Kinetics

AQA Year 1 · written to the mark scheme

In a nutshell

Kinetics is the study of how fast reactions go, and what you can change to speed them up or slow them down.

The one idea that ties the whole topic together: a reaction only happens when particles collide with enough energy (and the right orientation), so anything that increases the frequency of successful collisions increases the rate.

Assumed knowledge: Energetics for energy profiles and the activation energy.

Core content

Collision theory and activation energy

Reactions can only occur when particles collide with each other.

But a collision only leads to a reaction if two conditions are met:

the particles collide with energy greater than or equal to the activation energy, and
the particles collide with the correct orientation.

The energy is needed to start breaking the bonds in the reactant particles.

Activation energy (Eₐ) is the minimum energy that colliding particles must have for a reaction to occur.

Most collisions do not lead to a reaction, because most colliding particles do not have energy greater than or equal to the activation energy. They simply bounce apart unchanged.

You can see the activation energy as the "hump" on a reaction energy profile: the particles must get over it before they can turn into products.

Figure not available yet (chem-exothermic-profile)

The Maxwell-Boltzmann distribution

At any temperature the particles in a gas do not all have the same energy: they have a huge spread of energies, because collisions constantly transfer energy between them.

The Maxwell-Boltzmann distribution shows this spread: how many molecules have each energy.

The features you must be able to label and explain:

The x-axis is (kinetic) energy and the y-axis is the number of molecules with that energy. It is not time and not temperature.
The curve starts at the origin, because no molecules have zero energy.
The peak is the most probable energy (E_mp), the energy more molecules have than any other. This is not the mean energy (the mean lies a little to the right of the peak).
The curve never meets the x-axis, because there is no upper limit to the energy a molecule can have.
The area under the curve is the total number of molecules.
Only the molecules in the shaded tail, to the right of Eₐ, have enough energy to react.

Still don't get it? · how to read the Maxwell-Boltzmann curve

Imagine plotting the exam marks of a whole year group: along the bottom, the mark you could score; up the side, how many students got that mark. Almost nobody scores zero, most people cluster around the middle, and there is a long thin tail of a few very high scorers. The graph can never dip below "zero students".

Now swap "mark" for "energy" and "students" for "molecules":

the bottom axis is energy, not time and not temperature (this is the single most common labelling error);
the curve starts at the origin because, just as no student scores literally nothing, no molecule has zero energy;
the top of the hump is the energy the largest number of molecules happen to have, the most probable energy, which is not quite the average;
the tail never touches the bottom axis because there is no maximum energy, in the same way there is always the theoretical chance of one more top scorer.

Back to the exam wording: the curve "starts at the origin because no molecules have zero energy" and "approaches the energy axis asymptotically (never meets it)". Only molecules out in the tail, with energy greater than or equal to Eₐ, can react.

Rate of reaction and how it is measured

The rate of reaction is the change in concentration of a reactant or product per unit time. Its usual unit is mol dm^-3 s^-1.

In practice you follow a measurable quantity against time, for example:

the volume of gas produced (gas syringe),
the loss of mass as a gas escapes (balance),
a colour change (colorimeter),
the time for a precipitate to hide a mark (a "clock" reaction).

If you plot the quantity against time you get a curve that is steepest at the start and then levels off as reactants are used up.

The rate at any instant is the gradient of the tangent drawn to the curve at that time.
The initial rate is the gradient of the tangent at t = 0, where the reaction is fastest.
The rate falls over time because the reactant concentration falls, so collisions become less frequent.

Effect of temperature on reaction rate

Increasing the temperature increases the rate. The Maxwell-Boltzmann distribution explains why.

When the temperature rises:

the molecules have a wider range of energies, so the curve flattens and shifts to the right (the peak is lower and at a higher energy),
the area under the curve stays the same, because the number of molecules is unchanged,
a much larger proportion of molecules now have energy greater than or equal to Eₐ.

So there is a greater frequency of successful collisions, and the rate increases.

A small rise in temperature gives a large rise in rate because the number of molecules with energy greater than or equal to Eₐ increases sharply. The particles do collide slightly more often, but that increase in collision frequency is small and is not the main reason.

