Equilibrium — IB Chemistry SL SL Study Guide

For: IB Chemistry SL candidates sitting IB Chemistry SL.

Covers: reversible reactions, dynamic equilibrium, Le Chatelier's principle, equilibrium constant $K_c$ calculations, and the effects of temperature, pressure, and concentration on equilibrium position, with exam-aligned worked examples and common mistake guidance.

You should already know: IGCSE Chemistry, basic algebra.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Chemistry SL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Is Equilibrium?

Equilibrium is a state in a closed chemical system where the rate of the forward reaction equals the rate of the reverse reaction, so macroscopic properties like concentration, color, and pressure stay constant over time, even as molecular-level reactions continue to occur. It is tested across both Paper 1 (multiple choice) and Paper 2 (structured response) of the IB Chemistry SL exam, and forms the foundational logic for acid-base chemistry, redox reactions, and industrial process optimization questions later in the syllabus. Common synonyms include chemical equilibrium and steady-state equilibrium, though the latter only applies strictly to closed, isolated systems.

2. Reversible reactions and dynamic equilibrium

A reversible reaction is a chemical reaction that can proceed in both the forward (reactants → products) and reverse (products → reactants) directions, denoted with the double harpoon symbol $\rightleftharpoons$ instead of a single reaction arrow. Unlike irreversible reactions that run to completion with 100% product yield, reversible reactions never fully consume reactants, as products continuously break back down into starting materials.

Dynamic equilibrium is the specific state reached by reversible reactions in closed systems (no exchange of matter with the surroundings, e.g., a sealed glass flask, not an open beaker where gas can escape). The term "dynamic" confirms that reactions have not stopped: forward and reverse reactions continue occurring at equal rates, so there is zero net change in the concentration of reactants and products over time.

Worked Example: Dynamic Equilibrium Visualization

The decomposition of dinitrogen tetroxide follows the reaction: $$N_2{O}_4(g)\rightleftharpoons 2N{O}_2(g)$$ $N_2{O}_4$ is colorless, while $N{O}_2$ is a dark brown gas. If you seal pure $N_2{O}_4$ in a flask at 25°C, you will first observe the brown color deepen as the forward reaction proceeds faster than the reverse reaction, producing more $N{O}_2$. After 2-3 minutes, the color stops changing: you have reached dynamic equilibrium. The forward reaction ($N_2{O}_4$ splitting) and reverse reaction ($N{O}_2$ combining) are still occurring, but at equal rates, so the concentration of both gases stays constant.

Exam tip: You will lose 1 mark if you define dynamic equilibrium as "equal concentration of reactants and products". Always specify equal forward/reverse reaction rates and constant (not equal) concentrations, plus the closed system requirement.

3. Le Chatelier's principle

Le Chatelier's principle is a predictive rule that lets you determine how an equilibrium system will respond to an external stress (change in conditions) without doing complex calculations. The formal statement is: If a change in concentration, temperature, or pressure is applied to a system at dynamic equilibrium, the position of equilibrium will shift in the direction that counteracts the applied change, to establish a new equilibrium state.

The "position of equilibrium" refers to the relative ratio of products to reactants at equilibrium: a shift right means more products are formed, while a shift left means more reactants remain.

Worked Example: Le Chatelier's Concentration Stress Test

The iron(III) thiocyanate equilibrium is a common lab demonstration: $$F{e}^{3+}(aq)+SC{N}^{-}(aq)\rightleftharpoons [Fe(SCN){]}^{2+}(aq)$$ Reactants are pale yellow/colorless, while the product is a deep blood red. If you add extra $F{e}^{3+}$ ions to the equilibrium mixture (increasing reactant concentration), the equilibrium shifts right to consume the excess $F{e}^{3+}$, so the solution becomes significantly darker red. If you add hydroxide ions ($O{H}^{-}$) which precipitate $F{e}^{3+}$ as insoluble $Fe(OH{)}_3$, the equilibrium shifts left to replace the lost $F{e}^{3+}$, so the red color fades almost completely.

Exam tip: Le Chatelier's principle questions make up ~3-4 marks of every Paper 1 multiple choice exam. Always identify the specific stress first, then select the direction that reduces that stress.

4. Equilibrium constant $K_c$

The equilibrium constant $K_c$ is a fixed value for any reversible reaction at a constant temperature, that describes the ratio of product concentration to reactant concentration at equilibrium. It is derived from experimental observation: for any balanced reversible reaction, the ratio of products to reactants (each raised to the power of their stoichiometric coefficient) is constant at a fixed temperature.

For a general reaction: $$aA+bB\rightleftharpoons cC+dD$$ The $K_c$ expression is: $$K_c=\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}$$ Where $[X]$ = equilibrium concentration of substance $X$ in $mold{m}^{-3}$.

