Catalysis — homogeneous, heterogeneous, enzyme catalysis: Worked examples

This page is a drill ground. The parent note built the ideas; here we hit every kind of number the topic can throw at you — every sign, every extreme, the degenerate cases, a word problem, and an exam twist. Work each Forecast line before you read on.

Everything here rests on two tools you already met:

the Arrhenius Equation $k = A e^{- E_{a} / R T}$ — turns an Activation Energy $E_{a}$ into a speed,
the Michaelis–Menten rate law $v = \frac{V _{ma x} [ S ]}{K _{M} + [ S ]}$ from Rate Law and Order of Reaction logic applied to Enzymes and Proteins (Biomolecules).

Some symbols to keep straight before we start:

Definition The letters we will use

$E_{a}$ = activation energy = height of the energy hill (in $J/mol$ or $kJ/mol$ ). Lower hill → faster.
$A$ = the pre-exponential factor (or "frequency factor") in the Arrhenius Equation — roughly how often molecules attempt to cross the hill (collision frequency × orientation). It multiplies the exponential. When we compare a catalysed and uncatalysed path at the same conditions we keep $A$ the same so the only difference is $E_{a}$ ; that lets $A$ cancel when we divide.
$A_{f}, A_{b}$ = the pre-exponential factors for the forward and backward directions of a reaction (used in Ex 7).
$R = 8.314 J mol^{- 1} K^{- 1}$ , $T$ = temperature in kelvin (never °C in these formulas).
$R T$ = the "thermal energy budget" per mole. The combination $E_{a} / R T$ asks "how many budgets tall is the hill?" — a pure number with no units.
$[S]$ = substrate concentration; $V_{ma x}$ = the top speed when every enzyme site is busy; $K_{M}$ = the $[S]$ giving half of $V_{ma x}$ .
We write $k_{a f t er}$ for the rate constant with the catalyst/inhibitor and $k_{b e f or e}$ for the one without, and use this same before/after naming throughout.

The scenario matrix

Every catalysis calculation you can be asked is one of these cells. The examples below are tagged with the cell they cover.

Cell	What varies / the trap	Example
A. Sign check	Does $E_{a}$ go down (positive catalyst) or up (inhibitor)? Sign of the exponent.	Ex 1, Ex 2
B. Speed-up magnitude	Compute $k_{a f t er} / k_{b e f or e}$ from an $E_{a}$ drop.	Ex 1
C. Degenerate: $Δ E_{a} = 0$	No change in $E_{a}$ → ratio must be exactly $1$ .	Ex 3
D. Temperature limit	$T \to \infty$ and $T \to 0$ : what happens to the speed-up?	Ex 3
E. Enzyme low $[S]$	$[S] ≪ K_{M}$ → first order, $v \propto [S]$ .	Ex 4
F. Enzyme high $[S]$ / saturation	$[S] ≫ K_{M}$ → $v \to V_{ma x}$ (plateau, the "more is not more" trap).	Ex 4, Ex 5
G. Enzyme at $[S] = K_{M}$	exactly half $V_{ma x}$ (the defining point).	Ex 5
H. Word problem	Real industrial process, pick the catalysis type + reason.	Ex 6
I. Exam twist	Equilibrium / $Δ H$ invariance under catalysis.	Ex 7
J. Back-solve	Given the speed-up, find the $E_{a}$ drop (invert the exponential).	Ex 8

Cells A & B — sign and magnitude of the speed-up

Worked example Ex 1 — positive catalyst, "how many times faster?" (A, B)

A catalyst lowers $E_{a}$ from $80$ to $60 kJ/mol$ at $T = 350 K$ . By what factor does the rate constant rise?

Forecast: a $20 kJ/mol$ cut at a warm temperature — guess whether it's tens, thousands, or millions before computing.

Convert to joules and find the drop. $Δ E_{a} = E_{a, b e f or e} - E_{a, a f t er} = (80 - 60) \times 1 0^{3} = 20000 J/mol$ . Why this step? The Arrhenius exponent needs $E_{a}$ in $J/mol$ so it cancels the $J/mol$ inside $R T$ , leaving a pure number.
Compute the thermal budget. $R T = 8.314 \times 350 = 2909.9 J/mol$ . Why? $R T$ is the yardstick; $E_{a} / R T$ measures the hill in these units.
Form the ratio. From dividing the two Arrhenius expressions (same $A$ , same $T$ cancel), $\frac{k _{a f t er}}{k _{b e f or e}} = e^{(E_{a, b e f or e} - E_{a, a f t er}) / R T} = e^{20000/2909.9} = e^{6.873} .$ Why the $A$ 's cancel: we hold the pre-exponential factor $A$ and $T$ fixed so the only difference is the hill height — isolating the catalyst.
Evaluate. $e^{6.873} \approx 966$ .