Still don't get it? · why a small temperature rise gives a big rate rise

Picture a high-jump bar set almost at the ceiling. When the jumpers warm up, their average jump goes up only a little. But the number of people who can now clear that very high bar shoots up, because almost nobody could clear it before. A tiny improvement near the limit makes a huge difference to how many get over.

Step by step for molecules:

Only molecules with energy greater than or equal to Eₐ can react, and Eₐ sits far out in the sparse tail of the distribution where there are very few molecules.
A small temperature rise stretches and shifts that tail to the right.
Because the tail started so empty, even a small shift multiplies the number of molecules now sitting beyond Eₐ.
Meanwhile the molecules only speed up a little, so they collide only slightly more often.

So in the exam, write that "many more molecules now have energy greater than or equal to Eₐ, giving many more successful collisions per unit time". Do not explain the big rate rise by saying "they collide more often", that effect is far too small.

Required practical 3: investigating how rate changes with temperature

A standard way to show the effect of temperature uses the reaction of sodium thiosulfate with hydrochloric acid, which slowly produces a cloudy precipitate of sulfur:

Na₂S₂O₃(aq) + 2HCl(aq) ⟶ 2NaCl(aq) + SO₂(g) + S(s) + H₂O(l)

Method (an initial-rate, "disappearing cross" method):

Measure a fixed volume of sodium thiosulfate solution into a conical flask standing on paper marked with a cross.
Warm it in a water bath to a chosen temperature and record that temperature.
Add a fixed volume of hydrochloric acid and start the timer at the same moment.
Look down through the mixture and stop the timer when the cross can no longer be seen.
Repeat at a range of temperatures, keeping all volumes and concentrations the same.

Because the amount of sulfur needed to hide the cross is the same every time, the rate is taken as 1 / time. Plotting 1/t against temperature shows the rate rising steeply as temperature increases.

Control variables: total volume, concentrations, depth of liquid, and the same cross and observer each time.
Safety: sulfur dioxide is toxic and an irritant, so work in a well-ventilated room.

Effect of concentration and pressure

Increasing the concentration of a solution, or the pressure of a gas, increases the rate.

The reason is the same for both: there are more particles per unit volume, so the particles collide more frequently. A higher frequency of collisions means a higher frequency of successful collisions, so a faster rate.

For a gas, increasing the pressure squeezes the same number of particles into a smaller volume. This is the same effect as increasing concentration, so pressure and concentration are not two separate factors for a gas.
Concentration and pressure do not change the energy of the particles or the shape of the Maxwell-Boltzmann distribution. They only change how often particles meet.
If you double the concentration, you double the number of particles per unit volume, which doubles the frequency of collisions, so the rate roughly doubles.

Catalysts

A catalyst is a substance that increases the rate of a chemical reaction without being changed in chemical composition or amount at the end of the reaction.

A catalyst works by providing an alternative reaction route with a lower activation energy.

Figure not available yet (chem-catalyst-profile)

On the Maxwell-Boltzmann distribution, lowering the activation energy moves the Eₐ threshold to the left:

With the lower, catalysed Eₐ, a greater proportion of molecules now have energy greater than or equal to the activation energy.
So there are more successful collisions per unit time, and the rate increases.
The temperature is unchanged, so the distribution itself does not change: only the Eₐ threshold moves.

Worked examples

Example 1: finding rate from a concentration-time graph

Using the concentration-time graph above, calculate the mean rate of reaction over the first 20 s and over the following 20 s.

$\text{mean rate} = \frac{\text{change in concentration}}{\text{time taken}}$

From 0 to 20 s the concentration falls from 1.00 to 0.60 mol dm^-3:

$\text{rate} = \frac{1.00 - 0.60}{20} = \frac{0.40}{20} = 0.020\ \text{mol dm}^{-3}\,\text{s}^{-1}$

From 20 to 40 s the concentration falls from 0.60 to 0.36 mol dm^-3:

$\text{rate} = \frac{0.60 - 0.36}{20} = \frac{0.24}{20} = 0.012\ \text{mol dm}^{-3}\,\text{s}^{-1}$

The rate has fallen, because as the reactant is used up its concentration drops, so collisions become less frequent.

