The Maxwell-Boltzmann distribution — Chemistry

Before this

This leads to

Catalysts
The Arrhenius equation · coming soon
The Maxwell speed distribution · coming soon

No such thing as a typical molecule

The air around you is mostly nitrogen. At 25 °C, a nitrogen molecule moves, on average, at about 475 metres per second. That is faster than most passenger aircraft.

But "on average" hides something important. The molecules do not all move at 475 m/s. They crash into each other billions of times a second, and every collision changes their speeds. At any moment, about 1 molecule in 100 is moving slower than 100 m/s. About 1 in 100 is moving faster than 1000 m/s. A very few are moving much faster still.

Those few fast ones matter more than all the others.

The last lecture ended with a promise: "The next lecture explains why temperature changes rates so strongly, by looking at how the energies of particles are shared out." The factors lecture found that warming a reaction by just 10 °C can nearly double its rate, although the particles move only about 2% faster. This lecture explains that puzzle with one curve. The collision theory lecture promised to draw it. Here it is.

How the energy is shared out

Imagine you could measure the kinetic energy of every particle in a gas at one instant. Then you count how many particles have each energy, and plot the count against the energy. The result is the Maxwell–Boltzmann distribution.

The Maxwell–Boltzmann distribution for a gas at one temperature. The shaded area is the fraction of particles with at least the activation energy Eₐ. (This Eₐ is drawn much closer to the peak than real ones, so that the area can be seen.)

Do not learn the features of this curve as a list. Each one has a reason.

It starts at zero. To have almost no energy, a particle would have to be almost still in every direction at once. Collisions keep knocking particles about, so this almost never happens. That is why the curve starts at the origin and rises steeply. It is not that a zero energy is forbidden. It is just extremely unlikely.
It has a peak. The peak marks the most probable energy: the energy that more particles have than any other. It is not the energy that "most particles" have. Most particles have energies spread on both sides of it.
It has a long tail. A particle can be lucky in several collisions in a row, each one giving it more energy. So a few particles have energies many times the most probable value. The tail gets thinner and thinner, but it never quite reaches the axis.
It is lopsided. It rises steeply on the left and falls slowly on the right. Because the tail is long, the mean energy is to the right of the peak. For a gas, the mean energy is three times the most probable energy.
The area means something. The total area under the curve is the total number of particles. The area between two energies is the number of particles with energies in that range.

Now add the activation energy, $E_{a}$ , from the collision theory lecture: the minimum energy a collision needs to react. Draw it as a vertical line. The area to the right of it is the fraction of particles with at least $E_{a}$ . These are the particles that can react if they collide the right way round.

That is it. That is the whole idea. The rate depends on a small area in the tail, not on the energy of a typical particle. Everything else in this lecture follows from looking at that area.

What heating does to the curve

When a gas is heated, the particles move faster on average. The curve changes in four ways at once:

The peak moves to the right, to a higher energy.
The peak becomes lower.
The curve becomes wider and flatter, with a fatter tail.
The total area stays the same, because the number of particles has not changed.

The last two are linked. If the curve spreads wider but keeps the same area, it must become lower. The heated curve is lower near the peak and higher in the tail, so the two curves cross once.

Try it below. The grey dashed curve is the starting temperature, 25 °C. Move the temperature and the activation energy and watch the shaded area.

Explore — temperature, Eₐ and the tail

Raise the temperature and watch the shaded area beyond Eₐ. Then move Eₐ to the right and raise the temperature again. Which reactions are most sensitive to heating? The activation energies here are much smaller than real ones, so that the area is visible.

Temperature:

Activation energy Eₐ:

The same gas at 298 K and 400 K, with an activation energy of 10 kJ/mol (much smaller than a real one, so that the area can be seen).

At 400 K the peak is lower and further right, and the tail is fatter. The fraction with at least 10 kJ/mol rises from about 4.5% to about 11.1%, about 2.5 times as large.

Here is what you should have seen. The peak moves only a little. The shaded area beyond $E_{a}$ grows a lot. And the further $E_{a}$ is out in the tail, the more strongly heating multiplies the shaded area.

This is the answer to the puzzle. A 10 °C rise makes particles only about 2% faster on average, so collisions become about 2% more frequent. But the fraction of collisions with enough energy can nearly double. The factors lecture found these numbers with a Pre-University formula. For a reaction with $E_{a} = 50$ kJ/mol, warmed from 25 °C to 35 °C, the fraction rose from about $1.74 \times 1 0^{- 9}$ to about $3.34 \times 1 0^{- 9}$ , almost twice as many. Now you can see why. A real activation energy sits so far out in the tail that a small widening of the curve multiplies the area there.