Verify: exponent is positive (hill got lower ⇒ faster ⇒ ratio $> 1$ ) ✓. A ~20 kJ/mol cut at ~350 K giving ~10³ is in the right ballpark for the exponential rule of thumb. Answer ≈ 966× faster.

Worked example Ex 2 — negative catalyst / inhibitor (A, sign flips)

An inhibitor raises the effective barrier from $50$ to $70 kJ/mol$ at $T = 300 K$ . By what factor does the rate change?

Forecast: the hill got higher — do you expect a number bigger or smaller than $1$ ?

Signed drop. $E_{a, b e f or e} - E_{a, a f t er} = 50 - 70 = - 20000 J/mol$ (negative — the barrier grew). Why keep the sign? The whole point: a negative catalyst makes the exponent negative.
Budget. $R T = 8.314 \times 300 = 2494.2 J/mol$ .
Ratio. $\frac{k _{a f t er}}{k _{b e f or e}} = e^{- 20000/2494.2} = e^{- 8.019} \approx 3.3 \times 1 0^{- 4}$ . Why the $A$ 's still cancel here: exactly as in Ex 1, we keep the pre-exponential factor $A$ and $T$ the same for the with- and without-inhibitor paths, so dividing wipes out $A$ and only the barrier difference survives in the exponent.

Verify: ratio $< 1$ ⇒ slower ⇒ consistent with "inhibitor" ✓. It is the reciprocal-style partner of a speed-up: raising the hill by 20 kJ/mol shrinks the rate by ~3000×. Answer ≈ 0.00033× (about 3000× slower).

Cells C & D — degenerate ( $Δ E_{a} = 0$ ) and temperature limits

Worked example Ex 3 — no change, and the two temperature extremes (C, D)

(a) A "catalyst" is added but $E_{a}$ stays at $60 kJ/mol$ (it did nothing). Speed-up? (b) For a real drop of $Δ E_{a} = 20 kJ/mol$ , what does the speed-up ratio approach as $T \to \infty$ and as $T \to 0^{+}$ ?

Forecast: guess the number for (a). For (b), picture the fraction $Δ E_{a} / R T$ as $T$ balloons vs. shrinks.

(a) Substitute $Δ E_{a} = 0$ . $\frac{k _{a f t er}}{k _{b e f or e}} = e^{0/ R T} = e^{0} = 1$ . Why this matters: it's the sanity anchor — no change in the hill means exactly no change in rate, for any $T$ . A formula that gave anything but $1$ here would be wrong.
(b) Hot limit. As $T \to \infty$ , $\frac{Δ E _{a}}{R T} \to 0$ , so $e^{Δ E_{a} / R T} \to e^{0} = 1$ . Why? When thermal energy is enormous, molecules clear any hill easily; a shortcut helps almost nothing — the catalyst's advantage vanishes.
(b) Cold limit. As $T \to 0^{+}$ , $\frac{Δ E _{a}}{R T} \to + \infty$ , so $e^{Δ E_{a} / R T} \to \infty$ . Why? When there is barely any thermal energy, a lower hill is everything — the catalyst's relative advantage explodes.

Verify: the curve of speed-up vs $T$ (below) starts astronomically high on the cold left, decays smoothly toward $1$ as it heats. At $300 K$ it reads ~ $3000$ for $Δ E_{a} = 20$ kJ/mol, matching Ex 1-style numbers ✓.

Figure — Catalysis — homogeneous, heterogeneous, enzyme catalysis

Cells E, F, G — enzyme kinetics across all $[S]$

We now switch tools: the Rate Law and Order of Reaction result $v = \frac{V _{ma x} [ S ]}{K _{M} + [ S ]}$ . Take $V_{ma x} = 100 μ M/s$ and $K_{M} = 5 μ M$ throughout.

Worked example Ex 4 — low, equal, and high substrate (E, F, G)

Find $v$ at (a) $[S] = 0.5 μ M$ (low, $≪ K_{M}$ ), (b) $[S] = 5 μ M$ ( $= K_{M}$ ), (c) $[S] = 500 μ M$ (high, $≫ K_{M}$ ).

Forecast: which case gives roughly a straight-line dependence on $[S]$ , and which one plateaus?

