Catalysis — homogeneous, heterogeneous, enzyme catalysis: Question bank

Definition Every symbol used on this page (read this first)

$E_{a}$ = activation energy — the height of the energy hill a reaction must climb.
$Δ H$ = enthalpy change — the height difference between the product valley and the reactant valley ( $H_{p r o d} - H_{r e a c t}$ ).
$Δ G$ = Gibbs free-energy change — the thermodynamic quantity whose sign decides spontaneity: $Δ G < 0$ means the reaction can go on its own. Like $Δ H$ , it depends only on the valleys, not the hill.
$k$ = a rate constant — the number in $rate = k [\dots]$ that says how fast a given reaction runs. Two special ones:
- $k_{f}$ = forward rate constant, $k_{b}$ = backward rate constant, and $K_{e q} = k_{f} / k_{b}$ is the equilibrium constant telling us where the reaction settles.
When we compare the same reaction with and without a catalyst we tag its forward $k$ two ways: $k_{c a t}$ = rate constant with catalyst, $k_{u n}$ = rate constant without (uncatalysed). So $k_{c a t}$ and $k_{u n}$ are just the same $k$ measured on the two different roads.
$T$ = temperature, $R$ = the gas constant, $A$ = the pre-exponential (frequency) factor in $k = A e^{- E_{a} / R T}$ — it counts how often molecules collide with the right orientation.
Enzyme symbols: $[S]$ = substrate concentration; $[E]_{0}$ = total enzyme concentration; $v$ = reaction rate (velocity); $k_{2}$ = the rate constant of the step $E S \to E + P$ ; $V_{ma x} = k_{2} [E]_{0}$ = the maximum possible rate when every enzyme is busy.

True or false — justify

A catalyst is consumed in the reaction it speeds up.

False. It is regenerated — it enters the mechanism and leaves it, so its total amount is unchanged at the end even though it is chemically active midway.

A catalyst lowers the activation energy of the forward reaction only.

False. It lowers the hill itself, which is shared by both directions; the backward

E_{a}

drops by exactly the same amount, which is precisely why

K_{e q}

stays fixed.

A catalyst makes an otherwise non-spontaneous (

Δ G > 0

) reaction spontaneous.

False. Spontaneity is set by

Δ G

, a thermodynamic quantity fixed by the valleys; a catalyst only changes the hill (kinetics), so a "won't-go" reaction still won't go — just faster nowhere.

A negative catalyst (inhibitor) speeds up the reverse reaction.

False. It slows the reaction by blocking or raising the effective

E_{a}

of the path; it does not selectively accelerate the reverse direction — it hinders passage over the hill either way.

Adding a catalyst increases the yield of product at equilibrium.

False. Yield at equilibrium is governed by

K_{e q}

, which the catalyst cannot touch; you get the same final amount, just reached sooner.

In heterogeneous catalysis, doubling the reactant pressure always doubles the rate.

False. Once all surface active sites are occupied the rate saturates; beyond that point extra reactant just waits its turn, so the rate plateaus instead of doubling.

Enzymes work equally well at any temperature because catalysts are unchanged.

False. Enzymes are proteins; raising temperature past their optimum denatures them (the folded shape unravels), destroying the active site and the catalysis.

A promoter is itself a catalyst for the reaction.

False. A promoter (e.g. Mo in Haber) has little or no catalytic action alone; it enhances the true catalyst's activity, so it is a helper, not the catalyst.

The Michaelis constant

K_{M}

has the same units as a rate.

False.

K_{M}

equals the substrate concentration

[S]

at which

v = V_{ma x} /2

, so it carries units of concentration (e.g. mol/L), not rate.

Spot the error

"The reaction

2 S O_{2} + O_{2} V_{2} O_{5} 2 S O_{3}

is homogeneous catalysis because everything is in the gas region of the equation."

Error:

V_{2} O_{5}

is a solid while the reactants are gases — different phases — so this is heterogeneous catalysis on a surface.

"NO catalysing

2 S O_{2} + O_{2} \to 2 S O_{3}

in the lead chamber is heterogeneous because there are several substances."

Error: "different substances" is not "different phases." NO(g) and the gas reactants are all in the same (gas) phase, so this is homogeneous catalysis.

"Since the catalyst lowers

E_{a}

, the energy diagram shows the product valley pulled downward."

Error: Only the peak (transition state) is lowered. Reactant and product valleys are fixed, so

Δ H = H_{p r o d} - H_{r e a c t}

is unchanged — the hill drops, the house does not.

"In

E + S ⇌ E S \to E + P

, the enzyme

E

is used up making product."

Error:

E

reappears on the right of the final step; it is regenerated and free to bind the next substrate, which is why a tiny amount of enzyme processes vast amounts of substrate.

"An enzyme catalyses a huge range of reactions because it is such a good catalyst."

Error: Enzymes are famous for specificity — the lock-and-key active site fits essentially one substrate shape, so one enzyme drives roughly one reaction.