Stop and think

A student says: "Heating a reaction lowers its activation energy, and that is why it goes faster." Use the curve to explain what is wrong.

Show the answer

Heating does not move the $E_{a}$ line at all. The activation energy belongs to the reaction: it is set by the bonds that must break and the way the particles must meet. Heating changes the particles. It reshapes the curve, so a larger fraction of them reaches the same, fixed $E_{a}$ . The only common way to change $E_{a}$ itself is to give the reaction a different route, which is what a catalyst does.

A catalyst moves the line, not the curve

A catalyst gives the reaction a different route with a lower activation energy. On the diagram, the curve stays exactly the same, because the temperature and the particles are the same. Only the $E_{a}$ line moves to the left.

One curve, two activation energies. With the catalyst, the line moves left and the shaded area grows (here from about 4.5% to about 18% of the particles, with invented, very small values of Eₐ).

So heating and a catalyst both increase the shaded area, but in different ways. Heating changes the curve and leaves the line alone. A catalyst changes the line and leaves the curve alone. The next lecture explains how a catalyst provides its lower route.

Worked examples

Worked Example 1 — Writing the explanation for temperature

Question: "Use the Maxwell–Boltzmann distribution to explain why increasing the temperature increases the rate of a reaction."

Build the answer in the order of the cause, one link per sentence:

1. At a higher temperature, the particles have more kinetic energy on average, so the distribution moves to higher energies. Its peak is lower and further to the right.
2. The total area under the curve stays the same, because the number of particles is the same.
3. The area to the right of $E_{a}$ is larger, so a greater fraction of particles have at least the activation energy.
4. So a greater fraction of collisions are successful, and there are slightly more collisions each second too.
5. So the rate increases. The energy effect is much larger than the effect of more frequent collisions.

Check: every sentence names something on the diagram (the peak, the area, $E_{a}$ ) or a link in collision theory. An answer that only says "particles move faster and collide more" misses the main reason.

Worked Example 2 — Two reactions at the same temperature

Reaction P has $E_{a} = 50$ kJ/mol and reaction Q has $E_{a} = 100$ kJ/mol (invented reactions). Both happen at 25 °C, with the same number of collisions each second. Compare their rates.

Both reactions sit under the same curve, because the temperature is the same. Q's line is twice as far out along the energy axis, deep in the thin end of the tail. The Pre-University formula used in the earlier lectures gives the fraction of collisions with at least $E_{a}$ :

P: about 1.74 \times 1 0^{- 9} Q: about 3.02 \times 1 0^{- 18}

P's fraction is about $5.8 \times 1 0^{8}$ times larger, so with the same orientation fraction, P is hundreds of millions of times faster.

Check: doubling $E_{a}$ did not halve the fraction. It shrank it enormously, because the tail thins out faster and faster. This matches the explorer: moving $E_{a}$ right makes the shaded area fall steeply.

Worked Example 3 — Which reaction feels the heat more?

Warm reactions with $E_{a} = 20$ , 50 and 100 kJ/mol from 25 °C to 35 °C. By what factor does the fraction with enough energy grow in each case?

Using the same formula, the fraction grows by about × 1.30, × 1.92 and × 3.70. (With the 2% rise in collisions, the rates grow by about × 1.32, × 1.96 and × 3.76, the numbers in the factors lecture.)

The larger the activation energy, the further out in the tail the line sits, and the more a small widening of the curve multiplies the area there. That is why "a 10 °C rise doubles the rate" is only roughly true, for activation energies near 50 kJ/mol.

Check: try it in the explorer with small activation energies: from 298 K to 308 K, Eₐ = 5 kJ/mol gives a small increase, and Eₐ = 14 kJ/mol a larger one.

Three traps

Trap 1 — Drawing the hotter curve with more area or a higher peak

The hotter curve must enclose the same area as the cooler one, so its peak is lower. Many lost marks come from a hotter curve drawn taller, or from two curves that do not both start at the origin.

Trap 2 — A symmetrical hill, or a tail that lands on the axis

The curve rises steeply and falls slowly. Its tail approaches the energy axis but never touches it, because there is no largest possible energy. Draw it lopsided and leave the tail just above the axis.

Trap 3 — Moving the Eₐ line when the temperature changes

Temperature changes the curve. A catalyst changes the line. If your diagram moves both, it describes two changes at once.

Spot the error — a student's description of heating

A student compared the Maxwell–Boltzmann curves for a gas at 25 °C and at 45 °C. One line of the answer is wrong. Tap it.

A student's five-line description of the curves at 25 °C and 45 °C. Four lines are correct. The wrong line is: "The total area under the curve is larger, because the particles have more energy." The area is the number of particles, which does not change. The extra energy shows as a shift to the right, not as more area.