(a) Low $[S]$ . $v = \frac{100 \times 0.5}{5 + 0.5} = \frac{50}{5.5} = 9.09 μ M/s$ . Why this is "first order": since $[S] ≪ K_{M}$ , the denominator $\approx K_{M}$ , so $v \approx \frac{V _{ma x}}{K _{M}} [S]$ — rate is proportional to $[S]$ .
(b) $[S] = K_{M}$ . $v = \frac{100 \times 5}{5 + 5} = \frac{500}{10} = 50 μ M/s$ . Why exactly half: by construction $[S] = K_{M}$ makes the denominator $= 2 [S]$ , so $v = V_{ma x} /2$ . This is the definition of $K_{M}$ .
(c) High $[S]$ . $v = \frac{100 \times 500}{5 + 500} = \frac{50000}{505} = 99.01 μ M/s$ . Why it nears $V_{ma x}$ : $[S] ≫ K_{M}$ makes denominator $\approx [S]$ , the $[S]$ 's cancel, $v \to V_{ma x} = 100$ . Every active site is busy — saturation.

Verify (units + trend): all three are $μ M/s$ ✓. Values climb $9.09 \to 50 \to 99.01$ but flatten toward the ceiling $100$ — never exceeding it ✓. That plateau is the "more catalyst/substrate is NOT proportionally more" trap made numerical.

Worked example Ex 5 — back-solve the saturation trap (F, G)

How much $[S]$ do you need to reach $90%$ of $V_{ma x}$ ? Then comment on the "just double it" intuition.

Forecast: to go from $50%$ (at $[S] = K_{M} = 5$ ) to $90%$ , do you think $[S]$ merely doubles?

Set the target. Want $v = 0.9 V_{ma x}$ , i.e. $\frac{V _{ma x} [ S ]}{K _{M} + [ S ]} = 0.9 V_{ma x}$ . Why cancel $V_{ma x}$ : it appears on both sides, so the answer won't depend on it.
Cancel and solve. $\frac{[ S ]}{K _{M} + [ S ]} = 0.9 \Rightarrow [S] = 0.9 K_{M} + 0.9 [S] \Rightarrow 0.1 [S] = 0.9 K_{M} \Rightarrow [S] = 9 K_{M}$ . Why algebra not guessing: the relationship is a hyperbola, not a line — intuition fails.
Numbers. $[S] = 9 \times 5 = 45 μ M$ .

Verify: plug back: $v = \frac{100 \times 45}{5 + 45} = \frac{4500}{50} = 90 μ M/s = 90%$ of $V_{ma x}$ ✓. Lesson: to climb the last stretch you needed 9× the $K_{M}$ , not a mere doubling — saturation makes the top of the curve expensive. This is the exam trap in cell F.

Cell H — real-world word problem

Worked example Ex 6 — pick the type and justify (H)

In the Contact process, sulfur dioxide gas and oxygen gas pass over solid $V_{2} O_{5}$ pellets to make $S O_{3}$ . Classify the catalysis and state the mechanism steps.

Forecast: are the catalyst and reactants in the same phase or different phases?

Identify phases. Reactants $S O_{2}, O_{2}$ are gas; catalyst $V_{2} O_{5}$ is solid. Different phases. Why phase is the key: the classification homogeneous vs heterogeneous is defined purely by whether catalyst and reactants share a phase (see Adsorption).
Classify. Different phase ⇒ heterogeneous catalysis.
Mechanism (surface, four explicit steps).
- (i) Diffusion — the gas molecules $S O_{2}$ and $O_{2}$ must first travel from the bulk gas to the solid surface. This is a transport step, not a chemical one; if the gas is stagnant or the pellet pores are clogged, this step alone can throttle the whole reaction (a "diffusion-limited" regime). See the transport picture in Adsorption.
- (ii) Adsorption — molecules stick onto active sites on the $V_{2} O_{5}$ surface; the surface bonds stretch and weaken the reactant bonds, which is precisely what lowers $E_{a}$ .
- (iii) Reaction — the adsorbed, weakened species react on the surface to form $S O_{3}$ .
- (iv) Desorption — $S O_{3}$ leaves the surface, freeing the active site so the catalyst is regenerated for the next cycle.

Verify: matches the parent's heterogeneous examples ( $V_{2} O_{5}$ for Contact) ✓ and the four-step Adsorption sequence, with the diffusion (transport) step made explicit ✓. Compare: the Lead-chamber version uses $N O$ gas with gas reactants — same phase — that one is homogeneous. Same overall reaction ( $S O_{2} \to S O_{3}$ ), different catalysis class because of phase.

Cell I — the equilibrium / $Δ H$ exam twist

Worked example Ex 7 — does the catalyst move the equilibrium? (I)

A reaction has forward and reverse activation energies $E_{a, f} = 100$ and $E_{a, b} = 60 kJ/mol$ . A catalyst lowers both by $30 kJ/mol$ at $T = 300 K$ . Show $K_{e q}$ and $Δ H$ are unchanged.

Forecast: if both directions get $30 kJ/mol$ faster... does their ratio change?

