A glow stick glows brighter when you warm it in your hands and slower when you put it in the freezer — the same chemical reaction, running at different rates, because temperature changes how often and how hard particles collide.
Understand the core concepts covered in this lesson.
Apply your knowledge to solve problems and explain phenomena.
Evaluate and analyse scientific information and data.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
You’ve probably snapped a glow stick and seen it start glowing. If you put it in a cup of hot water, it glows intensely for a short time then fades quickly. If you put it in the freezer, it glows dimly for a very long time.
The chemical reaction inside is identical in both cases — same chemicals, same amount. (1) If the same reaction is occurring, why does temperature change how fast it happens? (2) Why does the hot glow stick run out faster? Write your predictions before reading on.
Type your predictions below — you will revisit them at the end of the lesson.
Write your predictions in your book. You will revisit them at the end.
📚 Content
The rate of a chemical reaction is defined as the change in concentration of a reactant or product per unit time. In practical terms, rate can be measured by monitoring any observable quantity that changes as the reaction proceeds.
Reaction rate is not constant throughout a reaction — it is typically fastest at the start (when reactant concentrations are highest) and slows progressively as reactants are consumed. A rate versus time graph shows a decreasing curve that approaches zero as the reaction nears completion.
Chemical reactions don’t happen just because reactant particles are present — they happen only when particles collide in exactly the right way with exactly enough energy.
Collision theory states: for a chemical reaction to occur, reactant particles must collide. However, not every collision results in a reaction. An effective collision is one that results in the formation of products. For a collision to be effective, two conditions must both be met simultaneously:
The proportion of collisions that are effective is small for most reactions at room temperature — which is why reactions that are thermodynamically favourable can still be slow.
The activation energy (Eₐ) is the minimum kinetic energy that colliding particles must have for the collision to be effective. It represents the energy required to break the bonds in the reactants sufficiently to allow atoms to rearrange into products.
An energy diagram (reaction progress diagram) plots the energy of the reacting system against the progress of the reaction from reactants to products:
Key features: reactants at their initial energy level (left); the transition state (activated complex) at the peak; products at their final energy level (right). Eₐ is measured from the reactant energy level to the peak. ΔH is the difference between reactant and product energy levels. For an exothermic reaction, products are lower than reactants (ΔH < 0). For endothermic, products are higher (ΔH > 0).
Every variable that affects reaction rate does so by changing either the frequency of collisions, the proportion of collisions with sufficient energy, or both. These four factors are explored in full in L12:
| Variable | Effect on Collision Frequency | Effect on Proportion Exceeding Eₐ | Net Effect on Rate |
|---|---|---|---|
| Increase temperature | Increases | Increases (more exceed Eₐ) | Rate increases |
| Increase concentration | Increases | No change | Rate increases |
| Increase surface area | Increases (at surface) | No change | Rate increases |
| Add catalyst | No significant change | Increases (lower Eₐ) | Rate increases |
A glow stick contains two chemical compartments: an outer tube containing hydrogen peroxide (H₂O₂) and a fluorescent dye, and an inner glass vial containing a phenyl oxalate ester. When snapped, the two solutions mix and a chemiluminescent reaction produces light.
At higher temperature (warm water): particles have greater average kinetic energy; a larger proportion of collisions exceed Eₐ; more effective collisions occur per second; reaction rate is faster; the glow is brighter but shorter (reactants consumed more quickly).
At lower temperature (freezer): particles have lower average kinetic energy; fewer collisions exceed Eₐ; fewer effective collisions per second; reaction rate is slower; the glow is dimmer but lasts longer.
Crucially, the total amount of light produced is approximately the same at both temperatures (same amount of reactant) — only the rate of production differs.
📐 Worked Examples
Rate = Δvolume ÷ Δtime = (32 − 0) mL ÷ (60 − 0) s = 32 ÷ 60 = 0.53 mL/s
Rate = Δvolume ÷ Δtime = (52 − 48) mL ÷ (180 − 120) s = 4 ÷ 60 = 0.067 mL/s
Between 120 and 180 seconds, the reaction is nearly complete — the concentration of HCl has decreased significantly. Lower concentration means fewer H⁺ ions per unit volume, decreasing the frequency of collisions between H⁺ and CaCO₃ surface particles. Fewer collisions per second means fewer effective collisions per second (the proportion of effective collisions is unchanged since T is constant and Eₐ is unchanged). Therefore reaction rate decreases.
(a) 0.53 mL/s. (b) 0.067 mL/s. (c) Decreased HCl concentration lowers collision frequency between H⁺ and CaCO₃, reducing effective collisions per second and slowing the rate.
The reaction releases 40 kJ/mol (ΔH = −40 kJ/mol) → exothermic. Products are lower in energy than reactants.
