Factors affecting rate — concentration temperature surface area — Chemistry

Before this

Collision theory
The particle model

This leads to

The same reaction, at very different speeds

Drop a lump of marble into dilute hydrochloric acid. It fizzes steadily, and the lump takes many minutes to disappear. Now grind the same mass of marble to a powder and add it to the same acid. It froths up at once and is gone in moments. Same marble, same acid, same products:

CaCO₃(s) + 2HCl(aq) \to CaCl₂(aq) + H₂O(l) + CO₂(g)

A log in a fireplace burns quietly for an hour. Wood dust floating in the air of a sawmill can explode. The chemistry is the same. Only the speed has changed.

The last lecture gave you one sentence that explains every case like this. It said that rate depends on three things: how many collisions happen each second, what fraction of them have at least the activation energy, and what fraction meet the right way round. It promised that the next lectures would use that sentence "to explain each way of speeding a reaction up." This lecture keeps that promise for concentration, surface area and temperature. It also introduces the fourth way, catalysts.

The sentence from collision theory

Rate ∝ (collisions per second) × (fraction with energy ≥ $E_{a}$ ) × (fraction with the right orientation)

For every factor below, ask one question: which of these three does it change?

Concentration: more particles in the same space

The concentration of a solution tells you how many particles of the solute are in each cubic decimetre. Doubling the concentration of an acid puts twice as many acid particles into every cubic decimetre.

Why does that matter? A reaction between A and B needs an A particle to meet a B particle. Look at the figure. With 4 A and 4 B in a box, there are $4 \times 4 = 16$ possible A–B pairs. With 8 A and the same 4 B, there are 32. Twice as many A particles means twice as many A–B meetings every second.

Two boxes of the same volume at the same temperature. Doubling the number of A particles doubles the number of A–B pairs that can meet, so A and B collide twice as often. The energy and orientation of each collision do not change.

Notice what does not change. Each collision is just as energetic as before, and just as likely to be the right way round, because the temperature and the particles are the same. Only the first factor, collisions per second, goes up. So the number of successful collisions per second goes up, and the rate rises.

For gases, pressure does the same job. Squeeze a gas into half the volume and its particles are twice as crowded: the pressure doubles, and so does the number of particles in each cubic decimetre. A higher pressure of a reacting gas is a higher concentration.

Be careful about one word. A higher concentration usually raises the rate, but not always by the same factor. For many reactions, doubling one concentration roughly doubles the rate. For some it quadruples it, and for a few it changes nothing at all, because of the way the reaction happens step by step. At Pre-University, rate equations measure this exactly. At this level, say "more collisions per second", and leave the exact factor to experiment.

Stop and think

A student writes: "In a more concentrated acid, the particles collide harder, so more collisions have enough energy." What is wrong with this?

Show the answer

Concentration does not change how hard particles collide. Their energies depend on the temperature, which has not changed. So the fraction of collisions with at least $E_{a}$ stays the same. What changes is how many collisions happen each second. More collisions per second, with the same fraction successful, gives more successful collisions per second. Mixing these two ideas up is one of the most common ways to lose marks on this topic.

Surface area: only the outside can react

When one reactant is a solid, the other reactant can reach only the particles on its surface. The particles inside the lump are hidden. They wait their turn until the layers above them have reacted away.

So breaking a solid into smaller pieces exposes more particles and gives more collisions per second. The mass stays the same, but the area grows. Take a 1 cm cube. Its surface area is $6 \times 1 \times 1 = 6 cm^{2}$ . Cut it into 1 mm cubes. There are now $10 \times 10 \times 10 = 1000$ cubes, each with an area of $6 \times 0.1 \times 0.1 = 0.06 cm^{2}$ . The total is $1000 \times 0.06 = 60 cm^{2}$ , ten times as much. Cut each edge into $n$ pieces and the area is multiplied by $n$ . A fine powder can have hundreds or thousands of times the area of the lump it came from.

Again, ask which factor changed. The energy of each collision and its orientation are unchanged. Only the number of collisions per second rises. Surface area matters only when the reactants are in different states, such as a solid in a solution or a solid in a gas. If both reactants are dissolved in the same solution, every particle is already exposed, and there is no surface to enlarge.

Temperature: the factor that does two things

Heating changes two of the three factors at once. This is the hard part of the lecture. Spend time here.

First, more collisions. Hotter particles move faster, so they meet more often. But the effect is small. The average speed of particles grows with the square root of the kelvin temperature. Warming from 25 °C (298 K) to 35 °C (308 K) raises it by a factor of $308 \div 298 \approx 1.017$ : less than 2% more collisions.

