrium position shifts towards the side with the larger number of gas molecules. Producing more gas molecules increases the total volume occupied, which opposes the reduction in pressure. To apply the rule, count the gas molecules on each side of the symbol equation using the balancing numbers, ignoring solids and liquids. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the left side has 1 + 3 = 4 molecules and the right side has 2, so lowering pressure shifts the equilibrium position to the left, reducing the yield of ammonia. If both sides have equal gas molecule counts, pressure changes have no effect on the equilibrium position. A strong answer earns marks for these points. State that decreasing pressure shifts the equilibrium position towards the side with more gas molecules. Count gas molecules on each side of the symbol equation using the balancing numbers, ignoring solids and liquids. Apply the rule to a named equation, for example N₂(g) + 3H₂(g) ⇌ 2NH₃(g), where 2 molecules on the right shift to 4 on the left. Explain that the shift opposes the decrease in pressure by producing more gas molecules. A common slip: Counting atoms rather than molecules: for example treating 3H₂ as six particles. Instead, count the balancing number of each gaseous species, so 3H₂ contributes three molecules. Exam tip: Write the gas molecule totals above each side of the equation before deciding the direction of the shift.
Key words
Say what each word means before you turn the card.
Effect of pressure on equilibrium position
Try answering aloud or in your head before turning the card.
Counting gas molecules from a symbol equation
Try answering aloud or in your head before turning the card.
See every definition
- Effect of pressure on equilibrium position
- A decrease in pressure causes the equilibrium position to shift towards the side with the larger number of molecules as shown by the symbol equation for that reaction.
- Counting gas molecules from a symbol equation
- Count the gas molecules on each side of the symbol equation using the balancing numbers, ignoring solids and liquids.
There is more to this lesson
The full lesson goes on to check your understanding and help you earn the marks. Create a free account to carry on, with your progress saved.
A week of Premium from your first lesson · Then the free plan continues · No card, no charge
Test your understanding of The effect of pressure changes on equilibrium (HT only): a decrease in pressure causes the equilibrium position to shift towards the side with the larger number of molecules as shown by the symbol equation for that reaction.
Active recall beats re-reading. Take an instant quiz now to test what you’ve learned and lock it into long-term memory.
A week of Premium from your first lesson · Then the free plan continues · No card, no charge