QCE Chemistry - Unit 3 - Chemical equilibrium systems

Factors That Affect Equilibrium | QCE Chemistry

Understand how concentration, pressure, volume and temperature change equilibrium position in QCE Chemistry.

Updated 2026-05-18 - 5 min read

When an equilibrium system is disturbed, it shifts in the direction that partially opposes the disturbance. This is Le Chatelier's principle. The system does not usually return every concentration to its original value; it simply responds in the direction that reduces the stress.

Concentration changes

For:

$ aA + bB \rightleftharpoons cC + dD $

• adding $A$ or $B$ favours the forward reaction
• adding $C$ or $D$ favours the reverse reaction
• removing $A$ or $B$ favours the reverse reaction
• removing $C$ or $D$ favours the forward reaction

The equilibrium constant does not change when concentration changes, as long as temperature stays the same. The mixture shifts until the same $K_c$ value is restored.

You can also explain this with collision theory. Adding a reactant increases the frequency of successful forward-reaction collisions at that moment, so the forward rate becomes greater than the reverse rate. As products build up, the reverse rate increases until the two rates become equal again at a new equilibrium position.

Removing a species does the same idea in reverse. If product is removed, the reverse reaction has fewer successful collisions available, so the forward reaction is temporarily favoured and more product forms.

Pressure and volume changes

Pressure and volume changes mainly matter for gases. A smaller volume means a higher pressure, so the system favours the side with fewer moles of gas particles.

Example:

$ 2\mathrm{SO_2(g)} + \mathrm{O_2(g)} \rightleftharpoons 2\mathrm{SO_3(g)} $

The left side has 3 mol of gas. The right side has 2 mol of gas. Increasing pressure favours the product side, so the yield of $\mathrm{SO_3}$ increases.

If both sides have the same number of gaseous particles, changing pressure has little effect on equilibrium position.

Diluting an aqueous equilibrium is similar in style to a volume increase. The system tends to favour the side with more dissolved particles because that partially opposes the dilution. This rule is less common than the gas-pressure rule, but it is useful when a question explicitly talks about dilution of an aqueous equilibrium mixture.

Temperature changes

Temperature is different from concentration and pressure because it changes the value of $K_c$.

Treat heat like a chemical species:

• exothermic forward reaction: heat is a product
• endothermic forward reaction: heat is a reactant

Increasing temperature favours the endothermic direction because that direction absorbs heat. Decreasing temperature favours the exothermic direction because that direction releases heat.

For $K_c$:

• if the forward reaction is exothermic, increasing temperature decreases $K_c$
• if the forward reaction is endothermic, increasing temperature increases $K_c$

That is because $K_c$ describes the product-to-reactant concentration ratio at equilibrium. If heat favours the reverse direction, the product-heavy ratio becomes smaller. If heat favours the forward direction, the product-heavy ratio becomes larger.

Catalysts

A catalyst provides an alternative pathway with lower activation energy. It speeds up both the forward and reverse reactions. That means equilibrium is reached faster, but the final equilibrium position is unchanged.

On a concentration-time graph, a catalyst changes how quickly the curves level off. It should not change the final plateau values for the same starting mixture at the same temperature. On a rate-time graph, both forward and reverse rates become equal sooner.

Writing full explanations

A high-quality equilibrium answer usually has three parts:

1. State the disturbance.
2. State which direction is favoured and why that direction partially opposes the disturbance.
3. State the effect on yield, concentration, pressure, or $K_c$ only if the question asks for it.

For example, do not stop at "the equilibrium shifts right." A stronger answer is: "Increasing pressure favours the side with fewer gas moles, so the equilibrium shifts right and the equilibrium yield of ammonia increases."

Worked example

Exam traps

Other traps:

• applying pressure rules to solids and liquids
• forgetting to count only gaseous moles
• saying the system completely cancels the change instead of partially opposing it
• forgetting that temperature changes $K_c$

Quick check

Sources

• QCAA Chemistry subject page
• QCAA Chemistry 2025 syllabus
• OpenStax Chemistry 2e: Shifting Equilibria
• OpenStax Chemistry 2e: Chemical Equilibria