Second, more collisions with enough energy. At any temperature, a few particles have far more energy than average. They are the ones whose collisions can succeed. A small rise in temperature adds a lot to that small group. The last lecture used a Pre-University formula for the fraction of collisions with at least $E_{a}$ . For $E_{a} = 50$ kJ/mol, it gives about $1.7 \times 1 0^{- 9}$ at 25 °C and about $3.3 \times 1 0^{- 9}$ at 35 °C. That fraction has almost doubled.

Put the two together: $1.017 \times 1.92 \approx 1.96$ . The rate nearly doubles, and almost all of that comes from the second effect. That is it. That is the whole idea. Heating works mainly by giving more collisions the energy to get over the barrier, not by making more collisions.

You may have read that "a 10 °C rise doubles the rate". That is true only for reactions with an activation energy near 50 kJ/mol. A higher barrier makes the rate more sensitive to temperature, and a lower barrier makes it less sensitive. Try it on the graph below.

Graph builder — how temperature changes the rate

The solid line shows the rate at each temperature, compared with the rate at 25 °C. The dashed line shows what the rate would be if heating only made collisions more frequent. Choose an activation energy, then move the temperature and read the two factors.

Activation energy, Eₐ: 50 kJ/mol

Temperature: 35 °C

How much faster a reaction goes than at 25 °C, using the collision-theory model. The first factor is the rise in collisions per second. The second is the rise in the fraction of collisions with at least the activation energy.

Eₐ in kJ/mol	Change	More collisions	More with enough energy	Rate
20	25 → 35 °C	× 1.017	× 1.30	× 1.32
50	25 → 35 °C	× 1.017	× 1.92	× 1.96
100	25 → 35 °C	× 1.017	× 3.70	× 3.76
50	25 → 55 °C	× 1.049	× 6.32	× 6.63
100	25 → 55 °C	× 1.049	× 40.0	× 41.9

All values are rounded. The collision factor hardly changes with $E_{a}$ . The energy factor depends strongly on it.

What should you have seen? The dashed line hardly rises: more frequent collisions alone would make a 10 °C rise worth less than 2%. The solid line climbs steeply, and it climbs more steeply for a higher activation energy. Cooling works the same way in reverse. Below 25 °C the rate falls, and it falls fastest for reactions with high barriers.

Trap 1 — "Heating lowers the activation energy"

It does not. The activation energy is set by the reaction: the height of the barrier between reactants and products. Heating gives more particles enough energy to get over the same barrier. Only a catalyst changes the barrier, by opening a different route.

Catalysts: a lower barrier

The fourth way to speed up a reaction is different. A catalyst provides a different route from reactants to products, with a lower activation energy. On the energy profile from the last lecture, the hump is lower, while the reactants and products stay at the same levels, so $Δ H$ is unchanged. With a lower $E_{a}$ , a larger fraction of collisions have enough energy, so the rate rises. The catalyst is not used up: it is made again by the end of the reaction. The last lecture of this topic explains how catalysts do this.

Sort the changes below by which factor each one uses.

Sorter — which factor?

Each card changes the rate of a reaction. Tap a card, then tap the factor it uses. Read the reason after each one.

Nine changes to the rate of a reaction, sorted by the factor each one uses.

Concentration or pressure. Using 2.0 mol/dm³ acid instead of 0.5 mol/dm³. Squeezing a mixture of reacting gases into half the volume. Adding water to dilute an acid (slower).
Surface area. Crushing marble chips into powder before adding acid. Chewing food well before swallowing it.
Temperature. Warming the acid from 20 °C to 40 °C. Keeping milk in a refrigerator (slower).
Catalyst. Adding manganese(IV) oxide to hydrogen peroxide solution. Using enzymes in a washing powder.

What should you have learned? Every change fitted one part of the sentence from collision theory. Three of the factors give more collisions, and two give more collisions with enough energy. Temperature does both, but mostly the second.

Working with the factors

Worked Example 1 — Magnesium and two acids

Equal pieces of magnesium ribbon are added to 1.0 mol/dm³ and 2.0 mol/dm³ hydrochloric acid at the same temperature. Explain why bubbles form faster in the stronger acid.

The reacting particles are magnesium atoms and hydrogen ions: Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g). The chloride ions are spectators. They are in the solution, but they do not react.

The 2.0 mol/dm³ acid has twice as many H⁺ ions in each cubic decimetre. So twice as many H⁺ ions hit each square millimetre of magnesium surface every second. The temperature is the same, so the fraction of collisions with at least $E_{a}$ is the same. More collisions per second with the same fraction successful gives more successful collisions per second, so hydrogen is made faster.