Reactant energy level set as reference. The curve rises from reactant level to the transition state peak (85 kJ above reactant level). The curve then falls to the product level, which is 40 kJ below the reactant level. Labels: Eₐ = 85 kJ/mol (arrow from reactant level to peak); ΔH = −40 kJ/mol (arrow from reactant level down to product level); “Transition state / Activated complex” at the peak; “Reactants” on the left; “Products” on the right.
A catalyst lowers the activation energy. On the same diagram, draw a second curve with a lower peak (e.g. at 55 kJ above reactant level instead of 85 kJ). The reactant and product energy levels are unchanged — the catalyst does not change ΔH. Label the lower peak “Eₐ (catalysed)” and the original peak “Eₐ (uncatalysed).”
✏️ Activities
A student investigates the decomposition of hydrogen peroxide (H₂O₂ → H₂O + ½O₂) by measuring the volume of oxygen gas produced over time at 25°C.
| Time (s) | 0 | 20 | 40 | 60 | 80 | 100 |
|---|---|---|---|---|---|---|
| Volume O₂ (mL) | 0 | 14 | 24 | 30 | 33 | 34 |
(a) Calculate the average rate of O₂ production between 0 and 40 seconds (mL/s).
(b) Calculate the average rate between 60 and 100 seconds (mL/s).
(c) Explain why the rate in (b) is lower than in (a), using the concept of collision frequency.
Type your calculations and explanation.
Show calculations in your workbook.
Consider the exothermic reaction: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). This reaction has a high activation energy (around 400 kJ/mol) but releases approximately 890 kJ/mol of energy.
Question A: Sketch and label the energy diagram for this reaction. Mark Eₐ, ΔH, the transition state, reactants, and products.
Question B: Without a catalyst, natural gas (methane) does not ignite at room temperature despite being thermodynamically favourable (ΔH very negative). Use collision theory and your energy diagram to explain why.
Question C: A spark is used to ignite natural gas. Explain what the spark is doing in terms of activation energy.
Type your descriptions and explanations for A, B, and C.
Sketch the diagram and write responses in your workbook.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Wrong: A catalyst increases the amount of product formed at equilibrium.
Right: A catalyst speeds up both forward and reverse reactions equally, so equilibrium is reached faster but the equilibrium position and product yield do not change. Catalysts lower activation energy but do not alter ΔH or the thermodynamic favourability of a reaction.
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
8. (4 marks) State the two conditions required for an effective collision. For each condition, explain what happens when that condition is not met (i.e., what happens to the particles). (2 marks each)
Type your response — state each condition and what happens without it.
Write your response in your book.
9. (4 marks) A reaction has an activation energy of 60 kJ/mol and produces 35 kJ/mol of energy. (a) Is the reaction exothermic or endothermic? (b) On a labelled energy diagram, identify and give numerical values for Eₐ and ΔH. (c) Explain whether the reaction would proceed faster or slower if the activation energy were 30 kJ/mol instead of 60 kJ/mol (assume the same temperature). (1 + 2 + 1 marks)
Type your answer and describe the energy diagram labels.
Draw the diagram and write your response in your book.
10. (5 marks) A glow stick is cracked and placed in a water bath at 5°C. An identical glow stick is placed in a water bath at 45°C. (a) Predict which glow stick will be brighter and which will last longer. (b) Using collision theory, explain why the glow stick at 45°C glows more brightly. Your explanation must reference activation energy, kinetic energy of particles, and the proportion of effective collisions. (c) The total amount of light produced by both glow sticks is the same. Explain this observation in terms of the amount of reactant and reaction rate. (1 + 3 + 1 marks)
Type your full extended response.
Write your full response in your book.
Go back to your Think First response. Can you now explain precisely why hot water makes the glow stick brighter (using activation energy and effective collision language)? And why does the hot glow stick run out faster?
(a) Rate = (24 − 0) mL ÷ (40 − 0) s = 24 ÷ 40 = 0.60 mL/s
(b) Rate = (34 − 30) mL ÷ (100 − 60) s = 4 ÷ 40 = 0.10 mL/s
(c) Between 60 and 100 seconds, the concentration of H₂O₂ has decreased significantly. Fewer H₂O₂ molecules per unit volume means that collisions between H₂O₂ molecules (or between H₂O₂ and the catalyst surface) occur less frequently. Fewer collisions per second means fewer effective collisions per second (the proportion of effective collisions is unchanged since temperature and Eₐ are unchanged). Therefore reaction rate decreases.
A. Energy x-axis = reaction progress; y-axis = energy. Reactants (CH₄ + 2O₂) at a reference energy level on the left. The curve rises steeply to the transition state peak (400 kJ above reactant level — this is Eₐ). The curve falls to the product level (CO₂ + 2H₂O), which is 890 kJ below the reactant level (ΔH = −890 kJ/mol). Labels: Eₐ = 400 kJ/mol (arrow from reactant level to peak); ΔH = −890 kJ/mol (arrow from reactant level to product level); “Transition state / Activated complex” at peak.
