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How can the yield of a chemical product be optimised?
the factors that affect the rate of a chemical reaction (concentration, surface area, temperature and the presence of a catalyst) explained using collision theory and the Maxwell-Boltzmann distribution of kinetic energies, including the representation of these effects on energy profile diagrams
A focused VCE Chemistry Unit 3 answer on rate of reaction. Covers collision theory, the four factors that affect rate (concentration, surface area, temperature, catalyst), the Maxwell-Boltzmann distribution, activation energy, and energy profile diagrams.
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What this dot point is asking
VCAA wants the collision theory model, the four factors that affect reaction rate (concentration, surface area, temperature, catalyst), the Maxwell-Boltzmann distribution of kinetic energies, and how each factor is represented on an energy profile diagram or Maxwell-Boltzmann diagram.
The answer
Collision theory
Collision theory says that for a reaction to occur, reactant particles must:
- Collide with each other.
- Collide with enough kinetic energy to overcome the activation energy barrier (Ea).
- Collide with the correct orientation so that bonds can break and re-form.
Only a small fraction of collisions meet all three conditions; these are called successful collisions or fruitful collisions. The rate of a reaction is proportional to the frequency of successful collisions per unit time.
The four factors
| Factor | What changes | Why rate increases |
|---|---|---|
| Concentration (or pressure for gases) | More particles per unit volume | More frequent collisions per second |
| Surface area (solids) | More exposed particles per unit mass | More frequent collisions at the interface |
| Temperature | Higher mean kinetic energy | More particles exceed Ea; also slightly more frequent collisions |
| Catalyst | Lower activation energy | A larger fraction of collisions has enough energy to react |
Concentration. Doubling the concentration of a reactant roughly doubles the rate (for a first-order dependence). More particles in a given volume means more collisions per unit time.
Surface area. Only the particles at the surface of a solid can collide with the other reactant. Grinding a solid into powder increases the surface area dramatically and so increases the rate. This is why a flour mill is more explosive than a flour sack (huge surface area exposed to air).
Temperature. Raising the temperature has two effects: (1) particles move faster so collide more often (small effect), and (2) more particles have enough energy to overcome Ea (large effect). Effect (2) dominates because the Maxwell-Boltzmann fraction above Ea increases exponentially with temperature. Hence the rule of thumb that a 10°C rise roughly doubles rate.
Catalyst. A catalyst provides an alternative pathway with a lower Ea. The reactants bind to the catalyst surface or active site, the bonds rearrange, and the products leave. The catalyst is regenerated. It does not change ΔH (the reactants and products are unchanged) and does not shift the equilibrium position (it speeds up both forward and reverse reactions equally).
The Maxwell-Boltzmann distribution
The Maxwell-Boltzmann distribution plots the number of particles (y-axis) against kinetic energy (x-axis) for a sample at a given temperature.
Key features:
- The curve starts at the origin (no particles with zero kinetic energy).
- It rises to a peak (the most probable kinetic energy).
- It tails off to the right with a long high-energy tail.
- The area under the curve is the total number of particles (constant for a given sample).
The activation energy Ea is a vertical line. The area to the right of Ea is the number of particles with enough kinetic energy to react on collision.
How the distribution changes
- Higher temperature: peak shifts right (higher mean energy) and flattens (broader spread). The high-energy tail above Ea grows substantially. The area under the curve stays the same because particle count does not change.
- Adding a catalyst: the curve itself does not change (same temperature, same particles). The Ea line shifts left to a lower value, so more particles now lie to the right of it.
A common Section B question asks for a sketch of the Maxwell-Boltzmann curve at two temperatures with the Ea line marked, or for the same curve with two Ea lines (catalysed and uncatalysed).
Energy profile diagrams
An energy profile diagram plots potential energy (y-axis) against reaction progress (x-axis).
Features:
- The reactants sit on the left at one energy level.
- The products sit on the right at another energy level.
- A peak in between is the transition state.
- The height from reactants to peak is the activation energy (Ea) of the forward reaction.
- The vertical difference between reactants and products is ΔH (negative for exothermic, positive for endothermic).
- A catalysed pathway shows a lower peak (lower Ea) but the same reactant and product energies (so the same ΔH).
A catalyst draws a smaller hill in front of the same valley. ΔH does not change.
How each factor shows up on diagrams
| Factor | Effect on energy profile diagram | Effect on Maxwell-Boltzmann diagram |
|---|---|---|
| Concentration | No change (energies unchanged) | No change |
| Surface area | No change | No change |
| Temperature | No change (Ea and ΔH unchanged) | Peak shifts right, curve flattens, more area above Ea |
| Catalyst | Lower peak (lower Ea); ΔH unchanged | No change in curve; Ea line shifts left, more area to the right of it |
Note that concentration and surface area change the collision frequency, which is not visible on either of these diagrams. They are best discussed in words.
