Collision theory and the energy barrier

Why most collisions do nothing

Strike a match gently against the box and nothing happens. Strike it hard and it lights. The match has the same chemicals each time. In the second case the particles meet with more energy.

Chemical reactions are like this too. Before molecules can react, they must meet. This is the idea of collision theory. A collision leads to a reaction only if it passes two tests.

• Enough energy. The colliding molecules must carry at least a minimum energy, the threshold energy. Below it, they bounce off each other unchanged.
• The right orientation. The molecules must hit each other in the right way round, so that the atoms that form new bonds are the ones that touch.

A collision that passes both tests is an effective collision.

How many collisions are there? In a gas at ordinary pressure, one molecule collides billions of times in a second. If every collision reacted, reactions in gases would finish in a tiny fraction of a second. They do not. Most collisions are ineffective, and this is what controls the speed.

The number of collisions

The number of collisions in every second in a given volume is the collision frequency, $Z$. It grows when the concentration of the reactants grows. This is why the rate depends on concentration, as you saw in the rate law.

It also grows with temperature, but only a little. The collision frequency is proportional to $\sqrt{T}$. A rise from $300\ \text{K}$ to $310\ \text{K}$ raises $Z$ by $\sqrt{310/300} = 1.017$, which is $1.7$ percent. This is much too small to explain a doubling of the rate. Something else must be going on.

Energy: the long tail

The molecules in a gas do not all have the same energy. A few have little, most have an average amount, and a few have a lot. The picture of this spread is a curve called the distribution of molecular energies.

The area under a curve is the total number of molecules. The activation energy $E_a$ is marked by the red line. Molecules to the right of it have enough energy to react. This is the shaded part.

Now see what a rise in temperature does.

• The whole curve moves to the right, because the average energy is higher.
• The peak becomes lower and the curve becomes wider.
• The shaded area under the tail grows by a large factor.

The average energy goes up by a few percent, but the tail grows a great deal, because it is the far end of the curve. The exact share of molecules beyond $E_a$ is $e^{-E_a/RT}$. This is the same exponential factor as in the Arrhenius equation.

Example: the tail at two temperatures

Let $E_a = 75\ \text{kJ/mol}$.

At $300\ \text{K}$. $\dfrac{E_a}{RT} = \dfrac{75\,000}{8.314 \times 300} = 30.1$, and $e^{-30.1} = 8.7 \times 10^{-14}$. About one molecule in $10^{13}$ has enough energy.

At $310\ \text{K}$. $\dfrac{E_a}{RT} = 29.1$, and $e^{-29.1} = 2.30 \times 10^{-13}$.

The ratio. $\dfrac{2.30 \times 10^{-13}}{8.7 \times 10^{-14}} = 2.6$. The share of molecules that can react grows by $2.6$ times, though the average energy grew by only $3$ percent.

This is the answer to the puzzle. The rate depends on the tail, and the tail is very sensitive to temperature.

Orientation: the factor $P$

Think about the reaction between bromomethane and hydroxide ions, $\text{CH}_3\text{Br} + \text{OH}^- \rightarrow \text{CH}_3\text{OH} + \text{Br}^-$. The hydroxide ion must attack the carbon atom, from the side opposite to bromine. If it hits the hydrogen side, or the bromine, it bounces off, however hard it hits.

So only a part of the collisions with enough energy have the right orientation. We call this fraction the probability factor or steric factor, $P$. It is a number below 1. For simple atoms, $P$ is near 1. For large molecules it can be tiny.

Putting all this together:

Compare this with the Arrhenius equation. They match if $A = P \times Z$. So the pre-exponential factor $A$ holds the collision frequency and the orientation.

Example: the orientation factor

For a reaction in solution the collision factor is calculated as $Z = 4.0 \times 10^{11}\ \text{L mol}^{-1}\ \text{s}^{-1}$. The measured $A$ is $2.0 \times 10^{8}\ \text{L mol}^{-1}\ \text{s}^{-1}$.

Only about one collision in $2000$ has the right orientation. The others fail even when they have enough energy.

The simulation counts both tests together. A collision must pass the energy test and the orientation test, so the powers of ten add. A smaller $P$ moves the solid line up. A higher temperature moves it down.

The energy profile of a reaction

A graph of potential energy against the progress of the reaction is a potential energy profile.

Read the left graph from left to right.

• The reactants start at an energy we call zero.
• As they come together, the energy rises, because bonds start to stretch and the electrons repel one another.
• At the top of the hill the particles form an activated complex. This is an arrangement in which old bonds are partly broken and new bonds are partly formed. It has the highest energy on the path and is very short lived.
• Then the energy falls to the level of the products.

The height of the hill above the reactants is the activation energy of the forward reaction, $E_a$. This is the same $E_a$ as in the Arrhenius equation. Climbing the hill is the "enough energy" test.

The reverse reaction starts at the products and must climb the same hill from the other side. Its barrier is $E_a'$. The difference between the two levels is the enthalpy change:

So we want the reverse barrier: $E_a' = E_a - \Delta H$.

• Exothermic: $\Delta H$ is negative, so $E_a'$ is larger than $E_a$. For $E_a = 96\ \text{kJ/mol}$ and $\Delta H = -42\ \text{kJ/mol}$, $E_a' = 96 - (-42) = 138\ \text{kJ/mol}$.
• Endothermic: $\Delta H$ is positive, so $E_a'$ is smaller than $E_a$. For $E_a = 105\ \text{kJ/mol}$ and $\Delta H = +58\ \text{kJ/mol}$, $E_a' = 105 - 58 = 47\ \text{kJ/mol}$.

A reaction can be strongly exothermic and still be slow, because the barrier on the way is high. The energy change tells where a reaction ends. The barrier tells how fast it gets there.

An activated complex is not the same as an intermediate. An intermediate is a real species. It sits in a small dip on the path, and in principle it can be detected. The activated complex sits at the top of a peak. It is only a moment in the change.

Practise with new numbers

Try it in Python

The program turns an activation energy, a temperature and a probability factor into the number of collisions needed for one reaction.

Common mistakes

• Saying that more collisions explain the rise in rate. The collision frequency grows by under 2 percent for $10\ \text{K}$. The share of molecules with enough energy grows by a factor of two or more.
• Thinking that a collision with enough energy always reacts. The orientation also has to be right.
• Calling $\Delta H$ the activation energy. $\Delta H$ is the difference between the products and the reactants. $E_a$ is the height of the hill.
• Taking the reverse barrier as $E_a + \Delta H$. It is $E_a - \Delta H$. Keep the sign of $\Delta H$.
• Mixing up the activated complex and an intermediate. One is a peak and the other is a dip.

Remember these

• A reaction needs a collision, enough energy and the right orientation.
• The share of molecules with enough energy is $e^{-E_a/RT}$. It rises steeply with temperature.
• $k = P\,Z\,e^{-E_a/RT}$, so $A = PZ$.
• $\Delta H = E_a - E_a'$, so $E_a' = E_a - \Delta H$.
• The activated complex is at the top of the barrier. An intermediate is in a dip.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. The average kinetic energy of molecules grows only $3$ percent for a $10\ \text{K}$ rise. Why does the rate grow so much more?

2. A collision has more than the threshold energy, and it still does not react. Give a reason.

3. A reaction has $E_a = 60\ \text{kJ/mol}$ and $\Delta H = -35\ \text{kJ/mol}$. What is the activation energy of the reverse reaction?

4. Does a rise in temperature help the forward and the reverse reaction equally?

5. Is an activated complex the same as an intermediate?