What a catalyst does
A catalyst催化剂 is a substance that speeds up a reaction without being used up (OpenStax 12.7). U5.7 and U5.10 already showed how it works:
- a catalyst provides a new pathway (a new mechanism) whose rate-determining step has a lower activation energy;
- the catalyst is used up in one step of the mechanism and re-formed in a later step;
- the two pathways start and end at the same energies.
What a catalyst doesn’t change
Because the start and end are the same, a catalyst:
- doesn’t change ΔH: an endothermic reaction stays endothermic, and an exothermic one stays exothermic;
- doesn’t change K or the equilibrium position: it lowers the forward and reverse activation energies by the same amount, so both directions speed up equally and the system just reaches the same equilibrium sooner (2.11.5);
- doesn’t change the amount of product, which depends on the amounts of reactants and on K;
- doesn’t supply energy, and doesn’t change the temperature, the particles’ average kinetic energy, or the energy distribution.
A catalyst on the distribution curve
In U5.5, raising the temperature shifted the distribution curve to the right, so more particles had energy at least Ea. A catalyst works differently: the curve stays the same, and the Ea line moves to the left, so a larger fraction of particles has enough energy. Turn on “Add a catalyst” below to compare:
Schematic: the activation energies are far smaller than in real reactions so the shaded areas are visible. The fractions are calculated exactly from the energy distribution.
Three kinds of catalysis
Acid–base catalysis. A reactant gains or loses a proton (H⁺) and becomes a form that reacts more easily; the proton is released again in a later step. Take this schematic mechanism: S + H⁺ → SH⁺ (fast); SH⁺ + W → P + H⁺ (slow). H⁺ is used up and then re-formed, so it is the catalyst, and the rate depends on [H⁺].
Surface (heterogeneous) catalysis. The catalyst is in a different phase from the reactants, usually a solid catalyst with reactants in a gas or solution. OpenStax uses the reaction of C₂H₄ with H₂ on a nickel surface to show four processes. First the reactants are adsorbed onto the surface, and the surface weakens or breaks their bonds (activation). Then the adsorbed atoms react on the surface, and finally the products are desorbed. The surface also holds reactants in positions that favor reaction, improving the orientation of collisions. Because the reaction happens on the surface, the same mass of catalyst works better as finer particles with more surface area. Metals such as Ni, Pd, Pt, and Fe are common surface catalysts; the catalytic converters in car exhausts use Pt and Rh.
Enzyme catalysis. Enzymes are the catalysts of living things, usually proteins. A substrate molecule binds to the enzyme’s active site活性部位, forming an enzyme–substrate complex酶–底物复合物, which is a reaction intermediate. The substrate is converted into products in the active site, the products leave, and the enzyme is unchanged.
Step through how each kind of catalyst takes part in a reaction and is re-formed:
O₃ + O → 2O₂ (Cl as catalyst)
C₂H₄ + H₂ → C₂H₆ (Ni as catalyst)
Substrate → products (enzyme as catalyst)
Schematic drawings. What all three have in common: the catalyst takes part in the reaction and is then re-formed, so it doesn’t appear in the overall reaction.
Example 1: A catalyst’s effect on the energy profile
A one-step reaction has Ea(forward) = 120 kJ/mol and ΔH = −40 kJ/mol. With a catalyst, the activation energy of the rate-determining step drops to 70 kJ/mol. Find the reverse activation energy before and after adding the catalyst, and state whether K changes.
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Write the relationship. On the energy profile, the reverse reaction climbs from the products to the same peak:
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Before the catalyst. All values in kJ/mol:
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With the catalyst. A catalyst doesn’t change the energies of the reactants or products, so ΔH is still −40 kJ/mol:
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Compare the two directions.
Both activation energies drop by the same amount, so both directions speed up equally.
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Does K change? The energies of the reactants and products are unchanged, ΔH is the same, and both directions speed up equally, so K doesn’t change; the catalyst only makes equilibrium arrive sooner.
Example 2: Why use a powdered catalyst?
Pt catalyzes a gas-phase reaction. The same mass of Pt works far better as a fine powder than as a single lump. Explain using the steps of surface catalysis.
- Identify: the steps of surface catalysis. Reactants adsorb onto active sites on the Pt surface, react there, and the products desorb. The reaction happens only at the surface.
- Compare powder with a lump. For the same mass, finer particles have more total surface area. For example, cutting a 1 cm cube into 1 mm cubes makes 1000 times as many pieces and 10 times the total surface area.
- Connect to rate. More surface area → more active sites → more molecules adsorb and react at the same time → a higher rate. The Pt atoms inside a lump never touch the reactants, so they are wasted.
- What doesn’t change. The Pt isn’t used up; ΔH and K are unchanged, and so is the final amount of product. The reaction is just faster.
- Activation energy of the rate-determining step
The catalyst provides a pathway with a lower activation energy.
- Rate of the forward reaction
The activation energy is lower.
- Rate of the reverse reaction
The reverse activation energy drops by the same amount.
- ΔH
The start and end don’t change.
- Equilibrium constant K
Both directions speed up equally.
- Amount of product at equilibrium
The equilibrium position doesn’t change.
- Time needed to reach equilibrium
The reaction is faster.
- Rate constant k
A lower activation energy means a larger k.
