Homogeneous and heterogeneous catalysts

1. Overview

A catalyst speeds up a reaction and is still there, chemically unchanged, when the reaction is over. It works by opening a different route, with a different mechanism and a lower activation energy. This section shows how to explain that with the Boltzmann distribution, and how to draw and read reaction pathway diagrams for a reaction with and without a catalyst. The same ideas return in equilibrium (7.1) and at A Level (26.2).

Key Definitions

• Catalyst: a substance that increases the rate of a reaction by providing an alternative route with a lower activation energy, and is chemically unchanged at the end of the reaction.
• Catalysis: the speeding up of a reaction by a catalyst; the catalyst takes part in the reaction and is regenerated, so it is not used up.
• Activation energy (EA): the minimum energy required for a collision between particles to be effective, that is, to lead to a reaction.
• Effective collision: a collision between particles that leads to a reaction; the particles must collide with at least the activation energy.
• Reaction mechanism: the sequence of steps by which reactants are turned into products.
• Intermediate: a species that is made in one step of a mechanism and used up in a later step, so it does not appear in the overall equation.
• Boltzmann distribution: the spread of energies among the molecules of a sample at one temperature, shown as a curve of the number of molecules against energy.
• Reaction pathway diagram: a diagram of enthalpy against progress of reaction that shows the reactants, the products, the activation energy and the enthalpy change, ΔH.
• Homogeneous catalyst: a catalyst in the same phase as the reactants (usually the same physical state), for example a gas that speeds up a reaction between gases.
• Heterogeneous catalyst: a catalyst in a different phase from the reactants, usually a solid with gaseous reactants or reactants in solution.

Content

What a catalyst is

A catalyst increases the rate of a reaction and is chemically unchanged at the end. The process is called catalysis.

• It takes part. It joins in at one stage, for example by bonding to a reactant, and is re-formed (regenerated) at a later stage.
• It is not used up, so a small amount can convert a large amount of reactant.
• Its mass and chemical composition are the same at the end. Its physical form can change: a solid's surface may roughen, or lumps may break up.
• It does not appear in the overall equation. It is named beside or above the arrow.

Spotting a catalyst. Suppose A + B → AB goes by this route when C is added:

Step	Equation
1	A + C → AC
2	AC + B → AB + C
Overall (add the steps)	A + B → AB

C is used in step 1 and re-formed in step 2, so C is the catalyst. AC is made and then used up, so it is an intermediate. Neither appears in the overall equation. A substance that is used up does appear in the overall equation, so it is a reagent, not a catalyst.

A different mechanism with a lower activation energy

Without C, A and B must react directly, in one step with a high activation energy, $E_{\text{A}}$. With C, the reaction follows a different mechanism: the two steps above. For an effective catalyst, each step of the new route has a lower activation energy than the single uncatalysed step.

The sentence to learn: a catalyst provides an alternative route (a different mechanism) with a lower activation energy.

• The original route still exists, with the same $E_{\text{A}}$. Most of the particles that react simply go by the easier new route.
• The catalyst does not give the particles more energy. At the same temperature they have the same spread of energies and the same average speed.

Homogeneous and heterogeneous catalysts

These two words are in the section title, but the AS outcome itself is about catalysts and catalysis in general: it does not ask you to use them. They are here so that you know what they mean when you meet them. A homogeneous catalyst is in the same phase as the reactants (for these examples, the same physical state). A heterogeneous catalyst is in a different phase, usually a solid in contact with gaseous reactants or reactants in solution.

Catalyst	Reaction it speeds up	States	Type
iron, Fe(s)	N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the Haber process	solid catalyst, gaseous reactants	heterogeneous
nickel, Ni(s)	CH₂=CH₂(g) + H₂(g) → CH₃CH₃(g), hydrogenation	solid catalyst, gaseous reactants	heterogeneous
concentrated H₂SO₄	carboxylic acid + alcohol ⇌ ester + water	one liquid mixture	homogeneous
NO and NO₂	oxidation of SO₂(g) to SO₃(g) in the atmosphere	all gases	homogeneous

Explaining the difference between the two types, and how each one works, is A Level content in topic 26.2. For AS, learn the general ideas in the rest of this page.

The catalytic effect and the Boltzmann distribution

The Boltzmann distribution (topic 8.2) shows how many molecules have each energy at one temperature. It starts at the origin, rises to one peak at the most probable energy, and has a long tail that approaches the energy axis without touching it. The area under the curve is the total number of molecules, so the area to the right of a line at $E_{\text{A}}$ is the number of molecules with energy equal to or greater than $E_{\text{A}}$.

The curve depends on the temperature, not on the route the reaction takes, so at the same temperature a catalyst leaves it unchanged. What changes is where the $E_{\text{A}}$ line goes: the catalysed route has a lower $E_{\text{A}}$, so its line is further to the left.

