Loav · VCE Chemistry · Unit 3 / Area of Study 2
U3A2-01 · 2026 curriculum · Expanded original teaching guide · Study guide
1. How quickly is chemical change happening?
A reaction rate describes change in a measurable quantity per unit time. We might follow the decrease in reactant concentration, increase in product concentration, gas volume produced or mass lost when a gas escapes. These measurements are related to reaction progress, but their units differ. A concentration rate may be in mol L⁻¹ s⁻¹; a gas-production rate may be in mL s⁻¹. State which species or observable your rate refers to.
Average product-formation rate is Δ[product]/Δt. A reactant-disappearance rate is −Δ[reactant]/Δt, with the minus sign making a positive rate from a decreasing concentration. Instantaneous rate is the gradient of a tangent at one time. The initial rate is the tangent gradient at the start, not the total product divided by the entire experiment time. Stoichiometric coefficients matter when comparing rates for different species: for 2H₂O₂(aq) → 2H₂O(l) + O₂(g), peroxide disappears twice as rapidly in moles as oxygen forms.
2. Three requirements for a successful collision
Reactant particles must encounter one another, collide in a suitable orientation and have sufficient collision energy for the reaction pathway. Activation energy, Eₐ, is the energy barrier between reactants and the transition state. A collision below this barrier does not lead to that reaction. A high-energy collision can also fail if the reacting parts of molecules are not appropriately positioned. Increasing all collisions and increasing the fraction that succeed are related but different explanations.
Higher concentration places more reactant particles in a given volume, generally increasing collision frequency and the frequency of successful collisions when other conditions remain unchanged. It does not raise their average kinetic energy. For gases at fixed temperature, reducing volume increases partial pressures and concentrations, so particles encounter each other more frequently. Merely adding an inert gas at fixed volume raises total pressure but does not increase the concentrations of the reacting gases.
Increasing temperature raises average particle kinetic energy. More importantly, a larger fraction of collisions can exceed Eₐ; collision frequency also generally increases. Thus explain the change in successful collisions per second, rather than saying only that particles move faster. There is no universal rule that every reaction doubles its rate for each 10°C rise. The magnitude depends on the reaction and temperature range.
Dividing a solid into smaller pieces increases exposed surface area for the same mass. More reactive sites contact the other reactant, increasing successful collisions at the interface. It does not change the amount of solid or increase the concentration inside each piece. Surface-area arguments are relevant to heterogeneous reactions involving surfaces, not automatically to two uniformly mixed solutions.
3. Catalysts change the route
A catalyst participates in reaction steps and is regenerated overall. It provides an alternative pathway with lower activation energy. At the same temperature, more collisions can therefore lead to reaction. The catalyst does not supply heat to every molecule or increase the particles’ average kinetic energy. Catalysts are reaction-specific: a surface or active site must interact appropriately with the reactants. A substance that accelerates one reaction need not accelerate another.
For a reversible reaction, a catalyst accelerates approach to equilibrium in both directions. It does not change Kc or the equilibrium composition at a fixed temperature. An energy-profile horizontal axis is reaction progress, not elapsed time: a narrower drawn peak does not mean a shorter measured reaction time. Real catalytic pathways can have multiple steps and peaks; this schematic emphasises the energy barrier.
4. Open and closed systems
An open system can exchange matter with its surroundings; a closed system cannot, although it may exchange energy. Gas escaping from an open flask can produce a measurable mass decrease and can prevent a reverse reaction involving that gas from establishing equilibrium. A closed container can retain products, allowing reverse reaction rates to become significant. Neither “closed” nor “open” specifies temperature. Control temperature explicitly when investigating another factor.
Worked example 1: compare like with like
Original simulated data: in a fixed-volume solution, [H₂O₂] falls from 0.800 to 0.680 mol L⁻¹ over 40.0 s. Average disappearance rate = −(0.680 − 0.800)/40.0 = 3.00 × 10⁻³ mol L⁻¹ s⁻¹. The equation above gives average O₂ formation rate per solution volume = half this value = 1.50 × 10⁻³ mol L⁻¹ s⁻¹, assuming all generated oxygen is counted. This is an average; it does not determine the initial tangent gradient.
Worked example 2: identify what an energy profile changes
A synthetic profile has reactants at 60 kJ mol⁻¹, products at 20, and an uncatalysed transition state at 150. Forward Eₐ = 150 − 60 = 90 kJ mol⁻¹; reverse Eₐ = 150 − 20 = 130; ΔH = 20 − 60 = −40. A catalytic peak at 110 makes the barriers 50 and 90 respectively. ΔH remains −40. Both barriers decrease by 40; the final equilibrium yield is unchanged at the same temperature.
5. Read a rate experiment without confusing its endpoint
Before comparing curves, name the vertical-axis quantity. A graph of collected gas starts near zero and rises; the mass of an open reacting flask can fall as gas leaves. A steeper downward mass curve means a larger positive rate of mass loss, not a “negative-speed” reaction. A plateau shows that the measured quantity has stopped changing detectably. It does not on its own prove dynamic equilibrium: the limiting reactant may be exhausted, the apparatus may be blocked, or change may be too slow to resolve.
