Reaction rates

In a nutshell

Reaction rate is how quickly reactants are used up or products are made. You measure it as the change in a quantity per unit time, read off the gradient of a graph.

This subtopic explains rates with collision theory: particles react only when they collide with at least the activation energy. Concentration, gas pressure, temperature and catalysts all change the rate by changing how often particles collide, or what proportion of collisions have enough energy. The Boltzmann distribution is the graph you use to explain the energy part.

Assumed knowledge: Amount of substance: moles and gas volumes at RTP; Enthalpy changes: enthalpy profile diagrams and activation energy.

Core content

Simple collision theory: why particles must collide

For particles to react, they must collide. Most collisions do not lead to a reaction.

A collision only leads to a reaction if the particles collide with at least the activation energy.

Activation energy, E_a: the minimum energy required for a reaction to take place.

The rate of reaction is the change in concentration of a reactant or product per unit time. In practice you can follow any quantity that changes as the reaction goes, such as gas volume or mass.

So there are only two ways to make a reaction faster:

1. Make particles collide more frequently.
2. Make a greater proportion of collisions have energy equal to or greater than E_a.

Simple collision theory: concentration and gas pressure

Increasing concentration increases the rate because of collision frequency. The chain the marks follow:

1. Higher concentration means more particles in the same volume (more particles per unit volume).
2. So particles collide more frequently.
3. So there are more frequent successful collisions (collisions with energy ≥ E_a), and the rate increases.

Increasing the pressure of a gas works the same way. The same number of gas molecules is squeezed into a smaller volume, so the concentration of the gas increases.

1. More gas molecules per unit volume.
2. Molecules collide more frequently.
3. More frequent successful collisions, so a faster rate.

Concentration and pressure do not change the energy of the particles. The proportion of collisions with energy ≥ E_a stays the same; there are simply more collisions per second.

Simple collision theory: rate from the gradient of a graph

To follow a reaction, measure a physical quantity that changes with time, for example:

• the concentration of a reactant or product
• the volume of gas produced
• the mass of the reaction mixture (when a gas escapes)

Plot the quantity against time. The rate is the gradient of the graph:

$\text{rate} = \frac{\text{change in quantity}}{\text{change in time}}$

• On a straight line, the gradient is the same everywhere, so the rate is constant.
• On a curve, the rate keeps changing. Draw a tangent to the curve at the time you want; the gradient of the tangent is the rate at that instant.
• The initial rate is the gradient of the tangent at t = 0. It is the fastest point, because the reactant concentration is highest.
• The curve gets less steep because reactants are used up, so their concentration falls and collisions become less frequent. It goes flat when the reaction has stopped.

The tangent at t = 0 is steeper than the tangent at 50 s, so the reaction is slowing down as H₂O₂ is used up.

How to draw and use a tangent:

1. Place a ruler so it touches the curve at that one point only, with the curve bending away equally on both sides.
2. Draw the line long, across most of the graph.
3. Pick two points on the tangent (not on the curve) that are far apart, so your reading errors are small compared with the differences.
4. Gradient = Δy ÷ Δx. For a reactant the gradient is negative; quote the rate as a positive value.
5. Give the units of the quantity divided by time.

Still don't get it? · rate from a tangent

Think about a car journey. If you drive 120 miles in 3 hours, your average speed is 40 mph. But at any moment the speedometer might read 60 or 15. The speedometer tells you how fast you are going right now, not on average.

Now rebuild it step by step:

1. A graph of distance against time is steep when you are going fast and flat when you are stopped.
2. If the graph is a curve, its steepness changes from moment to moment.
3. To find the steepness at one moment, you draw a straight line that just grazes the curve at that point, going the same direction as the curve there. That line is the tangent.
4. The steepness (gradient) of that straight line is the "speedometer reading" at that moment.

A reaction graph is the same idea. Swap distance for concentration, gas volume or mass. The gradient of a tangent at a chosen time is the rate of reaction at that time, and the tangent at t = 0 gives the initial rate. The gradient of a straight line joining two points on the curve only gives the mean (average) rate over that interval.

Catalysts: how a catalyst increases the rate

A catalyst increases the rate of a reaction without being used up by the overall reaction.

It does this by allowing the reaction to proceed via a different route (an alternative pathway) with a lower activation energy.

• The catalyst does take part in the reaction, but it is regenerated by the end, so there is the same amount of catalyst at the end as at the start.
• The enthalpy change, ΔH, is unchanged: the reactants and products are the same, only the route between them changes.
• With a lower E_a, a greater proportion of collisions have enough energy to react, so the rate increases (see the Boltzmann section below).