Key Rules for $K_c$ Expressions:

1. Only include gaseous ($g$) and aqueous ($aq$) species: solids ($s$) and pure liquids ($l$, e.g., water in dilute aqueous solutions) have constant concentrations, so they are omitted from the expression.
2. $K_c$ is only temperature dependent: it does not change with concentration, pressure, or addition of a catalyst.
3. Units of $K_c$ vary based on reaction stoichiometry: calculate them by substituting $mold{m}^{-3}$ into the $K_c$ expression and cancelling common terms.

Worked Example: $K_c$ Calculation

For the Haber process reaction: $$N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g)$$ At 400°C, equilibrium concentrations are measured as: $[N_2]=1.5mold{m}^{-3}$, $[H_2]=2.0mold{m}^{-3}$, $[N{H}_3]=0.4mold{m}^{-3}$. Calculate $K_c$ and its units at 400°C.

1. Write the $K_c$ expression (all species are gaseous, so include all): $$K_c=\frac{[N{H}_3{]}^2}{[N_2][H_2{]}^3}$$
2. Calculate units: $$\frac{(mold{m}^{-3}{)}^2}{(mold{m}^{-3})\times (mold{m}^{-3}{)}^3}=mo{l}^{-2}d{m}^6$$
3. Substitute values: $$K_c=\frac{(0.4{)}^2}{1.5\times (2.0{)}^3}=\frac{0.16}{12}\approx 0.013mo{l}^{-2}d{m}^6$$

Exam tip: If $K_c>1$, the reaction is product-favored (more products at equilibrium); if $K_c<1$, it is reactant-favored. Always cross-check state symbols before writing $K_c$ expressions to avoid losing marks for including solids.

5. Effects of T, P, concentration on equilibrium position

We can combine Le Chatelier's principle and $K_c$ rules to predict the effect of all common external stresses on equilibrium systems:

1. Concentration

Changes in concentration do not alter the value of $K_c$, only shift the position of equilibrium:

• Increase reactant concentration: shift right to consume excess reactant
• Increase product concentration: shift left to consume excess product
• Decrease reactant concentration: shift left to produce more reactant
• Decrease product concentration: shift right to produce more product Industrial application: For esterification reactions, removing the ester product as it forms shifts equilibrium right to maximize yield.

2. Pressure

Pressure changes only affect equilibrium systems with gaseous species, where the total number of moles of gas differs on the reactant and product side. Pressure changes do not alter $K_c$:

• Increase total pressure: shift to the side with fewer moles of gas to reduce overall pressure
• Decrease total pressure: shift to the side with more moles of gas to increase overall pressure Example: Haber process has 4 moles of gas on the reactant side, 2 moles on the product side: high pressure shifts equilibrium right to increase ammonia yield. If moles of gas are equal on both sides (e.g., $H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$), pressure changes have no effect.

3. Temperature

Temperature changes do alter the value of $K_c$, as they change the rate of forward and reverse reactions differently:

• First identify if the forward reaction is exothermic ($\Delta H<0$, releases heat) or endothermic ($\Delta H>0$, absorbs heat)
• Increase temperature: equilibrium shifts in the endothermic direction to absorb excess heat
• Decrease temperature: equilibrium shifts in the exothermic direction to release extra heat
• If equilibrium shifts right, $K_c$ increases; if it shifts left, $K_c$ decreases Example: Haber process forward reaction is exothermic ($\Delta H=-92kJmo{l}^{-1}$): increasing temperature shifts equilibrium left, so $K_c$ decreases and ammonia yield falls.

Note on catalysts: Catalysts increase the rate of forward and reverse reactions equally, so they do not change equilibrium position or $K_c$, they only reduce the time taken to reach equilibrium.

Exam tip: If a question asks which factor changes $K_c$, the only correct answer is temperature. This is one of the most common multiple choice traps on this topic.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Defining dynamic equilibrium as "equal concentration of reactants and products". Why students do it: Confusing equal reaction rates with equal concentrations. Correct move: Define it as equal forward and reverse reaction rates, constant (not equal) concentrations, and occurring in a closed system.
• Wrong move: Including solid or pure liquid species in $K_c$ expressions. Why students do it: Forgetting that only gaseous and aqueous species have variable concentrations. Correct move: Cross-check state symbols for every species before writing your $K_c$ expression, omit all ($s$) and ($l$) species except in rare pure liquid reaction mixtures.
• Wrong move: Stating that pressure or concentration changes alter the value of $K_c$. Why students do it: Confusing a shift in equilibrium position with a change in the equilibrium constant. Correct move: Remember only temperature changes alter $K_c$; all other factors only change the relative amounts of reactants and products at the new equilibrium, not the constant ratio.
• Wrong move: Applying Le Chatelier's pressure rule to reactions with equal moles of gas on both sides, or no gaseous species. Why students do it: Memorizing the rule without checking its applicability conditions. Correct move: First count total moles of gaseous reactants and products before predicting pressure effects; if they are equal, or no gases are present, pressure has no effect.
• Wrong move: Predicting temperature effects without checking the sign of $\Delta H$ for the forward reaction. Why students do it: Mixing up exothermic and endothermic shift directions. Correct move: Label the endothermic direction first: increasing temperature always favors the endothermic direction, decreasing temperature favors the exothermic direction.