- Shape of the Maxwell–Boltzmann curve
The temperature is unchanged, so the curve is too; the Ea line moves.
- How fast the reactant molecules move on average
A catalyst supplies no energy, and the temperature is unchanged.
- Fraction of molecules with energy at least the new Ea
Ea moves left, so the area to its right grows.
Common mistakes
Mistake 1: “A catalyst supplies energy to the reaction and makes molecules move faster.” A catalyst changes neither the temperature nor the energy distribution of the molecules; it lowers the activation energy, so more of the same molecules can reach the new, lower Ea.
Mistake 2: “A catalyst shifts the equilibrium to the right, giving more product.” A catalyst speeds up the forward and reverse reactions equally, so K and the equilibrium position don’t change; equilibrium is just reached sooner.
Mistake 3: “A catalyst doesn’t take part in the reaction.” A catalyst does react (it is used up in one step) and is re-formed in a later step. It also can’t turn an endothermic reaction into an exothermic one.
Quick check
Which explanation best describes how a catalyst speeds up a reaction?
A catalyst changes the pathway (mechanism); it doesn’t supply energy or change the temperature or the equilibrium.
A catalyst is added to a system already at equilibrium. Which statement is correct?
The forward and reverse reactions speed up equally, so the equilibrium doesn’t shift.
On a Maxwell–Boltzmann distribution graph, what is the effect of adding a catalyst?
The temperature is unchanged, so the curve is too; the catalyst lowers Ea, so the Ea line moves left and the area to its right grows. A is the effect of heating.
Schematic mechanism: S + H⁺ → SH⁺; SH⁺ + W → P + H⁺. Which statement is correct?
H⁺ is used up and then re-formed; SH⁺ is formed and then used up. This is acid–base catalysis.
Palladium (Pd) catalyzes the hydrogenation of alkenes. The same mass of Pd works much faster as a fine powder than as a piece of foil. Which explanation is best?
Surface catalysis happens on the catalyst’s surface, so more surface area means more active sites.
Which statement about enzyme catalysis is correct?
The enzyme–substrate complex forms and then breaks apart, so it is an intermediate; the enzyme itself reappears at the end.
A catalyst is added to an endothermic reaction. Which statement is correct?
The energies of the reactants and products don’t change, so ΔH doesn’t change.
A student says, “A catalyst speeds up a reaction by changing its order.” Which evaluation is best?
Explain catalysis as “a new pathway with a lower activation energy.” D is wrong too: the catalyst does change the pathway and the activation energy.
AP-style free response
When a little black MnO₂ powder is added to H₂O₂ solution, lots of bubbles form at once. After the reaction, the MnO₂ is filtered off and dried, and its mass is the same as the mass added.
(a) What role does MnO₂ play? Justify your answer with the experimental evidence. Is this homogeneous or heterogeneous catalysis?
(b) Describe how the energy profiles of 2H₂O₂(aq) → 2H₂O(l) + O₂(g) with and without MnO₂ are alike and how they differ.
(c) With the same amount of H₂O₂, how do the final amount of O₂ produced and the rate of O₂ production change when MnO₂ is added?
(d) Use the Maxwell–Boltzmann distribution to explain why MnO₂ speeds up the reaction.
(e) If the same mass of MnO₂ is used as a single lump instead, how does the rate change, and why?
Model answer and scoring
(a) It is a catalyst: adding it speeds up the reaction greatly, and its mass is unchanged afterward, so it isn’t used up. MnO₂ is a solid and H₂O₂ is in solution, so this is heterogeneous (surface) catalysis.
(b) Alike: the energies of the start (reactants) and end (products), and ΔH. Different: with MnO₂ the pathway is different, and its rate-determining step has a lower activation energy (a lower highest peak); there may be a different number of steps and different intermediates.
(c) The amount of O₂ is unchanged, because it depends on the amount of H₂O₂; the rate of O₂ production increases.
(d) The temperature is unchanged, so the molecules’ energy distribution curve is unchanged. MnO₂ lowers the activation energy, so the Ea line moves left; the area under the curve to the right of Ea grows, so a larger fraction of molecules has enough energy, there are more effective collisions, and the reaction speeds up.
(e) The rate decreases. A lump of MnO₂ has less surface area, so there are fewer active sites that H₂O₂ can reach.
Scoring: (a) catalyst with both pieces of evidence, and heterogeneous; (b) similarities and differences; (c) both conclusions; (d) the curve is unchanged, Ea moves left, and the fraction increases; (e) surface area.
Summary
- A catalyst provides a new pathway with a lower activation energy; it is used up and re-formed in the mechanism, so it doesn’t appear in the overall reaction.
- A catalyst doesn’t change ΔH, K, the equilibrium position, or the amount of product, and it supplies no energy; it speeds up the forward and reverse reactions equally, so equilibrium is reached sooner.
- On a distribution graph, the curve stays the same and the Ea line moves left, so a larger fraction of molecules has enough energy.
- The three kinds of catalysis are acid–base (gaining or losing a proton), surface (adsorption, activation, reaction, and desorption, where more surface area is better), and enzyme (with the enzyme–substrate complex as an intermediate).
AP tip: To explain what a catalyst does, say that it “provides a pathway (new mechanism) with a lower activation energy,” and say what stays the same: ΔH, K, and the amount of product. On a distribution graph, remember that a catalyst moves the Ea line, not the curve.