"Because

k = A e^{- E_{a} / R T}

, halving

E_{a}

merely halves the exponent, so the rate barely changes."

Error:

E_{a}

sits inside a negative exponential, so a change in the exponent is amplified exponentially into

k

— a modest

E_{a}

cut can multiply the rate thousands-fold.

"Poison and inhibitor are the same thing."

Error (subtle): a poison destroys a catalyst's activity (e.g. As on Pt), while a negative catalyst/inhibitor slows the reaction directly; they act on different targets even though both reduce the observed rate.

Why questions

Why is the same pre-exponential factor

A

used for both

k_{c a t}

(catalysed forward

k

) and

k_{u n}

(uncatalysed forward

k

) when deriving the speed-up ratio?

Because

A

counts collision frequency and orientation, which are set by temperature and molecular geometry — not by the height of the hill; holding

A

fixed leaves

E_{a}

as the only difference, so the

A

's cancel on division, isolating the catalyst's true effect as

e^{(E_{a, u n} - E_{a, c a t}) / R T}

Is it exactly true that

A

is unchanged by a catalyst?

Not perfectly — a catalyst can slightly alter the collision geometry — but the change in

A

is tiny compared with the exponential change from lowering

E_{a}

, so treating

A

as equal is an excellent, deliberately simplifying, approximation.

Why does adsorption onto a solid catalyst lower the activation energy?

When a molecule is adsorbed on an active site, its bonds are stretched and weakened, so less extra energy is needed to break them — the hill is effectively lowered right on the surface.

Why does the enzyme rate curve flatten to

V_{ma x}

at high substrate concentration

[S]

At high

[S]

virtually every enzyme is locked in the

E S

complex; adding more substrate can't create more enzyme, so the throughput hits its ceiling

V_{ma x} = k_{2} [E]_{0}

(

k_{2}

= the

E S \to E + P

rate constant,

[E]_{0}

= total enzyme).

Why can a catalyst never violate the second law of thermodynamics?

It only speeds the approach to equilibrium without shifting

K_{e q}

Δ G

; it changes the route, not the destination, so no free energy is created.

Why is desorption a necessary step in heterogeneous catalysis, not just a nicety?

If products never left the active site, the site would stay blocked; desorption frees the site, regenerating the catalyst so it can process the next reactant molecule.

Why does a catalyst speed up forward and backward reactions "equally," in a precise sense?

Both directions cross the same lowered peak; the drop in the forward

E_{a}

equals the drop in the reverse

E_{a}

, so both rate constants (

k_{f}

forward,

k_{b}

backward) multiply by the same factor and their ratio

K_{e q} = k_{f} / k_{b}

is untouched.

Why is the steady-state assumption

d [E S] / d t \approx 0

reasonable for enzymes?

E S

is a fleeting intermediate that is formed and consumed almost as fast as it appears, so its concentration barely changes over the timescale of interest, letting formation rate equal breakdown rate.

Edge cases

What happens to the speed-up ratio

k_{c a t} / k_{u n}

T \to \infty

The exponent

(E_{a, u n} - E_{a, c a t}) / R T \to 0

, so the ratio

\to e^{0} = 1

: at very high temperature thermal energy dwarfs the

E_{a}

difference, and the catalyst's relative advantage shrinks toward none.

What is the reaction order in

[S]

(substrate concentration) when

[S]

is very small compared to

K_{M}

With

[S] ≪ K_{M}

, the denominator

K_{M} + [S] \approx K_{M}

, so

v \approx (V_{ma x} / K_{M}) [S]

— the rate is first order in substrate.

What is the reaction order in

[S]

when

[S]

is very large compared to

K_{M}

With

[S] ≫ K_{M}

v \to V_{ma x}

, a constant independent of

[S]

— the reaction is zero order in substrate (saturation).

If a "catalyst" appears in the final products in a different amount than it started, is it a catalyst?

No — regeneration (same amount out as in) is part of the definition; if the amount changes it was a reactant or was consumed, not a true catalyst.

For a reaction already at equilibrium, what does adding a catalyst do?

Nothing observable to the composition: forward and reverse rates rise equally and remain equal, so the system stays exactly at the same equilibrium position.

An enzyme is placed in a strongly acidic solution far from its optimum pH — predict its behaviour.

The extreme pH disrupts the protein's folded structure (denaturation), deforming the active site, so catalytic activity collapses even though enzymes are otherwise "unchanged" catalysts.

A solid catalyst is ground into a much finer powder — what changes and what does not?

Surface area increases, so there are more active sites and the rate rises; but

Δ H

K_{e q}

, and the final equilibrium position are unchanged — more sites means faster, not further.

Recall One-line summary of every trap

A catalyst changes the hill (kinetics), never the valleys (thermodynamics): it lowers $E_{a}$ for both directions, is regenerated, cannot shift $K_{e q}$ / $Δ G$ / $Δ H$ , and — on surfaces or in enzymes — is capped by finite active sites, giving saturation.