7. Practice Questions (IB Chemistry SL Style)

Question 1

Consider the following equilibrium reaction at 298 K: $$2S{O}_2(g)+O_2(g)\rightleftharpoons 2S{O}_3(g)\Delta H=-197{\text{kJ mol}}^{-1}$$ a) Write the expression for the equilibrium constant $K_c$ for this reaction, and state its units. (2 marks) b) State and explain the effect on the position of equilibrium if the total pressure of the system is increased at constant temperature. (2 marks) c) State and explain the effect of increasing temperature on the value of $K_c$ for this reaction. (2 marks)

Worked Solution 1

a) $K_c$ expression: $$K_c=\frac{[S{O}_3{]}^2}{[S{O}_2{]}^2[O_2]}$$ (1 mark) Units calculation: $\frac{(mold{m}^{-3}{)}^2}{(mold{m}^{-3}{)}^2\times (mold{m}^{-3})}=mo{l}^{-1}d{m}^3$ (1 mark) b) Equilibrium shifts to the right (1 mark). Explanation: There are 3 moles of gas on the reactant side and 2 moles of gas on the product side. Increasing pressure shifts equilibrium to the side with fewer moles of gas to counteract the pressure increase (1 mark). c) $K_c$ decreases (1 mark). Explanation: The forward reaction is exothermic. Increasing temperature shifts equilibrium to the endothermic reverse direction, so product concentration falls and reactant concentration rises, leading to a lower $K_c$ value (1 mark).

Question 2

The following equilibrium is established when iodine monochloride reacts with iodine: $$ICl(l)+\frac{1}{2}{I}_2(s)\rightleftharpoons I_{3}Cl(s)$$ Which of the following changes will cause a shift in the position of equilibrium? A) Adding more $I_2(s)$ B) Removing $I_{3}Cl(s)$ C) Increasing the pressure D) Adding more $ICl(l)$ (1 mark)

Worked Solution 2

Correct answer: D (1 mark). Explanation: Solids have constant concentration, so adding or removing $I_2(s)$ or $I_{3}Cl(s)$ (options A and B) has no effect. There are no gaseous species in the reaction, so pressure change (option C) has no effect. Adding more liquid $ICl$ increases its concentration in the reaction mixture, so equilibrium shifts right to consume the excess $ICl$.

Question 3

At 500 K, the following reaction reaches equilibrium with concentrations: $[H_2]=0.20mold{m}^{-3}$, $[I_2]=0.20mold{m}^{-3}$, $[HI]=1.2mold{m}^{-3}$. $$H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$$ a) Calculate the value of $K_c$ at 500 K. (2 marks) b) A catalyst is added to the system at equilibrium. State the effect, if any, on the value of $K_c$ and the time taken to reach equilibrium. (2 marks)

Worked Solution 3

a) $K_c$ expression: $$K_c=\frac{[HI{]}^2}{[H_2][I_2]}$$ (1 mark for correct expression) Substitute values: $K_c=\frac{(1.2{)}^2}{0.20\times 0.20}=\frac{1.44}{0.04}=36$ (unitless, as moles of gas are equal on both sides) (1 mark for correct calculation). b) $K_c$ is unchanged (1 mark). Time taken to reach equilibrium decreases (1 mark). Explanation: Catalysts increase forward and reverse reaction rates equally, so they do not change equilibrium position or $K_c$, only reduce the time to reach equilibrium.

8. Quick Reference Cheatsheet

9. What's Next

This equilibrium topic forms the foundational logic for three later IB Chemistry SL core topics: Topic 8 (Acids and Bases, where you will use equilibrium to calculate pH, acid dissociation constant $K_a$, and base dissociation constant $K_b$ for weak acids and bases), Topic 9 (Redox Processes, where equilibrium determines the cell potential of electrochemical cells), and Option C (Energy, where you will apply equilibrium rules to industrial processes like the Haber and Contact processes to maximize yield and operating efficiency. A solid grasp of the content in this guide will cut your study time for these later topics in half, as all build directly on the rules and calculation methods covered here.

If you have any questions about tricky $K_c$ calculations, Le Chatelier's principle applications, or exam-specific mark scheme conventions for this topic, you can ask Ollie, our AI tutor, for personalized explanations, extra practice questions, or feedback on your worked answers at any time. You can also find more IB Chemistry SL study resources and past paper practice on the homepage.