The explanation, link by link:

1. The catalyst provides an alternative route with a lower activation energy.
2. The Boltzmann distribution is unchanged, because the temperature is the same.
3. A greater proportion of molecules have energy equal to or greater than the lower $E_{\text{A}}$ (a larger area to the right of the line).
4. So a greater proportion of collisions are effective, and the frequency of effective collisions increases.
5. So the rate of reaction increases.

Do not mix this up with the effect of temperature:

	Adding a catalyst	Raising the temperature (8.2)
Boltzmann curve	unchanged	peak lower and at a higher energy; same area for the same sample
Activation energy	lower, because the route is different	unchanged
Molecules with energy ≥ $E_{\text{A}}$	more	more

Reaction pathway diagrams with and without a catalyst

Start from the single-route diagram in topic 5.1, then:

1. Keep one reactants level and one products level. They are fixed by the substances, so both routes share them, and there is a single $\Delta H$ arrow.
2. Draw the uncatalysed route as one hump, and the catalysed route between the same two levels with a lower peak. Label them "without catalyst" and "with catalyst".
3. Draw an $E_{\text{A}}$ arrow from the reactants line up to each peak.

Interpreting the diagram:

• $\Delta H$ is unchanged. The catalyst does not change the enthalpy of the reactants or the products.
• Both directions speed up. The height of a peak above the products is the activation energy of the reverse reaction. Lowering the peak lowers the forward and reverse values by the same amount. This is why a catalyst does not move the position of an equilibrium but lets it be reached sooner (topic 7.1).
• Endothermic reactions have a limit. The route ends at the products, so its peak can never be below the products level. However good the catalyst, $E_{\text{A}}$ can never be less than $\Delta H$. In practice the peak is drawn above the products, so $E_{\text{A}}$ is larger than $\Delta H$.
• Two humps are possible. The catalysed route is sometimes drawn as two smaller humps, one per step; the dip between them is the intermediate. Each step's $E_{\text{A}}$ is measured from the level where that step starts.

Worked example 1: activation energies with and without a catalyst

Question: For X(g) + Y(g) → Z(g), $\Delta H = -69$ kJ mol⁻¹. Without a catalyst, $E_{\text{A}} = +199$ kJ mol⁻¹; with a catalyst, $E_{\text{A}} = +110$ kJ mol⁻¹. (a) By how much does the catalyst lower $E_{\text{A}}$? (b) Calculate $E_{\text{A}}$ for the reverse reaction with and without the catalyst. (c) State $\Delta H$ for the catalysed reaction.

Step 1: lowering. $$199 - 110 = 89\ \text{kJ mol}^{-1}$$

Step 2: reverse activation energies. The products sit 69 kJ mol⁻¹ below the reactants, so each reverse barrier is 69 kJ mol⁻¹ taller than the forward one. $$E_{\text{A}}(\text{reverse, uncatalysed}) = 199 - (-69) = +268\ \text{kJ mol}^{-1}$$ $$E_{\text{A}}(\text{reverse, catalysed}) = 110 - (-69) = +179\ \text{kJ mol}^{-1}$$

Step 3: check. $268 - 179 = 89$ kJ mol⁻¹, the same lowering as in the forward direction.

Answer: (a) 89 kJ mol⁻¹. (b) +268 kJ mol⁻¹ without the catalyst, +179 kJ mol⁻¹ with it. (c) $\Delta H = -69$ kJ mol⁻¹, unchanged, because the reactants and products levels do not move. The data are to the nearest kJ mol⁻¹, so the answers are too.

Worked example 2: building a pathway diagram for an endothermic reaction

Question: For P(g) → Q(g) + R(g), $\Delta H = +18$ kJ mol⁻¹. Without a catalyst, the reverse reaction has $E_{\text{A}} = +94$ kJ mol⁻¹. With a catalyst, the forward reaction has $E_{\text{A}} = +41$ kJ mol⁻¹. (a) Calculate the uncatalysed forward $E_{\text{A}}$ and the catalysed reverse $E_{\text{A}}$. (b) Describe the pathway diagram. (c) A student says a better catalyst could bring the forward $E_{\text{A}}$ down to +12 kJ mol⁻¹. Explain why this is impossible.

Step 1: uncatalysed forward $E_{\text{A}}$. Rearrange $E_{\text{A}}(\text{reverse}) = E_{\text{A}}(\text{forward}) - \Delta H$: $$E_{\text{A}}(\text{forward}) = 94 + 18 = +112\ \text{kJ mol}^{-1}$$

Step 2: catalysed reverse $E_{\text{A}}$. $$E_{\text{A}}(\text{reverse}) = 41 - (+18) = +23\ \text{kJ mol}^{-1}$$

Step 3: the diagram. The products line sits 18 kJ mol⁻¹ above the reactants line, with an upward $\Delta H$ arrow. The uncatalysed peak is 112 kJ mol⁻¹ above the reactants. The catalysed curve joins the same two levels, with its peak 41 kJ mol⁻¹ above the reactants, still 23 kJ mol⁻¹ above the products.