You only need this general idea. Details of industrial processes are not required.

Catalysts: homogeneous and heterogeneous

Catalysts are classified by their phase compared with the reactants.

• A homogeneous catalyst is in the same phase as the reactants. It reacts with a reactant to form an intermediate, which then breaks down to give the products and regenerate the catalyst.
• A heterogeneous catalyst is in a different phase from the reactants. It is usually a solid with gaseous or liquid reactants, and the reaction happens on its surface.

The ozone example shows how a homogeneous catalyst is regenerated. Cl• is used up in the first step and made again in the second:

• Cl•(g) + O₃(g) → ClO•(g) + O₂(g)
• ClO•(g) + O(g) → Cl•(g) + O₂(g)
• Overall: O₃(g) + O(g) → 2O₂(g)

Catalysts: economic importance and sustainability

Catalysts have great economic importance, and they make industrial processes more sustainable.

• A catalyst lets a reaction run at a useful rate at a lower temperature.
• Lower temperatures mean less energy demand, so less fossil fuel is burned to supply that energy.
• Burning less fossil fuel means lower CO₂ emissions.
• Less energy and a faster rate mean lower costs and more product made in a given time.
• The catalyst is not used up, so it can be reused many times.

These benefits are weighed against the fact that some catalysts are toxic, so they must be handled, contained and disposed of safely.

PAG9: investigating reaction rates

To find a rate, you monitor a quantity as the reaction proceeds, taking readings at regular time intervals. At AS the techniques are measuring gas volume, mass and time.

Gas syringe method, in outline:

1. Measure the acid into a conical flask and set up the gas syringe at zero.
2. Add the solid, immediately insert the bung and start the stopwatch.
3. Record the gas volume at regular intervals (for example every 10 s) until it stops changing.
4. Plot volume against time and use tangents to find rates.

Practical points that earn marks:

• Mass loss suits a heavy gas such as CO₂. Hydrogen is so light that the mass change is too small to measure accurately, so collect H₂ in a gas syringe instead.
• A cotton wool plug lets the gas escape but stops acid spray leaving the flask, which would add to the mass loss.
• For the disappearing cross, the rate is proportional to 1/time. This works because the same amount of sulfur must form each time to hide the cross.
• Control variables: temperature (use a water bath), volume and concentration of the solution you are not changing, and the mass and particle size of any solid.
• Errors to reduce: gas escaping before the bung is in, a sticking syringe plunger, and a delay in starting the stopwatch.

The Boltzmann distribution: shape and activation energy

The Boltzmann distribution shows how the energies of the molecules in a gas are spread out at a given temperature. Molecules do not all have the same energy.

Features you must be able to draw and explain:

• Axes: number of molecules (y) against energy (x).
• The curve starts at the origin: no molecules have zero energy.
• It is not symmetrical: it rises steeply to a peak at low energy, then falls with a long tail.
• The tail approaches the x-axis but never touches it: there is no maximum energy.
• The area under the curve equals the total number of molecules.
• The shaded area to the right of E_a is the molecules with energy ≥ E_a. Only these can react when they collide, and they are a small proportion of the total.

The Boltzmann distribution: effect of temperature

At a higher temperature, the distribution shifts to the right and flattens.

How the higher-temperature curve (T2) differs:

• The peak is lower and moves to the right (to higher energy).
• The area under the curve is the same, because the number of molecules has not changed.
• At high energies, the T2 curve lies above the T1 curve, so there is more area to the right of E_a.
• It still starts at the origin and still never touches the x-axis.

The chain the marks follow:

1. At a higher temperature, molecules have more energy on average, so the distribution shifts to higher energy.
2. A greater proportion of molecules have energy ≥ E_a.
3. So there are more frequent successful collisions, and the rate increases.

Molecules also move faster and collide a little more often, but the main reason is step 2: the larger proportion of molecules with energy ≥ E_a.

Still don't get it? · why temperature has such a big effect

Imagine a class of students each trying to throw a ball over a high wall. Most throws fall short. Only a few strong throws clear it.

Now give everyone a warm-up and an energy drink. Most throws get a bit stronger. Some still fall short, but many more throws now clear the wall. The wall did not move. The throws changed.

Step by step:

1. In any gas, the molecules have a spread of energies. Most have a middling amount; a few have a lot.
2. To react, two molecules must hit each other with at least a certain energy. That energy is the wall.
3. Heating the gas gives the molecules more energy, so the whole spread moves towards higher energy.
4. The number of molecules does not change, so the curve gets lower and wider rather than taller.
5. The bit of the curve beyond the wall grows a lot, so many more collisions can now lead to a reaction.

