5.6.2.7The effect of pressure changes on equilibrium (HT only)
AQA GCSE Combined Science (8464), Higher tier · Chemistry › The rate and extent of chemical change › Reversible reactions and dynamic equilibrium
Practise The effect of pressure changes on equilibrium (HT only). 13 exam-style questions on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Higher tier only. For reactions involving gases, changing the pressure can shift the position of equilibrium. You need to count the gas molecules on each side of the balanced equation and predict which way the position moves, or explain why pressure has no effect.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
5
Count the molecules on each sideAdd the numbers in front of each gas formula in the balanced equation (no number means 1).
6
Predict the effect of increasing the pressureThe position shifts towards the side with the smaller number of molecules.
6
Predict the effect of decreasing the pressureThe position shifts towards the side with the larger number of molecules.
7
Recognise when pressure has no effectIf both sides have the same number of gas molecules, changing the pressure does not move the position.
7
Explain pressure effects using Le ChatelierFewer molecules exert a lower pressure, so shifting to that side counteracts a pressure increase.
8
Use pressure data to test an equationE.g. if the yield rises with pressure, the product side must have fewer gas molecules.
Notes
The rules (reactions involving gases)
An increase in pressure shifts the position of equilibrium towards the side with the smaller number of molecules, as shown by the balanced symbol equation.
A decrease in pressure shifts the position of equilibrium towards the side with the larger number of molecules.
If both sides have the same number of gas molecules, changing the pressure does not change the position of equilibrium.
Counting molecules
Add up the numbers in front of the formulae of the gases on each side. A formula with no number in front counts as 1.
Count only the gases, marked (g). Do not count solids or liquids.
N2(g) + 3H2(g) ⇌ 2NH3(g): 1 + 3 = 4 molecules on the left and 2 on the right. Increasing the pressure shifts the position to the right, so more ammonia forms.
H2(g) + I2(g) ⇌ 2HI(g): 2 molecules on each side, so changing the pressure does not change the position.
Why it works
Gas pressure is caused by molecules colliding with the walls of the container. Fewer molecules in the same volume means a lower pressure.
When the pressure is increased, the system counteracts the change by shifting towards the side with fewer molecules, which reduces the pressure.
Increasing the pressure of reacting gases also increases the rates of both reactions (more frequent collisions), so equilibrium is reached faster.
Cheatsheet
Higher pressure → side with fewer gas molecules
Lower pressure → side with more gas molecules
Same number of gas molecules on each side → pressure does not change the position
Count molecules using the numbers in front of the formulae in the balanced equation (no number = 1)
N2 + 3H2 ⇌ 2NH3: 4 molecules → 2 molecules, so a higher pressure gives more NH3
A higher pressure also increases the rate of gas reactions
How to answer each type of question
Predict and explain the effect of changing the pressure
2 to 3 marks6
Count the gas molecules on each side of the equation and write the numbers down.
Higher pressure: shift to the side with fewer molecules. Lower pressure: shift to the side with more molecules.
Say what happens to the amount of the substance the question asks about.
Example. Methane reacts with steam in a reversible reaction. CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g) Predict and explain the effect of increasing the pressure on the amount of hydrogen at equilibrium.
Show the model answer
The amount of hydrogen decreases (1). There are 2 molecules on the left and 4 molecules on the right (1), so the position of equilibrium shifts to the left, towards the side with fewer molecules (1).
Explain why pressure has no effect on a particular equilibrium
2 marks7
Count the gas molecules on each side and show they are equal.
Conclude that shifting in either direction would not change the pressure, so the position does not change.
Example. At very high temperatures, nitrogen and oxygen react to form nitrogen monoxide. N2(g) + O2(g) ⇌ 2NO(g) Explain why changing the pressure does not change the amount of nitrogen monoxide at equilibrium.
Show the model answer
There are 2 molecules on each side of the equation (1), so a shift in either direction would not change the pressure, and the position of equilibrium does not change (1).
Use pressure data to evaluate an equation or prediction
3 to 4 marks8
Describe the trend in the data as the pressure increases.
Say which way the position must be shifting.
Count the molecules in the suggested equation and check whether they fit the trend.
Example. Gas P decomposes in a reversible reaction. The percentage of P decomposed at equilibrium was measured at different pressures, at the same temperature: 1 atmosphere: 40%; 5 atmospheres: 19%; 10 atmospheres: 14% A student suggests the equation is P(g) ⇌ Q(g) + R(g). Explain whether the data support the student's equation.
