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

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:

1. Collide with each other.
2. Collide with enough kinetic energy to overcome the activation energy barrier (Ea).
3. 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

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

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.

Examples in context

Example 1. Catalytic converter on a Toyota Camry built at Altona. Toyota's Altona plant (closed 2017) supplied Camry models with three-way catalytic converters. The platinum-rhodium-palladium catalyst lowers the activation energy for $2 \text{NO} \to \text{N}_2 + \text{O}_2$ from $\sim 280 \, \text{kJ/mol}$ to $\sim 60 \, \text{kJ/mol}$. At exhaust temperature $\sim 600 \, \text{K}$, the fraction of molecules with $E > E_a$ rises from $\sim 10^{-25}$ to $\sim 10^{-6}$, increasing rate by a factor of $\sim 10^{19}$. The same converter oxidises $\text{CO}$ to $\text{CO}_2$ and unburnt hydrocarbons to $\text{CO}_2 + \text{H}_2 \text{O}$. Without the catalyst, the engine's $\sim 800^{\circ}\text{C}$ exhaust would still emit toxic $\text{NO}_x$; with it, emissions meet Euro 5 standards.

Example 2. Yeast fermentation rate at Mornington Peninsula wineries. Winemakers control fermentation rate by managing yeast cell concentration and tank temperature. The rate of glycolysis: $\text{glucose} \to 2 \, \text{ethanol} + 2 \, \text{CO}_2$ has overall activation energy $\sim 50 \, \text{kJ/mol}$ from the enzyme catalysis of hexokinase, pyruvate decarboxylase and alcohol dehydrogenase. Raising temperature from $12^{\circ}\text{C}$ to $18^{\circ}\text{C}$ roughly doubles the rate (Arrhenius prediction). Above $30^{\circ}\text{C}$ the rate stops increasing because enzymes denature. Mornington Peninsula's cool nights and warm days suit a slow, controlled fermentation that preserves delicate aromas; tank-jacket cooling holds the must at $13$ to $15^{\circ}\text{C}$ for $14$ to $21$ days.

Try this

Q1. Using collision theory, explain why a powdered solid reacts faster than a single chunk with the same mass. [2 marks]

• Cue. Powder has greater surface area; more particles exposed for collision; rate of successful collisions per second increases.

Q2. A reaction at $25^{\circ}\text{C}$ takes $400 \, \text{s}$ to complete. At $45^{\circ}\text{C}$ it takes $50 \, \text{s}$. (a) Calculate the rate ratio. (b) Estimate the activation energy using the rule of thumb that rate roughly doubles every $10^{\circ}\text{C}$. (c) Sketch the Maxwell-Boltzmann curves at both temperatures. [3 marks]

• Cue. (a) Rate ratio $= 400 / 50 = 8 \approx 2^3$, consistent with two $\times 10^{\circ}\text{C}$ steps doubling. (b) About $E_a \sim 50 \, \text{kJ/mol}$ (from Arrhenius rough estimate). (c) Higher T curve broader and shifted right; same total area.

Q3. A catalyst is added to the reaction $2 \text{H}_2 \text{O}_2 \to 2 \text{H}_2 \text{O} + \text{O}_2$. (a) State the effect on $E_a$, rate and $\Delta H$. (b) Sketch the energy profile with and without catalyst. (c) Explain why MnO$_2$ powder works better than MnO$_2$ lumps. [2+2+2 marks]

• Cue. (a) $E_a$ decreases; rate increases; $\Delta H$ unchanged. (b) Two peaks: catalysed peak lower; reactant and product levels same. (c) More surface area; more active sites for $\text{H}_2 \text{O}_2$ adsorption.