The exam version: the wall is the activation energy, E_a. At a higher temperature the Boltzmann distribution shifts to the right, so a greater proportion of molecules have energy equal to or greater than E_a, giving more frequent successful collisions and a faster rate. Always mark E_a on your sketch and say "proportion" (or "more molecules") with energy above it.

The Boltzmann distribution: effect of a catalyst

A catalyst does not change the Boltzmann distribution. The temperature is the same, so the curve is exactly the same.

What changes is where E_a is. The catalysed route has a lower activation energy, so E_a moves to the left.

The chain the marks follow:

1. The catalyst provides an alternative route with a lower activation energy.
2. A greater proportion of molecules have energy ≥ the (lower) E_a (the whole shaded area, not just the part beyond the uncatalysed E_a).
3. So there are more frequent successful collisions, and the rate increases.

Worked examples

1. Mean rate from a mass-loss graph

Marble chips (CaCO₃) react with excess hydrochloric acid in a flask on a balance. The mass falls as CO₂ escapes.

Find the mean rate of mass loss over the first 60 s, and over 60 to 120 s.

A straight line (chord) between two points on the curve gives the mean rate over that interval.

First 60 s:

$\text{mass lost} = 150.00\ \text{g} - 149.08\ \text{g} = 0.92\ \text{g}$

$\text{mean rate} = \frac{0.92\ \text{g}}{60\ \text{s}} = 0.01533\ \text{g s}^{-1}$

$\text{mean rate} = 0.015\ \text{g s}^{-1}\ \text{(2 s.f.)}$

From 60 s to 120 s:

$\text{mass lost} = 149.08\ \text{g} - 148.80\ \text{g} = 0.28\ \text{g}$

$\text{mean rate} = \frac{0.28\ \text{g}}{60\ \text{s}} = 0.004667\ \text{g s}^{-1}$

$\text{mean rate} = 0.0047\ \text{g s}^{-1}\ \text{(2 s.f.)}$

The rate has fallen because the acid is being used up, so its concentration falls and collisions are less frequent.

Where marks go: the mass lost is the difference, not the mass on the balance. The mass differences (0.92 g, 0.28 g) have only 2 significant figures, so the answers do too. Rate is quoted as a positive value with units g s^-1.

2. Rate at a given time from a tangent (concentration)

Hydrogen peroxide decomposes: 2H₂O₂(aq) → 2H₂O(l) + O₂(g). A tangent has been drawn to the concentration curve at t = 50 s.

Find the rate of reaction at 50 s.

Read two points that are far apart on the tangent: (0 s, 0.70 mol dm^-3) and (120 s, 0.00 mol dm^-3).

$\text{gradient} = \frac{\Delta[\text{H}_2\text{O}_2]}{\Delta t} = \frac{0.00\ \text{mol dm}^{-3} - 0.70\ \text{mol dm}^{-3}}{120\ \text{s} - 0\ \text{s}}$

$\text{gradient} = \frac{-0.70\ \text{mol dm}^{-3}}{120\ \text{s}} = -5.83 \times 10^{-3}\ \text{mol dm}^{-3}\ \text{s}^{-1}$

The gradient is negative because H₂O₂ is a reactant being used up. The rate is the size of the gradient:

$\text{rate} = 5.8 \times 10^{-3}\ \text{mol dm}^{-3}\ \text{s}^{-1}\ \text{(2 s.f.)}$

Where marks go: the tangent must touch the curve at 50 s, and both points you read must be on the tangent. Using two points on the curve (for example at 0 s and 50 s) gives a mean rate, not the rate at 50 s. The answer needs its units, mol dm^-3 s^-1.

3. Initial rate from a gas volume graph, and converting to mol s^-1

Magnesium ribbon reacts with excess hydrochloric acid. The hydrogen is collected in a gas syringe. A tangent has been drawn at t = 0.

(a) Find the initial rate in cm³ s^-1.

Two points on the tangent: (0 s, 0 cm³) and (25 s, 60 cm³).

$\text{initial rate} = \frac{60\ \text{cm}^3 - 0\ \text{cm}^3}{25\ \text{s} - 0\ \text{s}} = 2.4\ \text{cm}^3\ \text{s}^{-1}$

(b) Convert this to mol of H₂ per second. The molar gas volume at RTP is 24.0 dm³ mol^-1 (given on the data sheet), which is 24 000 cm³ mol^-1.