Show the model answer
As the pressure increases, the percentage of P decomposed decreases (1), so a higher pressure shifts the position of equilibrium to the left (1). The data support the equation: there is 1 molecule on the left and 2 on the right (1), and a higher pressure shifts the position towards the side with fewer molecules (1).
Shortcuts and memory tricks
Squeeze it and it shrinks: higher pressure favours the side with fewer molecules.
Write the molecule count under each side of the equation before you answer.
Same number of molecules on both sides: pressure does not move the position (but it still changes the rate).
Where marks are lost
Counting atoms instead of molecules. Use the big numbers in front of the formulae, not the small subscripts.
Forgetting that a formula with no number in front counts as one molecule.
Writing 'towards the smaller side' or 'the side with less mass'. Say 'towards the side with fewer molecules'.
Applying the pressure rule to reactions in solution with no gases. Pressure does not affect their position of equilibrium.
Exam technique
Quote the numbers: 'there are 4 molecules on the left and 2 on the right' earns credit that 'fewer on the right' alone may not.
Use the balanced equation given in the question, and check it is balanced before you count.
State the effect on the product (increases or decreases) as well as the direction of the shift.
Quick recall
Cover the answers and test yourself. The app has these as flashcards that come back just before you'd forget them.
How many molecules of gas are shown on the left-hand side of the equation, and how many on the right-hand side?
3 on the left and 2 on the right
Sample questions
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy3 marks
Sulfur dioxide reacts with oxygen in a reversible reaction to form sulfur trioxide: 2SO2(g) + O2(g) ⇌ 2SO3(g)
(a) How many molecules of gas are shown on the left-hand side of the equation, and how many on the right-hand side?[1]
(b) The pressure is increased. What happens to the amount of sulfur trioxide at equilibrium? Tick (✓) one box.[1]
It decreases.
It increases.
It stays the same.
(c) Complete the sentence. Increasing the pressure shifts the position of equilibrium towards the side of the equation with .................... molecules of gas.[1]
Show the answer and mark scheme
(a)Answer: 3 on the left and 2 on the right
3 on the left and 2 on the right
(b)Answer: It increases.
(c)Answer: fewer
fewer / a smaller number of
Question 2Medium2 marks
(a) For which of these reactions at equilibrium does a change in pressure have no effect on the position of equilibrium? Tick (✓) one box.[1]
2SO2(g) + O2(g) ⇌ 2SO3(g)
N2(g) + 3H2(g) ⇌ 2NH3(g)
H2(g) + I2(g) ⇌ 2HI(g)
N2O4(g) ⇌ 2NO2(g)
(b) Explain your answer.[1]
Show the answer and mark scheme
(a)Answer: H2(g) + I2(g) ⇌ 2HI(g)
(b)Answer: This reaction has the same number of gas molecules (2) on each side of the equation.
it has the same number of gas molecules on each side of the equation (2 and 2)
Question 3Hard7 marks
Ethanol can be made by reacting ethene with steam: C2H4(g) + H2O(g) ⇌ C2H5OH(g) The reaction is carried out at 300 °C and a pressure of 60 to 70 atmospheres, using a catalyst.
(a) Explain the effect of increasing the pressure on the yield of ethanol.[3]
(b) Increasing the pressure also increases the rate of the reaction. Explain why.[2]
(c) At pressures above about 70 atmospheres, some of the ethene polymerises to form poly(ethene). Give two reasons why a pressure much higher than 70 atmospheres is not used.[2]
Show the answer and mark scheme
(a)Answer: There are 2 molecules of gas on the left and 1 on the right; a higher pressure shifts the equilibrium towards fewer molecules, so the yield of ethanol increases.
there are 2 molecules (of gas) on the left and 1 on the right
increasing the pressure shifts the equilibrium towards the side with fewer molecules
so the yield of ethanol increases
(b)Answer: There are more gas particles in the same volume, so collisions are more frequent.
there are more gas particles in the same volume
so collisions are more frequent
(c)Answer: Some ethene would be wasted by forming poly(ethene); higher pressures are more expensive and more dangerous.
some ethene would be wasted by forming an unwanted product (poly(ethene))
high pressures are expensive (stronger equipment / more energy for compressors)