Step 4: the limit. Whichever route the reaction takes, it has to end at the products, which are 18 kJ mol⁻¹ above the reactants. So the highest point on any route is at least 18 kJ mol⁻¹ above the reactants. A peak only 12 kJ mol⁻¹ above the reactants would sit below the products line, which is impossible. The numbers show the same thing: the reverse $E_{\text{A}}$ would be $12 - 18 = -6$ kJ mol⁻¹, and an activation energy cannot be negative.

Answer: (a) +112 kJ mol⁻¹ and +23 kJ mol⁻¹. (b) As in Step 3, with both curves and both $E_{\text{A}}$ arrows labelled. (c) The forward $E_{\text{A}}$ can never be less than $\Delta H$ = +18 kJ mol⁻¹, whatever the catalyst, so +12 kJ mol⁻¹ is impossible.

Key Equations

Enthalpy change (kJ mol⁻¹; the same with or without a catalyst): $$\Delta H = H_{\text{products}} - H_{\text{reactants}}$$

Reverse activation energy, for either route ($E_{\text{A}}$ in kJ mol⁻¹, always positive; $\Delta H$ with its sign): $$E_{\text{A}}(\text{reverse}) = E_{\text{A}}(\text{forward}) - \Delta H$$

Lowering by a catalyst (kJ mol⁻¹; the same for the forward and reverse reactions): $$\text{lowering} = E_{\text{A}}(\text{uncatalysed}) - E_{\text{A}}(\text{catalysed})$$

Limit for an endothermic reaction ($\Delta H$ positive), with or without a catalyst (in practice $E_{\text{A}}$ is larger than $\Delta H$): $$E_{\text{A}}(\text{forward}) \geq \Delta H$$

Common Mistakes to Avoid

• Mistake: writing only "a catalyst lowers the activation energy". Fix: say it provides an alternative route with a lower $E_{\text{A}}$. In Worked example 1, the catalysed route has +110 kJ mol⁻¹ while the uncatalysed route keeps +199 kJ mol⁻¹.
• Mistake: "the catalyst gives the particles more energy" or "makes them collide more often". Fix: at the same temperature the energies are unchanged; a greater proportion of collisions are effective. Write "the frequency of effective collisions increases".
• Mistake: drawing a new Boltzmann curve for the catalyst, or a curve that starts above zero or whose tail meets the axis. Fix: keep one curve, starting at the origin with a tail that never touches the axis, and add a second $E_{\text{A}}$ line to the left of the first. A new curve means a change of temperature.
• Mistake: a pathway diagram with one curve, unlabelled curves, or a catalysed curve ending at a different level. Fix: two labelled curves sharing both levels. In Worked example 1, $\Delta H$ is −69 kJ mol⁻¹ for both routes.
• Mistake: measuring $E_{\text{A}}$ from the wrong level. Fix: the forward value is measured from the reactants, the reverse from the products. In Worked example 1 the uncatalysed forward value is +199 kJ mol⁻¹ but the reverse is +268 kJ mol⁻¹.
• Mistake: "the catalyst does not react", or a vague "the catalyst does not change". Fix: say it takes part and is regenerated, so it is chemically unchanged at the end. Filter, wash and dry a solid catalyst after a reaction and its mass is the same as at the start.
• Mistake: calling any added reagent a catalyst. Fix: a catalyst is regenerated and absent from the overall equation. NaBH₄ used to reduce an aldehyde is used up, so it is a reducing agent.

Exam Tips

• Define a catalyst with both parts: it increases the rate, and it is chemically unchanged at the end. When asked how it works, add "alternative route with a lower activation energy".
• Explain with the Boltzmann distribution as a chain: alternative route, lower $E_{\text{A}}$, curve unchanged, more molecules with energy ≥ $E_{\text{A}}$, greater frequency of effective collisions, faster rate. Each link is a separate point.
• Sketch a Boltzmann distribution with labelled axes (number of molecules, energy). Mark both $E_{\text{A}}$ lines and say which is which.
• Construct a pathway diagram with labelled axes, two labelled curves, arrowheads on every arrow, $E_{\text{A}}$ arrows from the reactants line and one $\Delta H$ arrow.
• Calculate a reverse $E_{\text{A}}$ by writing the equation, substituting $\Delta H$ with its sign in brackets, and giving the + sign and kJ mol⁻¹.
• State a catalyst's effects precisely: rate increases; $E_{\text{A}}$ is lower (new route); $\Delta H$ and the Boltzmann curve are unchanged.