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Inquiry Question 2: What factors affect equilibrium and how?

Use collision theory to explain and predict the effect of concentration, temperature, surface area and catalysts on the rate at which a system approaches equilibrium

A focused answer to the HSC Chemistry Module 5 dot point on collision theory. How concentration, temperature, surface area and catalysts change reaction rate by changing collision frequency and the fraction of successful collisions, the Maxwell-Boltzmann link to activation energy, and why a faster approach to equilibrium does not change its final position.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-28

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
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In plain English

Imagine a crowded dance floor where people only "match up" (react) if they bump into each other AND they are both moving fast enough AND they happen to be facing the right way when they collide. Collision theory says a chemical reaction is exactly this: particles must collide, with enough energy, in the right orientation, or nothing happens. Cranking up the music and packing more dancers onto the floor (concentration) means more bumps per minute. Turning up the tempo (temperature) does something extra: dancers move faster AND a much bigger share of them are now moving fast enough to properly connect, not just bump and bounce off. Chopping a big block of ice into crushed ice before adding it to a drink (surface area) exposes way more of it to the liquid at once, so more contact happens even though nothing about the ice itself changed. A catalyst is like a shortcut side-door into the venue that avoids a steep, energy-draining staircase, more people get in successfully at the same tempo, but the party itself (the equilibrium position) ends up in exactly the same state, it just gets there faster.

What this dot point is asking

NESA wants you to use collision theory to explain and predict how concentration, temperature, surface area and catalysts change the rate at which a reaction proceeds and, specifically in this Module 5 context, the rate at which a system approaches equilibrium. This is a direct partner to Le Chatelier's principle: Le Chatelier tells you where an equilibrium ends up when disturbed, while collision theory tells you how fast the system gets there, and one of the most commonly tested ideas in Module 5 is that these two things are independent of each other. A catalyst is the clearest illustration: it changes the rate but never the final equilibrium position.

The answer

The three conditions for a reaction to occur

Collision theory states that a chemical reaction between two particles can only occur if their collision satisfies all three conditions:

The particles must collide. No contact, no reaction.
The collision must have sufficient energy at least equal to the activation energy, $E_a$ , the minimum energy needed to begin breaking existing bonds.
The particles must collide with the correct orientation so that the reacting parts of each particle are positioned to form new bonds.

A collision meeting all three conditions is called a successful (or effective) collision. The overall reaction rate depends on how many successful collisions occur per second, which is controlled by two separate quantities:

\text{rate} \propto (\text{collision frequency}) \times (\text{fraction of collisions that are successful})

Each of the four factors below changes the rate by acting on one or both of these quantities.

Concentration (and pressure, for gases)

Increasing the concentration of a solution, or the pressure of a gas mixture, packs more particles into the same volume. This increases the collision frequency because particles are closer together and encounter each other more often. The fraction of collisions that are successful is unchanged, since the average kinetic energy of the particles has not changed. The rate increase comes entirely from more collisions happening per second.

Temperature

Increasing temperature increases the average kinetic energy of the particles. This raises the rate through two compounding effects, not just one:

Collision frequency increases, because faster-moving particles collide more often.
The fraction of successful collisions increases, because the Maxwell-Boltzmann distribution of particle energies shifts to higher energies and broadens, so a much larger fraction of particles now exceeds the activation energy $E_a$ (see the shaded regions in the figure above).

Because both effects act together, a modest temperature rise (a commonly quoted approximation is that a 10 degrees C rise roughly doubles the rate for many reactions) typically increases rate far more than an equivalent percentage increase in concentration, which only increases collision frequency.

Surface area

For a reaction involving a solid, breaking the solid into smaller pieces (for example, crushing a lump into powder) increases its total exposed surface area without changing its mass, its concentration, or the concentration of the other reactant. Only particles at an exposed surface can collide with the other reactant, so a larger surface area increases the collision frequency between the solid and the other reactant. As with concentration, the fraction of successful collisions is unchanged, only how often a collision can happen at all.

Catalysts

A catalyst provides an alternative reaction pathway with a lower activation energy, $E_a$ . At the same temperature, this means a larger fraction of collisions now have enough energy to react, since fewer particles need to clear a lower energy barrier. A catalyst:

increases the rate of both the forward and reverse reactions equally,
is not consumed in the overall reaction (it is regenerated at the end of the mechanism),
does not change $\Delta H$ (the energy difference between reactants and products stays the same), and
does not shift the position of equilibrium or change the value of $K_{eq}$ , it only reduces the time taken to reach the same equilibrium.

