Position of equilibrium

IB Chemistry SL· Reactivity 3.2· 35 min read

1. What is the Position of Equilibrium?★★☆☆☆⏱ 10 min

📘 Definition

Position of equilibrium

The relative concentrations of reactants and products in a system at dynamic equilibrium, where forward and reverse reaction rates are equal.

Example:

If $K > 1$ , position of equilibrium lies right (favours products); if $K < 1$ , it lies left (favours reactants).

The position of equilibrium describes how far a reaction goes towards products before it reaches dynamic equilibrium, it does not describe how fast the reaction reaches equilibrium. A reaction can reach equilibrium quickly but still favour reactants.

📐 Worked Example

For the reaction $ ext{N}_2(g) + 3 ext{H}_2(g) ightleftharpoons 2 ext{NH}_3(g) $,$ K_c = 1.7 imes 10^{-4}$ at 500 K. Comment on the position of equilibrium.

1
Write the equilibrium constant expression:
2
$K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$
3
Compare $K_{c}$ to 1: $1.7 \times 1 0^{- 4} < 1$ .
4
When $K_{c} < 1$ , the denominator (reactant concentrations) is larger than the numerator (product concentrations).
5
Conclusion: The position of equilibrium lies far to the left, favouring reactants, so very little ammonia forms at equilibrium.

2. Comparing Q and K to Predict Reaction Direction★★★☆☆⏱ 15 min

📘 Definition

Reaction Quotient

Q

A value calculated using the same ratio as $K$ , but using current concentrations of reactants and products at any point in the reaction (not just at equilibrium).

To predict which direction a reaction will shift to reach equilibrium, compare $Q$ (current ratio) to $K$ (equilibrium ratio):

$Q < K$ : Reaction shifts right to make more products
$Q > K$ : Reaction shifts left to make more reactants
$Q = K$ : System is at equilibrium, no net change

📐 Worked Example

For the reaction $2 SO_{2} (g) + O_{2} (g) ⇌ 2 SO_{3} (g)$ , $K_{c} = 4.2$ at 600 °C. Current concentrations are $[SO_{2}] = 0.10$ M, $[O_{2}] = 0.20$ M, $[SO_{3}] = 0.30$ M. Predict the direction of reaction.

1
Write the expression for Q:
2
$Q = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]}$
3
Substitute current concentrations and calculate Q:
4
$Q = \frac{(0.30)^2}{(0.10)^2 (0.20)} = 45$
5
Compare Q to $K_{c}$ : $45 > 4.2$ , so $Q > K_{c}$ .
6
Conclusion: The product concentration is too high, so the reaction shifts left to consume products and form more reactants to reach equilibrium.

3. Le Chatelier's Principle for Equilibrium Shifts★★★☆☆⏱ 15 min

📘 Definition

Le Chatelier's Principle

If a change in condition is applied to a system at equilibrium, the position of equilibrium shifts in the direction that counteracts the applied change.

Concentration: Adding a substance shifts equilibrium to consume the added substance; removing shifts to produce more of it.
Pressure (gases only): Increasing pressure shifts to the side with fewer moles of gas; decreasing shifts to more moles of gas.
Temperature: Increasing temperature shifts in the endothermic direction to absorb added heat; decreasing shifts in exothermic direction.

📐 Worked Example

Predict how the position of equilibrium shifts when temperature is increased for this exothermic reaction: $N_{2} (g) + 3 H_{2} (g) ⇌ 2 NH_{3} (g) Δ H = - 92 kJ mol^{- 1}$

1
Identify the endothermic direction: forward reaction is exothermic (releases heat), so reverse reaction is endothermic (absorbs heat).
2
Apply Le Chatelier's principle: increasing temperature adds heat, so equilibrium shifts to absorb the extra heat.
3
Conclusion: Equilibrium shifts left (towards reactants), so ammonia concentration decreases, and reactant concentrations increase.

4. Effect of Changes on Equilibrium Constant K★★★★☆⏱ 10 min

Only changes in temperature alter the value of the equilibrium constant $K$ . Changes in concentration, pressure, or adding a catalyst do not change $K$ , because they do not alter the thermodynamics of the reaction, only the position of equilibrium shifts temporarily.

For exothermic forward reactions: increasing temperature decreases $K$ (equilibrium shifts left, so product ratio falls). For endothermic forward reactions: increasing temperature increases $K$ (equilibrium shifts right, product ratio rises).

📐 Worked Example

For the endothermic decomposition of calcium carbonate: $CaCO_{3} (s) ⇌ CaO (s) + CO_{2} (g) Δ H = + 178 kJ mol^{- 1}$ , what happens to $K_{p}$ when temperature is decreased?

1
Forward reaction is endothermic, so decreasing temperature removes heat. Equilibrium shifts in the exothermic direction (reverse) to counteract the change.
2
$K_{p}$ for this reaction is equal to the partial pressure of $CO_{2}$ , because solids are excluded from equilibrium expressions: $K_{p} = P_{CO_{2}}$ .
3
Shifting left reduces the partial pressure of $CO_{2}$ , so $K_{p}$ decreases.
4
Conclusion: Decreasing temperature for an endothermic reaction decreases the value of $K$ .

5. Common Pitfalls

Wrong move:

Claiming adding a catalyst shifts the position of equilibrium

Why:

Catalysts lower activation energy of both forward and reverse reactions equally, so they do not change equilibrium concentrations

Correct move:

Always state catalysts have no effect on position of equilibrium or the value of K

Wrong move:

Predicting all pressure changes shift equilibrium for gaseous reactions

Why:

If total moles of gas are equal on both sides, changing pressure changes all concentrations equally, so Q stays equal to K

Correct move:

First count moles of gas on each side; if equal, no shift occurs after pressure change

Wrong move:

Reversing the direction of shift when comparing Q and K

Why:

If Q < K, the product ratio is too small, so reaction shifts right to make more products, not left

Correct move:

Remember the rule: $Q < K \to$ shift right, $Q > K \to$ shift left

Wrong move:

Claiming changing concentration or pressure changes K

Why:

K is only temperature dependent for a given reaction, other changes do not alter its value

Correct move:

Only temperature changes change K; concentration/pressure changes only shift equilibrium

Wrong move:

Mixing up K change for exothermic reactions when temperature increases

Why:

Increasing temperature shifts exothermic reactions left, which lowers the product/reactant ratio, so K decreases, not increases

Correct move:

Recall: increasing T increases K for endothermic, decreases K for exothermic

6. Quick Reference Cheatsheet

Change	Effect on equilibrium position	Effect on K
Increase [reactant]	Shifts right	No change
Increase pressure (fewer moles right)	Shifts right	No change
Increase pressure (equal moles both sides)	No shift	No change
Increase T (exothermic forward)	Shifts left	Decreases K
Increase T (endothermic forward)	Shifts right	Increases K
Add catalyst	No shift	No change

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · 1
Compare Q and K to predict direction
2024 · 2
Predict shift after pressure change
2023 · 1
Relate K size to equilibrium position

Going deeper

syllabusIB SL Chemistry Equilibrium SyllabusOfficial syllabus reference

What's Next

Mastering position of equilibrium is a core foundation for all subsequent equilibrium topics in IB Chemistry SL, which make up a large portion of exam marks. You will apply the same Q vs K comparison and Le Chatelier's principle to acid-base equilibria, the most heavily tested sub-topic in this unit. This concept also extends to solubility equilibria, where you predict precipitation formation using the same reasoning. Building a strong understanding here will simplify all later equilibrium topics.