Try it yourself

Work each one on paper first. Sketch the curves where it helps. Open a hint only when you are stuck.

Sketch the Maxwell–Boltzmann distribution for a gas at one temperature. Label the axes, the most probable energy, the mean energy and an activation energy, and shade the particles that can react.

Hint 1
The axes are "number of particles" (up) and "kinetic energy" (along).

Hint 2
Start at the origin, rise steeply, fall slowly. The mean is to the right of the peak.

Full answer
A lopsided curve starting at the origin, with a peak near the left (the most probable energy), a long tail to the right that approaches but never meets the axis, the mean energy marked to the right of the peak, and a vertical $E_{a}$ line well out in the tail with the area to its right shaded. Check against the first figure in this lecture: same starting point, same lopsided shape, shading only beyond $E_{a}$ .
On one set of axes, a student draws curves for the same gas at 20 °C and 60 °C. The 60 °C curve has a higher peak, further to the right. Give two things that are wrong, or one thing that is wrong and one that is right, with reasons.

Hint 1
Think about the total area under each curve.

Hint 2
The same number of particles is spread over a wider range at 60 °C.

Full answer
Right: the 60 °C peak should be further to the right, because the particles have more energy on average. Wrong: its peak should be lower, not higher. The number of particles is the same, so the area must be the same. A curve that is wider and also taller would have more area, which would mean more particles. The 60 °C curve should be below the 20 °C curve near the peak and above it in the tail, and both must start at the origin.
Explain, using the Maxwell–Boltzmann distribution, why a catalyst and a rise in temperature both increase the rate, but in different ways.

Hint 1
Ask, for each change: does the curve change, or does the $E_{a}$ line move?

Hint 2
Both changes affect the same thing in the end: the area beyond the line.

Full answer
Heating changes the distribution: the curve moves right and flattens, so more of its area lies beyond the same $E_{a}$ . A catalyst leaves the distribution unchanged, because the temperature is unchanged, but provides a route with a lower $E_{a}$ , so the line moves left into a thicker part of the curve. In both cases a larger fraction of particles has at least the activation energy for the route the reaction takes, so a larger fraction of collisions succeed. Heating also slightly increases the collision rate. A catalyst does not.
Two reactions have activation energies of 30 kJ/mol and 90 kJ/mol. Warmed from 25 °C to 35 °C, the fraction of collisions with enough energy grows by about × 1.48 for one and × 3.25 for the other. Which is which? A cook says "turning up the oven by 10 °C speeds everything up by the same amount". Use the curve to explain why this is not true, and which reactions in a kitchen would be most sensitive to the oven setting.

Hint 1
Where is each $E_{a}$ line on the curve: in the fatter part or far out in the thin tail?

Hint 2
A small widening of the curve makes the biggest relative difference where the curve is thinnest.

Full answer
The 90 kJ/mol reaction grows by × 3.25, and the 30 kJ/mol reaction by × 1.48. Its line sits far out where the tail is very thin, so when heating fattens the tail, the area beyond it is multiplied by more. So reactions do not all speed up by the same factor. Reactions with large activation energies are the most sensitive to temperature, and those with small ones are the least. In a kitchen, the slow changes that need a lot of energy to start, such as browning, respond most strongly to the oven setting. (This kitchen example is general; real food chemistry involves many reactions at once.)

The bigger idea

James Clerk Maxwell worked out the distribution of molecular speeds in 1860. Ludwig Boltzmann extended it to energies and to gases in force fields in 1868, and in 1872 he showed how collisions drive any gas towards this distribution. At the time, many scientists still doubted that atoms and molecules were real. Their work treated a gas as a crowd too large to follow one by one, and described the crowd with statistics instead. That approach, now called statistical mechanics, became one of the foundations of physics and chemistry.

The same idea, that the rare particles in the tail decide what happens, reaches far beyond reactions. It explains why water evaporates below its boiling point (only the fastest molecules escape), why lizards and insects slow down on cold mornings, and why food lasts longer in a refrigerator. At Pre-University, the Arrhenius equation turns the shaded area into a formula.

Key concepts in this lesson

The Maxwell–Boltzmann distribution shows how many particles in a gas have each kinetic energy at one temperature.
The curve starts at the origin, rises to a peak at the most probable energy, and has a long tail, so the mean energy lies to the right of the peak.
The total area under the curve is the number of particles, and the area beyond Eₐ is the fraction of particles with enough energy to react.
Heating moves the curve right, lowers and widens it with the same area, and multiplies the area beyond Eₐ, most of all for large activation energies.
A catalyst does not change the curve; it moves the Eₐ line to the left, so a larger fraction of particles has enough energy.