Worked example
A student investigates the rate of reaction between calcium carbonate and dilute HCl, measuring the volume of CO2 produced per minute. Predict and explain the effect of (a) using powdered marble instead of marble chips, and (b) doubling the HCl concentration.
(a) Powdered marble: rate increases significantly. Powdering increases the surface area of CaCO3, so more carbonate ions are exposed to collisions with H^+ ions per unit time. More frequent collisions per unit time means a faster rate.
(b) Doubling [HCl]: rate roughly doubles (assuming first order in HCl, which is the case here). Doubling the concentration of H^+ ions doubles the collision frequency between H^+ and CaCO3 per unit time, so the rate doubles.
Neither change alters the activation energy or ΔH; both work by increasing collision frequency rather than by changing the energy barrier.
Common traps
Saying "particles move faster" as the only temperature effect. Markers want the Maxwell-Boltzmann distribution argument: the fraction of particles above Ea increases.
Drawing the Maxwell-Boltzmann curve starting on the y-axis. It must start at the origin (no particles with zero kinetic energy).
Drawing the high-temperature curve with a larger area. The area under the curve is the total number of particles, which is constant. The curve flattens, broadens and shifts right, but the area stays the same.
Saying a catalyst lowers ΔH or shifts equilibrium. A catalyst lowers Ea only. ΔH and equilibrium position are unchanged.
Confusing rate with extent. Rate is "how fast"; extent is "how far". A catalyst increases rate but does not change extent; raising temperature on an exothermic reaction increases rate but reduces extent.
Drawing the catalysed energy profile with different reactant or product levels. The catalyst lowers the peak only; the reactant and product energies stay the same.
In one sentence
The rate of a reaction is the frequency of successful collisions per unit time, increased by concentration, surface area and temperature (mostly via the rightward shift of the Maxwell-Boltzmann distribution above Ea) and by a catalyst (which provides an alternative pathway with a lower Ea without changing ΔH).
Past exam questions, worked
Real questions from past VCAA papers on this dot point, with our answer explainer.
2024 VCE4 marksUse collision theory and the Maxwell-Boltzmann distribution to explain why increasing temperature from 25°C to 35°C roughly doubles the rate of many reactions.Show worked answer →
A 4-mark answer needs the Maxwell-Boltzmann shape change, the fraction-above-Ea argument, the collision-frequency point and the rule of thumb.
- The Maxwell-Boltzmann distribution shows the spread of kinetic energies of particles in a sample. At 25°C, only a small fraction of particles have kinetic energy greater than the activation energy (Ea).
- Raising temperature to 35°C shifts the distribution to the right and flattens it, so the fraction of particles with energy greater than Ea increases sharply.
- There is also a small increase in the frequency of collisions (faster particles collide more often), but this effect is minor compared with the Ea effect.
- The "rule of thumb" that a 10°C rise roughly doubles rate reflects the exponential dependence: the Arrhenius-type behaviour means the fraction of successful collisions roughly doubles for many reactions, dwarfing the small change in collision frequency.
Markers reward the distribution shift and the Ea fraction as the dominant cause; an answer that says "particles move faster so they collide more" without the Ea argument scores at most 2 of 4.
2025 VCE2 marksExplain how a catalyst increases the rate of a chemical reaction without being consumed.Show worked answer →
A 2-mark answer needs the alternative-pathway point and the not-consumed point.
A catalyst provides an alternative reaction pathway with a lower activation energy (Ea). On a Maxwell-Boltzmann diagram, the fraction of particles with kinetic energy greater than the (lower) Ea is larger, so the fraction of successful collisions per unit time is larger and the rate increases.
The catalyst takes part in the reaction (often binding reactants and lowering the energy of the transition state) but is regenerated unchanged at the end, so it is not consumed and continues to catalyse further turnovers. A catalyst does not change ΔH or the equilibrium position; it only changes the rate.
Related dot points
- the writing of thermochemical equations to represent the energy released or absorbed in physical and chemical changes, including the sign convention for ΔH for exothermic and endothermic reactions, and the use of ΔH values with mole ratios to calculate the energy released or absorbed
A focused VCE Chemistry Unit 3 answer on thermochemical equations. Covers the meaning of ΔH, the sign convention for exothermic and endothermic reactions, the use of states in the equation, and how to scale ΔH using mole ratios.
- the concept of dynamic equilibrium for reversible reactions, the equilibrium law expression and equilibrium constant Kc (including the meaning of Q vs Kc and the units of Kc), and the qualitative application of Le Chatelier's principle to predict the effect on equilibrium of changes in concentration, gas pressure (volume), temperature and the addition of a catalyst
A focused VCE Chemistry Unit 3 answer on equilibrium. Covers dynamic equilibrium, the equilibrium expression and Kc, the role of the reaction quotient Q, and Le Chatelier's principle for changes in concentration, pressure, temperature, and the addition of a catalyst.