Rate factors at a glance

Concentration or pressure up -> collision frequency up. Fraction of successful collisions unchanged.
Temperature up -> collision frequency up AND fraction of successful collisions up (Maxwell-Boltzmann shift). The strongest single lever on rate.
Surface area up (smaller particle size) -> collision frequency up. Fraction of successful collisions unchanged. Total mass, moles and final yield are unaffected.
Catalyst added -> lowers $E_a$ for both forward and reverse pathways equally, raising the fraction of successful collisions at a fixed temperature. Rate up; position of equilibrium, $K_{eq}$ , and $\Delta H$ all unchanged.
Rate versus equilibrium position: collision theory (this dot point) explains how FAST equilibrium is reached; Le Chatelier's principle explains WHERE it ends up. Only a temperature change affects both.

Worked examples

Worked example 1: doubling concentration.

For a reaction $A + B \rightarrow \text{products}$ , doubling the concentration of $A$ (with $B$ and temperature unchanged) doubles the frequency of collisions involving $A$ particles. Since the fraction of those collisions that are successful is set by temperature and orientation requirements, which have not changed, the reaction rate increases (the exact factor depends on the rate law, but the mechanism increasing it is collision frequency alone).

Marker's note: full marks need the answer to explicitly separate "more collisions" from "no change to the fraction of successful collisions"; simply writing "more collisions, more reaction" without this distinction is treated as a partial (surface-level) answer at this level.

Worked example 2: the industrial choice of a catalyst over higher temperature alone.

In the contact process, $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ , $\Delta H = -198\ \text{kJ mol}^{-1}$ , operators want a fast rate without sacrificing the equilibrium yield of $SO_3$ . Simply raising temperature would speed up the reaction, but because the forward reaction is exothermic, Le Chatelier's principle predicts higher temperature also shifts the equilibrium left, lowering yield. Adding a $V_2O_5$ catalyst instead lowers $E_a$ for both directions, increasing the rate at a chosen (moderate) temperature without moving the equilibrium position or changing $K_{eq}$ . This is why the catalyst is chosen in addition to, not instead of, a moderate operating temperature (see Le Chatelier's principle for the full temperature/pressure trade-off).

Marker's note: top marks require explicitly stating that the catalyst leaves $K_{eq}$ and the equilibrium position unchanged, since this is the most commonly tested contrast between a catalyst and a temperature change in Module 5 extended response.

Worked example 3: reading a rate-vs-time graph for two particle sizes.

Two identical masses of the same solid are reacted with excess acid, one as a single lump and one as a fine powder, at the same temperature. A rate-vs-time (or volume-of-gas-vs-time) graph shows the powder's curve rising more steeply at first and reaching the same final (plateau) volume sooner than the lump's curve, but both curves plateau at the SAME final volume. The steeper initial gradient shows a faster rate (more collision frequency from greater surface area); the equal final volume shows the total amount of product is set by the (equal) moles of limiting reagent, not by surface area.

Marker's note: a common error is concluding the powder produces MORE gas. The correct reading is that it produces the SAME amount of gas FASTER; surface area affects only the gradient (rate), never the final height (yield) of the curve for identical starting amounts.

Common traps

Saying a catalyst "speeds up the reaction so more product forms": A catalyst changes the rate, not the equilibrium yield; more product forms sooner, not more product overall.
Describing temperature's effect using only collision frequency: This misses the Maxwell-Boltzmann effect (fraction of successful collisions) and typically loses a mark in a "using collision theory" question.
Confusing surface area with concentration: Crushing a solid changes surface area (collision frequency via exposed particles), not the concentration of dissolved reactant in solution; keep the two factors and their justifications separate.
Believing a catalyst lowers $\Delta H$: A catalyst lowers $E_a$ (the height of the energy barrier), not the energy difference between reactants and products.
Treating "successful collision" as only needing enough energy: Orientation is a separate, equally required condition; a fast, energetic collision at the wrong angle still fails.