Reaction enthalpy before. $Δ H = E_{a, f} - E_{a, b} = 100 - 60 = 40 kJ/mol$ . Why this formula: the barrier from reactant side minus barrier from product side equals the height difference of the two valleys — see Chemical Equilibrium.
After the catalyst. New barriers $70$ and $30$ . New $Δ H = 70 - 30 = 40 kJ/mol$ — identical. Why: subtracting the same $30$ from both leaves the difference untouched. The valleys (reactant, product) never moved; only the peak dropped.
Equilibrium constant. $K_{e q} = k_{f} / k_{b}$ . Write each with its own pre-exponential factor: $k_{f} = A_{f} e^{- E_{a, f} / R T}$ , $k_{b} = A_{b} e^{- E_{a, b} / R T}$ . The catalyst multiplies both rates by the same factor $e^{30000/ R T}$ , so $K_{e q}^{n e w} = \frac{A _{f} e ^{- (E_{a, f} - 30000) / R T}}{A _{b} e ^{- (E_{a, b} - 30000) / R T}} = \frac{A _{f}}{A _{b}} e^{- (E_{a, f} - E_{a, b}) / R T} = K_{e q}^{o l d} .$ Assumption made explicit: this cancellation needs the same catalyst factor on both directions (true, since the catalyst offers one shared lowered path) and that the pre-exponential factors $A_{f}, A_{b}$ are unchanged by the catalyst — an implicit but standard assumption. Only then do the $e^{30000/ R T}$ terms cancel cleanly.

Verify: the common factor $e^{30000/ R T} = e^{12.03} \approx 1.68 \times 1 0^{5}$ multiplies both $k_{f}$ and $k_{b}$ , so it cancels exactly in the ratio ✓. Both $Δ H$ (=40 kJ/mol) and $K_{e q}$ are invariant — the reaction reaches the same equilibrium, just sooner. "Lower the HILL, not the HOUSE."

Cell J — inverting the exponential

Worked example Ex 8 — given the speed-up, find the

$E_{a}$ drop (J) A catalyst makes a reaction $1 0^{4}$ times faster at $T = 310 K$ (body temperature). What $E_{a}$ drop does this require?

Forecast: we've seen ~20–25 kJ/mol give ~10⁴ near room temperature — expect something similar.

Write the ratio and take a log. $\frac{k _{a f t er}}{k _{b e f or e}} = e^{Δ E_{a} / R T} = 1 0^{4}$ . Take natural log of both sides: $\frac{Δ E _{a}}{R T} = ln (1 0^{4}) = 4 ln 10 = 9.210.$ Why $ln$ : it is the exact inverse of $e^{(\cdot)}$ — it "undoes" the exponential to free $Δ E_{a}$ .
Multiply out. $Δ E_{a} = 9.210 \times R T = 9.210 \times 8.314 \times 310$ . Why multiply by $R T$ : rearranging the equation isolates $Δ E_{a}$ .
Compute. $R T = 8.314 \times 310 = 2577.3 J/mol$ ; $Δ E_{a} = 9.210 \times 2577.3 = 23737 J/mol \approx 23.7 kJ/mol$ .

Verify: forward-check — $e^{23737/2577.3} = e^{9.210} = 1.0 \times 1 0^{4}$ ✓, units $J/mol$ ✓, and ~24 kJ/mol matches the forecast/parent's "22,000× from 25 kJ/mol" scale ✓.

Recall Cover-the-answers self-test (all cells)

Positive catalyst 80→60 kJ/mol at 350 K, speed-up? ::: $e^{20000/2909.9} \approx 966$ × Inhibitor 50→70 kJ/mol at 300 K, factor? ::: $e^{- 8.019} \approx 3.3 \times 1 0^{- 4}$ (slower) $Δ E_{a} = 0$ speed-up? ::: exactly $1$ (no change) Speed-up as $T \to \infty$ ? ::: approaches $1$ (catalyst advantage vanishes) $v$ at $[S] = K_{M}$ ? ::: $V_{ma x} /2$ $[S]$ for $90% V_{ma x}$ ? ::: $9 K_{M}$ Contact process ( $V_{2} O_{5}$ solid, gas reactants) type? ::: heterogeneous Does a catalyst change $K_{e q}$ ? ::: no — both directions scale equally Speed-up $1 0^{4}$ at 310 K needs $Δ E_{a} =$ ? ::: $\approx 23.7$ kJ/mol

The scenario matrix

Cells A & B — sign and magnitude of the speed-up

Cells C & D — degenerate (ΔEa​=0) and temperature limits

Cells E, F, G — enzyme kinetics across all [S]

Cell H — real-world word problem

Cell I — the equilibrium / ΔH exam twist

Cell J — inverting the exponential

Cells C & D — degenerate ( $Δ E_{a} = 0$ ) and temperature limits

Cells E, F, G — enzyme kinetics across all $[S]$

Cell I — the equilibrium / $Δ H$ exam twist