For the rate at a single instant you cannot use this method: you must draw a tangent to the curve at that time and find its gradient. The mark here is the working (substituting the readable values) plus the final answer with its unit. The data are given to 2 significant figures, so the answers are quoted to 2 significant figures.

Example 2: comparing rates using 1 / time

In a sodium thiosulfate and hydrochloric acid experiment, the cross disappeared after 68 s at 25 °C and after 33 s at 40 °C. Compare the rates.

Rate is proportional to 1 / time, so:

$\text{rate at 25 °C} = \frac{1}{68} = 0.015\ \text{s}^{-1}$

$\text{rate at 40 °C} = \frac{1}{33} = 0.030\ \text{s}^{-1}$

$\frac{0.030}{0.015} = 2.0$

A rise of 15 °C has roughly doubled the rate. This is the Maxwell-Boltzmann effect in numbers: a modest temperature rise produces a large increase in the proportion of molecules with energy greater than or equal to Eₐ. Both times are given to 2 significant figures, so the rates are quoted to 2 significant figures, with the unit s^-1.

Common exam mistakes

Explaining the large rate rise from a small temperature rise by saying the particles "collide more often" or there is "increased collision frequency". The mark needs many more molecules with energy greater than or equal to Eₐ, giving many more successful collisions per unit time. The collision-frequency effect is far too small to be the reason.
Labelling the Maxwell-Boltzmann axes wrongly. The commonest error is putting temperature or time on the x-axis. The x-axis is (kinetic) energy and the y-axis is the number of molecules.
Not knowing why the curve starts at the origin. It is because no molecules have zero energy, not because "the rate is zero", "the reaction has not started", or "the origin is absolute zero".
When sketching a higher-temperature curve, joining the new curve back to the original curve or to the x-axis. The curve must approach the energy axis as an asymptote and never touch it.
Confusing the most probable energy (E_mp, the peak) with the mean energy. They are not the same; the mean lies to the right of the peak.
Saying that increasing pressure or concentration speeds up a reaction by "giving the molecules more energy". They do not change the particles' energy or the distribution; they increase the number of particles per unit volume, so collisions are more frequent.
Giving increasing pressure and increasing concentration as two separate factors for a gas. For a gas they are the same effect.
In any "more collisions" explanation, dropping the words successful (or effective) and in a given time. The mark is for more successful collisions per unit time, not just "more collisions".
Defining a catalyst by describing how it works. A definition must say it increases the rate and is chemically unchanged / not used up at the end; the alternative-route and lower-Eₐ point is the explanation of how it works, asked by a different command.
Calculating rate as volume ÷ time or amount ÷ time. That gives only a mean; the rate at a point is the gradient of the tangent. Also, use volume (not "amount") for a gas, and remember to plot through the origin (0, 0).

Key definitions

Activation energy (Eₐ): the minimum energy that colliding particles must have for a reaction to occur (the minimum energy needed for a successful collision).
Rate of reaction: the change in concentration of a reactant or product per unit time (unit: mol dm^-3 s^-1).
Maxwell-Boltzmann distribution: the distribution (spread) of molecular energies in a gas at a given temperature.
Most probable energy (E_mp): the energy possessed by the greatest number of molecules (the energy at the peak of the distribution).
Catalyst: a substance that increases the rate of a chemical reaction without being changed in chemical composition or amount; it works by providing an alternative reaction route of lower activation energy.

Specification

I can define activation energy.
I can explain why most collisions do not lead to a reaction.
I can draw and label a Maxwell-Boltzmann distribution, and draw it for a higher and lower temperature.
I can state the meaning of rate of reaction and how it is measured.
I can use the Maxwell-Boltzmann distribution to explain why a small increase in temperature gives a large increase in rate.
I can describe how to investigate the effect of temperature on rate (Required practical 3).
I can explain how a change in concentration or pressure changes the rate, in terms of collision frequency.
I can define a catalyst and explain, using a Maxwell-Boltzmann distribution, how it increases the rate.

now prove it

Try Kinetics questions, marked against the full mark scheme.

Practice Kinetics