B. At room temperature, the average kinetic energy of methane and oxygen molecules is much less than the activation energy (400 kJ/mol). Only a tiny fraction of collisions have sufficient kinetic energy to reach the transition state and result in bond-breaking and rearrangement. Despite the reaction being thermodynamically favourable (very negative ΔH), the high energy barrier means almost no collisions are effective at room temperature. The reaction does not proceed at a measurable rate.
C. The spark provides a concentrated burst of energy to a small region of the gas mixture. This locally raises the temperature dramatically, increasing the average kinetic energy of nearby molecules so that a proportion of them exceed the activation energy (400 kJ/mol) and undergo effective collisions. These initial reactions release enough heat to ignite the surrounding gas — the process becomes self-sustaining because the heat released by each reaction provides energy for the next set of collisions to exceed Eₐ.
Q1 B: Both conditions required: energy ≥ Eₐ AND correct orientation. Option A omits both. Option C confuses Eₐ with ΔH.
Q2 B: Lower temperature → lower average KE → fewer particles exceed Eₐ → fewer effective collisions → slower rate → dimmer but longer glow. Option A incorrectly states Eₐ changes with temperature. Option D incorrectly predicts shorter duration.
Q3 C: Eₐ is measured from the reactant energy level to the transition state peak. Option A describes ΔH. Option B gives the reverse activation energy. Option D also describes ΔH.
Q4 A: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. CO₂ gas is produced and escapes the open vessel, reducing total mass. The decrease in mass directly measures the rate of CO₂ production.
Q5 B: A large activation energy means very few particles have sufficient kinetic energy to overcome the barrier at room temperature. The large negative ΔH is irrelevant to the rate — it tells you the energy released if the reaction proceeds, not how fast it proceeds. Eₐ controls rate; ΔH controls whether the reaction is favoured thermodynamically.
Q6 C: Even at very high temperatures where all particles exceed Eₐ, collisions with incorrect orientation cannot lead to products. The orientation condition is independent of energy. Both conditions must be met. The claim is incorrect.
Q7 D: The Arrhenius equation (k = A×e^(-Ea/RT)) describes the relationship between rate constant k and temperature T, with Eₐ and R as constants. By measuring the rate at multiple temperatures and applying the Arrhenius equation, Eₐ can be calculated. Option A confuses Eₐ with ΔH.
Q8: Condition 1 — Sufficient energy (kinetic energy ≥ activation energy Eₐ): if this condition is not met, the colliding particles do not have enough energy to break the bonds in the reactants and reach the transition state. The particles simply bounce apart elastically with no reaction. Condition 2 — Correct orientation: if this condition is not met, even if the particles have sufficient energy, the collision occurs between the wrong parts of the molecules (not the reactive sites). The atoms cannot rearrange into products, and the particles bounce apart without reacting.
Q9: (a) Exothermic (ΔH = −35 kJ/mol; energy released). (b) Energy reactants at reference level. Transition state peak at 60 kJ above reactant level (this is Eₐ = 60 kJ/mol). Product level at 35 kJ below reactant level (ΔH = −35 kJ/mol). Arrow from reactant level to peak labelled Eₐ = 60 kJ/mol. Arrow from reactant level down to product level labelled ΔH = −35 kJ/mol. (c) If Eₐ = 30 kJ/mol (lower barrier), a greater proportion of particles in the Maxwell-Boltzmann distribution would have kinetic energy ≥ 30 kJ/mol compared to ≥ 60 kJ/mol at the same temperature. More particles can undergo effective collisions per second → the reaction would proceed faster.
Q10: (a) 45°C glow stick is brighter; 5°C glow stick lasts longer. (b) At 45°C, the water bath increases the average kinetic energy of the molecules inside the glow stick (the chemiluminescent reactants). A larger proportion of these molecules now have kinetic energy equal to or greater than the activation energy (Eₐ) of the chemiluminescent reaction. This means a greater proportion of collisions are effective (have both sufficient energy and correct orientation). More effective collisions per second means more reactant is converted to product per second → more light is produced per second → the glow is brighter. (c) Both glow sticks contain the same amount of reactant. The total amount of light produced depends on the total number of product molecules formed, which depends only on the amount of reactant (not the rate). At 45°C, the reactant is consumed quickly at a high rate; at 5°C, the same total amount of reactant is consumed slowly at a low rate. In both cases, all the reactant is eventually consumed and the same total amount of light is produced — just over different time spans.
Climb platforms, hit checkpoints, and answer questions on Collision Theory & Reaction Rate. Quick recall from lessons 1–11.