$\text{initial rate} = \frac{2.4\ \text{cm}^3\ \text{s}^{-1}}{24\,000\ \text{cm}^3\ \text{mol}^{-1}}$

$\text{initial rate} = 1.0 \times 10^{-4}\ \text{mol s}^{-1}\ \text{(2 s.f.)}$

Where marks go: forgetting to change dm³ to cm³ (dividing 2.4 by 24.0) gives 0.10 mol s^-1, which is 1000 times too big. The tangent must start at the origin and follow the curve's starting slope, not join the origin to a later point on the curve.

Common exam mistakes

• Writing "more collisions" for the effect of concentration, pressure or temperature. Rate is about time, so the mark needs more frequent collisions (or more collisions per second).
• Saying a higher concentration gives particles more energy, or a greater proportion of collisions with energy ≥ E_a. Concentration only increases collision frequency; the energy distribution is unchanged.
• Explaining the pressure effect without saying the molecules are in a smaller volume / more molecules per unit volume. "Higher pressure means faster molecules" is wrong.
• Explaining the temperature effect only by "particles move faster and collide more often". The key mark is that a greater proportion of molecules have energy ≥ E_a.
• Drawing the higher-temperature Boltzmann curve with a taller peak, a peak that moves left, or a much larger area. The peak is lower and to the right, and the area stays the same.
• Letting a Boltzmann curve touch or cross the x-axis at high energy, or start above the origin. It starts at the origin and never meets the x-axis.
• Mislabelling the axes, for example "temperature" or "time" on the x-axis. It is energy (x) and number of molecules (y).
• Saying a catalyst "lowers the activation energy" with nothing else. The full point is that it provides an alternative route with a lower activation energy.
• Saying a catalyst gives the molecules more energy, or shifting the Boltzmann curve for a catalyst. The curve is unchanged; only E_a moves to the left.
• Saying a catalyst "does not take part in the reaction". It does take part, but it is not used up overall because it is regenerated.
• Mixing up the catalyst types: homogeneous = same phase as the reactants, heterogeneous = different phase. Learn the word phase.
• For sustainability, writing only "catalysts are good for the environment". Give the chain: lower temperature, less energy demand, less fossil fuel burned, less CO₂.
• Reading rate points off the curve instead of the tangent, using a tiny triangle, or giving a negative rate or no units.
• Using mass loss to follow a reaction that makes hydrogen: the change in mass is too small to measure accurately.

Key definitions

• Rate of reaction: the change in concentration of a reactant or product per unit time.
• Activation energy, E_a: the minimum energy required for a reaction to take place.
• Catalyst: a substance that increases the rate of a reaction without being used up by the overall reaction, by allowing the reaction to proceed via a different route with a lower activation energy.
• Homogeneous catalyst: a catalyst that is in the same phase as the reactants.
• Heterogeneous catalyst: a catalyst that is in a different phase from the reactants.
• Boltzmann distribution: the distribution of energies of the molecules in a gas at a particular temperature.
• Initial rate: the rate at the start of a reaction (t = 0), found from the gradient of a tangent to the curve at t = 0.

Specification

• I can explain the effect of concentration, and of the pressure of a gas, on rate in terms of the frequency of collisions. (3.2.2 a)
• I can calculate a rate from the gradient of a graph of concentration, gas volume or mass against time, including drawing a tangent to a curve and giving units. (3.2.2 b)
• I can explain that a catalyst increases the rate without being used up by the overall reaction. (3.2.2 c i)
• I can explain that a catalyst provides a different route with a lower activation energy, and show this on an enthalpy profile diagram. (3.2.2 c ii)
• I can explain the terms homogeneous and heterogeneous catalyst, with an example of each. (3.2.2 d i)
• I can explain the economic importance of catalysts and how they improve sustainability by lowering temperatures, reducing energy demand from burning fossil fuels and so reducing CO₂ emissions, and weigh this against the toxicity of some catalysts. (3.2.2 d ii)
• I can describe how to investigate reaction rates by measuring mass, gas volume and time (PAG9). (3.2.2 e)
• I can sketch and explain the Boltzmann distribution and mark the activation energy on it. (3.2.2 f)
• I can use Boltzmann distributions to explain how a temperature increase changes the proportion of molecules with energy above E_a, and so the rate. (3.2.2 g i)
• I can use a Boltzmann distribution to explain how a catalyst changes the proportion of molecules with energy above E_a, and so the rate. (3.2.2 g ii)

Related notes

• Enthalpy changes: enthalpy profile diagrams and the meaning of activation energy
• Amount of substance: gas volumes at RTP, used to convert a rate into mol s^-1
• Chemical equilibrium: how a catalyst affects the forward and reverse rates in an equilibrium
• Alkenes: hydrogenation with a nickel catalyst
• Haloalkanes: radicals catalysing the breakdown of ozone
• How fast? (A level): rate equations, orders and the rate constant