A reaction can be faster but produce less total product. Separate two decisions. First, use collision theory to compare an initial rate under controlled conditions. Second, use stoichiometry and the limiting reactant to predict the final amount if reaction proceeds essentially to completion. Changing acid concentration can affect both rate and available moles; changing particle size at unchanged mass can affect rate without changing the theoretical amount. Check both questions rather than assuming steeper curves must finish higher.
Activation energy is a pathway barrier to reaching a transition state. It should not be equated automatically to the sum of all reactant bond energies: breaking and making interactions can overlap during a reaction. Likewise, a catalyst can participate in intermediate steps even though it is regenerated in the overall process. Its absence from the net stoichiometric equation does not mean it never interacts with a reactant.
For a heterogeneous catalyst, adsorption brings suitable reactants onto active sites; surface interactions can weaken or reorient bonds and open another pathway. Products must leave sites so another catalytic cycle can occur. More exposed catalyst area can provide more accessible sites, but this is different from claiming that every catalyst increases bulk collision frequency. A blocked or poisoned surface may lose activity even when its total mass is unchanged.
Worked example 3: faster reaction, smaller final amount
Challenge, original synthetic comparison. Consider CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l). Run A uses 0.800 g CaCO₃ and 50.0 mL of 0.200 mol L⁻¹ HCl. Run B uses 0.400 g of the same-sized solid particles and 50.0 mL of 0.400 mol L⁻¹ HCl. Temperature is controlled; generated CO₂ is measured at SLC. Use M(CaCO₃) = 100.1 g mol⁻¹.
Amount: A contains 0.800/100.1 = 0.007992 mol carbonate and 0.200 × 0.0500 = 0.0100 mol acid. The acid can react with 0.00500 mol carbonate, so acid limits. V(CO₂) = 0.00500 × 24.8 = 0.124 L. B contains 0.003996 mol carbonate and 0.0200 mol acid; carbonate limits. V(CO₂) = 0.003996 × 24.8 = 0.0991 L. B’s final gas amount is smaller.
Rate: B’s higher acid concentration increases collisions per exposed surface area, but its smaller mass of same-sized particles also provides less total surface area. These changes compete, so the supplied information does not justify a definite overall initial-rate ranking. If a measured B curve is steeper initially, it can still plateau lower. To isolate concentration, keep the total solid mass and size distribution fixed. This is why a qualitative collision explanation must acknowledge all variables, not select the convenient one.
A threshold-time method, such as time to reach a fixed observable endpoint, sometimes uses 1/t as a relative-rate proxy. This is defensible only when each run reaches the same extent or amount at the threshold under comparable observation conditions. It is not automatically a concentration rate in mol L⁻¹ s⁻¹. Changing the viewing depth, endpoint criterion or amount needed to obscure a mark undermines the comparison. Repeats estimate scatter; an unusual reading needs investigation rather than automatic deletion.
Practise, then check
- Gas volume rises from 12.0 to 30.0 mL between 10.0 and 25.0 s. Calculate the average rate.
- Correct this claim: “Doubling concentration doubles the energy of collisions.”
- Two runs reach the same final product amount, but one reaches it sooner. What does this show, and what does it not prove?
- Design a fair comparison of surface area using equal masses of one solid and the same solution reactant. Explain one limitation.
- A monitored gas-producing reaction gives 20.0 mL in the first 10.0 s and another 8.0 mL in the next 10.0 s. Calculate the two average rates. Can either be identified exactly as the initial rate?
- Two runs reach the same defined colour threshold after 36.0 s and 24.0 s. Estimate their relative rates and identify one condition required.
- A catalyst comparison changes catalyst identity, mass and reaction temperature together. All three runs finish at the same gas volume. What conclusion can be drawn, and how would you redesign the comparison?
Worked answers
- ΔV/Δt = (30.0 − 12.0)/(25.0 − 10.0) = 1.20 mL s⁻¹. This is gas-volume production rate over that interval.
- At fixed temperature, concentration increases particles per volume and collision frequency. It does not change the collision-energy distribution in the way a temperature rise does. The rate need not exactly double without additional kinetic information.
- The earlier arrival supports a faster reaction over at least part of the run; equal final amounts support equal observed extent under those conditions. It does not by itself identify a catalyst or establish equal initial rates.
- Keep solid identity and mass, solution concentration/volume, temperature and mixing constant; vary particle size and measure an appropriate initial rate with repeats. Irregular particles make exact exposed area difficult to quantify. Practical work requires teacher-approved equipment and risk assessment.
- First interval: 20.0/10.0 = 2.00 mL s⁻¹; second: 8.0/10.0 = 0.80 mL s⁻¹. The average rate decreases. Neither is exactly the initial rate, which is a tangent gradient at t = 0. The first may approximate it only if the early curve is close to linear.
- Using the same-threshold proxy, rate₂/rate₁ = (1/24.0)/(1/36.0) = 1.50. The same amount or extent of reaction must produce the threshold in both runs. Consistent optical path length and endpoint judgement help make that assumption credible.
- The observed final gas volumes agree; different times would compare overall performance under those combined conditions. They do not isolate catalyst identity. Control initial reactant quantities, concentration, temperature, mixing and a justified catalyst loading; account for accessible surface or active-site differences when interpreting equal masses. Repeat trials and compare gradients or a defined threshold, not only endpoints.