Check the reasoning: it changes only the first factor of the sentence. ✓

Worked Example 2 — How much more surface?

A marble cube has sides of 2 cm. Find its surface area. Then find the total surface area when it is cut into (a) 1 cm cubes (b) 2 mm cubes.

One cube: $6 \times 2 \times 2 = 24 cm^{2}$ .

(a) Each edge is cut into 2, so there are $2^{3} = 8$ cubes, each with area $6 \times 1 \times 1 = 6 cm^{2}$ . Total $8 \times 6 = 48 cm^{2}$ : twice as much.

(b) Each edge is cut into 10, so there are $1 0^{3} = 1000$ cubes, each with area $6 \times 0.2 \times 0.2 = 0.24 cm^{2}$ . Total $1000 \times 0.24 = 240 cm^{2}$ : ten times as much.

Check with the rule "cut each edge into $n$ , multiply the area by $n$ ": $24 \times 2 = 48$ ✓ and $24 \times 10 = 240$ ✓. The mass did not change at all.

Worked Example 3 — Warming two reactions

Reaction P has $E_{a} = 20$ kJ/mol and reaction Q has $E_{a} = 100$ kJ/mol. Both are warmed from 25 °C to 35 °C. Use the model above to compare how much each speeds up.

Collisions per second rise by the same factor for both, about × 1.017, because that depends only on the temperature.

The fraction with enough energy rises by about × 1.30 for P and about × 3.70 for Q. So P's rate rises by about $1.017 \times 1.30 \approx 1.32$ , and Q's by about $1.017 \times 3.70 \approx 3.76$ .

The reaction with the higher barrier is far more sensitive to temperature. A 10 °C rise is worth about a third more for P, and nearly four times as much for Q. Check the direction: when the barrier is high, only the very fastest particles get over it, and their number grows fastest when you heat. ✓

Worked Example 4 — A real case: the sugar dust explosion

On 7 February 2008, sugar dust exploded at the Imperial Sugar refinery in Port Wentworth, Georgia, in the United States. Fourteen workers were killed and 36 were injured. The US Chemical Safety Board found that the first explosion began in an enclosed conveyor under the sugar silos. It also found that large amounts of sugar dust had built up around the plant (CSB report, September 2009). Sugar in a bowl does not explode. Explain why sugar dust can.

Surface area. Fine dust has an enormous surface area. Floating in the air, every grain is surrounded by oxygen, so almost every sugar particle can collide with oxygen at once.

Temperature. A spark or a hot surface heats a small region, so many collisions there have at least the activation energy. Burning sugar releases energy, which heats the dust next to it, and the reaction spreads, as with the spark in the last lecture.

Together. In a lump of sugar, the burning can only creep across the outside. In a dust cloud, the heat reaches fresh fuel in every direction, almost instantly. The whole cloud reacts in a fraction of a second, and the hot gases expand violently: an explosion. The CSB found that the first explosion was followed by further dust explosions through the building.

More classic mistakes

Trap 2 — "Hotter particles collide more often, so that is why the rate rises"

That effect is real but small: under 2% for a 10 °C rise near room temperature. The main reason is that many more collisions have at least the activation energy. An answer that gives only "more collisions" for temperature misses the important point.

Trap 3 — Giving surface area for a reaction in solution

Surface area matters only when a solid reacts with a liquid, a solution or a gas. If two dissolved substances react, every particle is already exposed, and stirring or concentration is what matters. Check the state symbols first: (s) with (aq), (l) or (g) means surface area counts.

Trap 4 — "Higher concentration means more energetic collisions"

Concentration changes how many collisions happen each second, not how energetic they are. Energy is set by temperature. Keep the two factors apart, because examiners ask you to explain each one in its own words.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Zinc granules react with dilute sulfuric acid, giving hydrogen. Use collision theory to explain the effect of each change on the rate. (a) using zinc powder instead of granules (b) using more dilute acid (c) warming the acid (d) adding a little copper(II) sulfate solution, which acts as a catalyst.

Hint 1
For each change, decide which part of the sentence it affects: collisions per second, the fraction with enough energy, or the barrier itself.

Hint 2
(b) makes the reaction slower. Explain it the same way as a faster one, in reverse.

Full answer
(a) Faster: more zinc surface is exposed, so H⁺ ions hit zinc atoms more often each second. (b) Slower: fewer H⁺ ions in each cubic decimetre, so fewer collisions each second. (c) Faster: mainly because more collisions have at least $E_{a}$ , and a little because collisions happen more often. (d) Faster: the catalyst gives a route with a lower activation energy, so a larger fraction of collisions succeed. Check: (a) and (b) change the number of collisions, (c) mainly the energy fraction, (d) the barrier. ✓
A sugar cube has sides of 1.5 cm. (a) Find its surface area. (b) It is ground into tiny cubes with sides of 0.15 mm. Find their total surface area, and how many times larger it is. (c) Explain why icing sugar dissolves in tea much faster than a sugar cube.