Examples in context

Example 1. Iron ore sintering feed preparation at Port Kembla. Before iron ore is fed into a blast furnace, it is often crushed and sintered into small, porous granules rather than left as large lumps. Increasing the exposed surface area increases the collision frequency between the ore and the reducing gases (CO) flowing through the furnace, speeding up the reduction reactions without needing any change in gas concentration or furnace temperature. This is a direct industrial application of the surface-area factor in collision theory.

Example 2. Vanadium(V) oxide catalyst beds in the contact process. Sulfuric acid manufacturers such as Incitec Pivot near Port Kembla pass $SO_2$ and $O_2$ over solid $V_2O_5$ catalyst beds rather than relying on temperature alone to achieve a workable rate. Because the catalyst lowers the activation energy for both the forward and reverse reactions equally, a commercially useful rate is reached at a moderate 450 degrees C, a temperature chosen mainly to protect the equilibrium yield (per Le Chatelier), with the catalyst supplying the additional rate that a colder, higher-yield equilibrium alone could not achieve quickly enough.

Try this

Q1. Using collision theory, explain why crushing a solid reactant into a powder increases reaction rate but does not increase the total mass of product formed. [3 marks]

Cue. Greater surface area increases collision frequency (more exposed particles), speeding the rate; the total moles of limiting reagent are unchanged, so the final yield is the same.

Q2. Explain, with reference to the Maxwell-Boltzmann distribution, why a 10 degree C rise in temperature typically has a much larger effect on reaction rate than a 10 percent increase in concentration. [4 marks]

Cue. Temperature increases both collision frequency and the fraction of particles exceeding $E_a$ (distribution shifts and broadens); concentration increases only collision frequency.

Q3. A catalyst is added to a reaction that has already reached equilibrium. State and justify the effect, if any, on (a) the forward and reverse reaction rates, (b) the position of equilibrium, (c) the value of $K_{eq}$ . [1+1+1 marks]

Cue. (a) No change, since the system is already at equilibrium (rates already equal); if disturbed and re-equilibrating, a catalyst speeds up both rates equally. (b) No change. (c) No change.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksUsing collision theory, explain why increasing the concentration of hydrochloric acid increases the rate of its reaction with magnesium ribbon.

Show worked solution →

A 3-mark explain needs the mechanism linking concentration to rate through collision frequency.

Increasing the concentration of $HCl$ increases the number of $H^+$ ions per unit volume of solution. With more acid particles present, the frequency of collisions between $H^+$ ions and the magnesium surface increases. Since the fraction of collisions that are successful (with energy at least $E_a$ and correct orientation) is unchanged, a higher collision frequency directly produces a higher rate of successful collisions per second, so the reaction proceeds faster.

Marking criteria: 1 mark for stating collision frequency increases with concentration, 1 mark for linking this to more particles per unit volume, 1 mark for explicitly noting the fraction of successful collisions is unchanged (only the frequency changes).

foundation4 marksTwo identical masses of calcium carbonate are reacted with excess dilute hydrochloric acid at the same temperature: one as a single large lump, one as a fine powder. (a) Which reacts faster? (b) Explain why, using collision theory. (c) State one variable that must be controlled for this to be a fair comparison.

Show worked solution →

(a) Faster reaction: The powdered calcium carbonate reacts faster.
(b) Explanation: Crushing the lump into powder greatly increases its total surface area exposed to the acid. Since acid particles can only collide with $CaCO_3$ particles that are at an exposed surface, the greater surface area increases the frequency of collisions between $H^+$ ions and $CaCO_3$ per second. The fraction of those collisions that are successful is unchanged, so the increase in rate comes entirely from more collisions occurring, not from each collision being more likely to succeed.
(c) Controlled variable: The mass of calcium carbonate, the concentration and volume of acid, and the temperature must all be kept identical between the two trials (any one is an acceptable answer), so surface area is the only variable being tested.

Marking criteria: 1 mark for correctly identifying the powder as faster, 2 marks for the collision-theory explanation (surface area to collision frequency, with the fraction of successful collisions stated as unchanged), 1 mark for a valid controlled variable.

core5 marksThe Maxwell-Boltzmann distribution below is an owned illustrative graph comparing the same reaction mixture at

T_1

and a higher temperature

T_2

, with the activation energy

E_a

marked. (a) Describe two differences between the

T_2

curve and the

T_1

curve. (b) Explain, using the shaded regions beyond

E_a

, why increasing temperature increases the reaction rate more than the collision-frequency effect alone would predict.