Hint 1
Change 0.15 mm into centimetres first: 0.015 cm.

Hint 2
How many small cubes fit along one edge? Then use "cut each edge into $n$ , multiply the area by $n$ ".

Full answer
(a) $6 \times 1.5 \times 1.5 = 13.5 cm^{2}$ . (b) $1.5 \div 0.015 = 100$ small cubes fit along each edge, so the area is multiplied by 100: $1350 cm^{2}$ . (c) Dissolving also happens at the surface of the solid. The fine powder exposes far more sugar to the water, so water particles reach more sugar particles every second. Check (b) the long way: $10 0^{3} = 1 0^{6}$ cubes, each $6 \times 0.01 5^{2} = 0.00135 cm^{2}$ , total $1350 cm^{2}$ . ✓
For a reaction with $E_{a} = 50$ kJ/mol, the fraction of collisions with at least $E_{a}$ is about $1.74 \times 1 0^{- 9}$ at 25 °C and about $3.34 \times 1 0^{- 9}$ at 35 °C. The collisions per second rise by a factor of 1.017. (a) By what factor does the energy fraction rise? (b) By what factor does the rate rise? (c) What percentage of the increase in rate comes from more frequent collisions?

Hint 1
(a) Divide the new fraction by the old one.

Hint 2
(b) Multiply the two factors. (c) Compare the rise from the collision factor alone (0.017) with the total rise.

Full answer
(a) $3.34 \div 1.74 \approx 1.92$ . (b) $1.017 \times 1.92 \approx 1.95$ , so the rate almost doubles. (c) The rate rises by about 0.95, that is 95%. The collision factor alone would give only 0.017, about 1.7%. So only about $0.017 \div 0.95 \approx 2%$ of the increase comes from more frequent collisions. The other 98% comes from more collisions having enough energy.
A food scientist models the reactions that spoil milk with an activation energy of 80 kJ/mol (an invented value). Using the model in this lecture, these reactions are about 12 times slower at 4 °C, inside a refrigerator, than at 25 °C. (a) Milk left out at 25 °C goes sour in about 1 day. Estimate how long it lasts in the refrigerator. (b) Which part of the collision-theory sentence explains most of this? (c) A friend says: "Freezing food stops all reactions, so frozen food lasts for ever." Use collision theory to judge this.

Hint 1
Twelve times slower means it takes about twelve times as long.

Hint 2
(c) A lower temperature cuts the fraction of collisions with enough energy. Does it cut it to exactly zero?

Full answer
(a) About 12 days, if nothing else changes. Real milk also depends on how clean it is and how often it is opened, so this is a rough estimate. (b) The fraction of collisions with at least $E_{a}$ : it falls by a factor of about 11.5, while collisions per second fall by only about 4%. (c) The friend is wrong. Freezing makes the fraction with enough energy very small, but not zero, so reactions continue very slowly. This is why frozen food has a "best before" date too.

The bigger idea

Industry uses every factor in this lecture at once. The Haber process, which you met in the Gibbs lecture, runs at about 400–450 °C to make enough collisions energetic. It runs at about 200 atmospheres to crowd the gas particles together, and over an iron catalyst to lower the barrier. Each choice is one part of the collision-theory sentence. Engineers must also balance speed against cost and safety, which is why the conditions are a compromise.

Safety rules use the same ideas in reverse. Factories that handle flour, sugar, wood or metal powders control dust and remove sources of sparks, because a fine powder plus a hot spot is a reaction waiting to happen. Cooking, preserving food, digesting a meal and keeping medicines on a cool shelf all come down to the same sentence. The next lecture shows how chemists measure a rate in the laboratory, so that these effects become numbers.

Key concepts in this lesson

Every factor that changes the rate does so by changing the collisions per second, the fraction with at least the activation energy, or the activation energy itself.
A higher concentration of a solution, or a higher pressure of a gas, puts more particles in each cubic decimetre, so there are more collisions per second.
A larger surface area exposes more particles of a solid, so there are more collisions per second, and it matters only when a solid reacts with a liquid, solution or gas.
A higher temperature gives slightly more collisions and many more collisions with at least the activation energy, and the second effect is much larger.
How much a 10 °C rise speeds up a reaction depends on its activation energy, so "doubling" is only a typical value, not a rule.
A catalyst provides a route with a lower activation energy and is not used up.