Show worked solution →

(a) Two differences. The $T_2$ curve is shifted to the right (its peak sits at a higher kinetic energy) and is flatter and broader than the $T_1$ curve, reflecting a wider spread of particle energies at the higher temperature. The area under both curves is the same, since the total number of particles has not changed.

(b) Explanation. The shaded area beyond $E_a$ under the $T_2$ curve is noticeably larger than the shaded area beyond $E_a$ under the $T_1$ curve, even though only a modest temperature rise has occurred. This shows that a larger fraction of particles now has energy at least equal to $E_a$ at $T_2$ . Because faster-moving particles also collide more frequently, temperature increases the rate through two compounding effects at once, more collisions per second AND a greater fraction of those collisions being energetic enough, whereas a concentration or surface-area increase raises only the collision frequency. This is why a modest temperature rise (for example 10 degrees C) typically increases rate far more than a similarly modest increase in concentration.

Marking criteria: (a) 1 mark each for the rightward shift and the flattening/broadening (max 2). (b) 1 mark for correctly reading that the shaded area beyond $E_a$ is larger at $T_2$ , 1 mark for stating this means a greater fraction of particles can now react, 1 mark for explicitly contrasting this "two compounding effects" mechanism with the single collision-frequency effect of concentration or surface area.

core4 marksA 5.00 g sample of zinc granules reacts completely with excess dilute sulfuric acid, releasing hydrogen gas. If the same mass of zinc were used as zinc powder instead, the reaction is observed to reach completion in one quarter of the time, releasing the same total volume of gas measured at the same temperature and pressure. Calculate the volume of hydrogen gas released at that temperature and pressure, given the molar volume of a gas is

24.79\ \text{L mol}^{-1}

at 25 degrees C and 100 kPa, and

M(\text{Zn}) = 65.38\ \text{g mol}^{-1}

. State why the volume of gas produced is the same for both the granules and the powder.

Show worked solution →

Step 1: write the balanced equation.

\text{Zn}_{(s)} + \text{H}_2\text{SO}_{4(aq)} \rightarrow \text{ZnSO}_{4(aq)} + \text{H}_{2(g)}

One mole of zinc produces one mole of hydrogen gas.

Step 2: moles of zinc.

n(\text{Zn}) = \frac{m}{M} = \frac{5.00\ \text{g}}{65.38\ \text{g mol}^{-1}} = 0.076476\ \text{mol}

Step 3: moles of hydrogen gas (1:1 ratio).

n(\text{H}_2) = 0.076476\ \text{mol}

Step 4: volume of hydrogen gas at 25 degrees C, 100 kPa.

V = n \times V_m = 0.076476\ \text{mol} \times 24.79\ \text{L mol}^{-1} = 1.8961\ \text{L}

Step 5: round to 3 significant figures (matching 5.00 g).

V(\text{H}_2) = 1.90\ \text{L}

Why the volume is the same for both forms of zinc. The total amount (moles) of zinc reacted is identical in both cases, since the mass and the acid (in excess) are the same, so by the 1:1 stoichiometry the total moles, and hence the total volume, of hydrogen produced is identical. Powdering the zinc only increases its surface area, which increases the collision frequency between zinc and acid particles and so increases the RATE at which the gas is produced; it does not change the total amount of gas produced once the reaction has gone to completion.

Marking criteria: 1 mark for the correct 1:1 mole ratio from the balanced equation, 1 mark for correct moles of zinc, 1 mark for the volume calculation to 3 significant figures with units, 1 mark for correctly explaining that surface area affects rate (collision frequency) but not the total yield of gas.

exam6 marksCompare and contrast the effect of increasing temperature with the effect of adding a catalyst on the rate of approach to equilibrium for the reaction

2\text{SO}_{2(g)} + \text{O}_{2(g)} \rightleftharpoons 2\text{SO}_{3(g)}

\Delta H = -198\ \text{kJ mol}^{-1}

, using collision theory and an energy profile diagram in your answer.

Show worked solution →

This is a 6-mark COMPARE AND CONTRAST: markers reward correctly identifying what the two disturbances share and where they genuinely differ, using collision theory throughout.

Band 6 plan.

Similarity: both increasing temperature and adding a catalyst increase the RATE at which equilibrium is reached, and both do so by increasing the number of successful collisions per second.
Mechanism of temperature: raising temperature increases the average kinetic energy of $SO_2$ and $O_2$ particles, increasing both the collision frequency and (via the Maxwell-Boltzmann distribution) the fraction of collisions with energy at least $E_a$ . Both the forward and reverse reaction rates increase, but because the forward reaction is exothermic, the reverse (endothermic) rate increases more, so the equilibrium POSITION shifts left and Keq decreases.
Mechanism of catalyst: a catalyst (here $V_2O_5$ ) provides an alternative reaction pathway with a lower $E_a$ , so a larger fraction of collisions at the SAME temperature now have enough energy to react. Crucially, it lowers $E_a$ equally for the forward and reverse reactions, so both rates increase by the same factor; the equilibrium position and Keq are unchanged, only the time to reach equilibrium is reduced.
Contrast, stated explicitly: temperature changes both the RATE and the POSITION (and Keq); a catalyst changes only the RATE, never the position or Keq.
Energy profile diagram reference: on the profile, raising temperature does not change the height of the activation-energy peak or the height difference between reactants and products ( $\\Delta H$ ); it only changes how many particles have enough energy to climb the (unchanged) peak. A catalyst lowers the height of the peak itself (for both directions) without changing the reactant or product energy levels, so $\\Delta H$ is unchanged.

Model paragraph (excerpt). Both raising temperature and adding a vanadium(V) oxide catalyst increase the rate at which the contact-process equilibrium is reached, but they act on collisions in fundamentally different ways. Raising temperature increases the average kinetic energy of $SO_2$ and $O_2$ particles, increasing both collision frequency and the Maxwell-Boltzmann fraction of collisions exceeding the fixed activation energy; because the forward reaction is exothermic, its reverse (endothermic) rate is boosted more than the forward rate, so the equilibrium position shifts left and Keq falls. A catalyst instead lowers the activation energy of both the forward and reverse pathways by the same amount, at a fixed temperature, so a larger fraction of the same population of collisions becomes successful in both directions equally; the position of equilibrium and the value of Keq are therefore unaffected, and only the time taken to reach that unchanged position is reduced.

Marker's note: top-band answers (1) explicitly state the shared outcome (both increase rate), (2) correctly explain temperature acting on the energy DISTRIBUTION while surface area/concentration act only on collision FREQUENCY, (3) state the key contrast that temperature alone changes Keq and equilibrium position while a catalyst changes neither, and (4) refer concretely to the energy profile (peak height for $E_a$ , unchanged $\\Delta H$ for a catalyst) rather than describing it only in words.

exam5 marksOn the energy profile diagram below (owned illustrative figure showing the reaction pathway with and without a catalyst), a student claims: 'The catalyst lowers the energy of the products, which is why the reaction becomes faster.' Assess the accuracy of this claim, referring to specific features of the diagram.

Show worked solution →

The claim is incorrect, and a strong answer must identify exactly where the reasoning goes wrong using the diagram.

What the diagram actually shows: Both the catalysed (dashed) and uncatalysed (solid) pathways start at the same reactant energy level and end at the same product energy level; the vertical drop from reactants to products, representing $\Delta H$ , is identical for both curves. The only difference between the two curves is the HEIGHT OF THE PEAK: the catalysed pathway's peak is lower, meaning a smaller activation energy, $E_a$ , for the catalysed route.
Why the reaction is faster: The reaction is faster because a smaller $E_a$ means a larger fraction of collisions, at the same temperature, now has enough energy to react (via the Maxwell-Boltzmann distribution), not because the products themselves have lower energy. If the catalyst genuinely lowered the product energy without changing the peak height in the same way, $\Delta H$ would have changed, which never happens for a true catalyst.
Corrected statement: "The catalyst lowers the activation energy of the reaction pathway (the height of the peak), not the energy of the products, which increases the fraction of successful collisions at a given temperature and so increases the rate."

Marking criteria: 1 mark for identifying the claim as incorrect, 2 marks for correctly locating the actual change as the peak height ( $E_a$ ) rather than the product energy level, 1 mark for linking the lower $E_a$ to a larger fraction of successful collisions, 1 mark for explicitly noting $\Delta H$ (and hence product energy) is unchanged by a catalyst.