In a series of runs of a reaction with a single reactant A, halving the initial concentration [A]₀ doubles the half-life. What is the order of the reaction?

Second order, whose half-life is 1/(k[A]₀)

Zero order, whose half-life is [A]₀/2k

First order, whose half-life is 0.693/k

Third order, whose half-life is 3/(2k[A]₀²)

Second order, whose half-life is 1/(k[A]₀)

The acid-catalysed hydrolysis of an ester in dilute aqueous solution follows first-order kinetics in the ester, although water is consumed in the reaction. What is the best explanation?

Water is in such large excess that its concentration stays constant and is absorbed into k

Water acts only as a catalyst, so it is regenerated and does not enter the rate law

Water is in such large excess that its concentration stays constant and is absorbed into k

The rate-determining step is unimolecular, so its molecularity fixes the order at one

The H⁺ catalyst is consumed at the same rate as the ester, which cancels its effect

Which statement correctly describes the effect of a catalyst on a reversible reaction at a fixed temperature?

It lowers the barrier for the forward and reverse reactions equally, so K is unchanged

It lowers the activation energy of only the forward reaction, so the equilibrium shifts towards products

It makes ΔG° of the reaction more negative, and that is why the reaction becomes faster

It lowers the barrier for the forward and reverse reactions equally, so K is unchanged

It raises the average kinetic energy of the molecules, so more collisions exceed Ea

Reactions P and Q have the same pre-exponential factor; their activation energies are 40 kJ mol⁻¹ (P) and 80 kJ mol⁻¹ (Q), both independent of temperature. The temperature is raised from 300 K to 310 K. Which statement is correct? (R = 8.314 J mol⁻¹ K⁻¹)

k of Q rises by the larger factor, although k of Q remains far smaller than k of P

k of P rises by the larger factor, because a lower barrier is crossed more easily

k of P and k of Q rise by the same factor, because the temperature rise is the same

k of Q rises by the larger factor and overtakes k of P at 310 K

For an enzyme obeying Michaelis–Menten kinetics, what does the Michaelis constant Km represent operationally?

The substrate concentration at which the initial rate is exactly half of Vmax

The dissociation constant of the ES complex, whatever the relative sizes of k₋₁ and kcat

The substrate concentration needed to saturate every active site of the enzyme

The substrate concentration at which the initial rate is exactly half of Vmax

Lineweaver–Burk plots of an enzyme measured with and without a fixed concentration of a reversible inhibitor are parallel straight lines, the inhibited line lying above the other. What does this indicate?

Uncompetitive inhibition: apparent Vmax and Km both fall by the same factor

Competitive inhibition: the apparent Km rises while Vmax is unchanged

Uncompetitive inhibition: apparent Vmax and Km both fall by the same factor

Pure noncompetitive inhibition: apparent Vmax falls while Km is unchanged

Mixed inhibition: the inhibitor binds E and ES with different affinities

An enzyme is preincubated with compound X. Samples taken at intervals are passed through a gel-filtration desalting column to remove free X and then assayed at saturating substrate. Activity falls steadily with preincubation time and is not recovered after X is removed; including a high substrate concentration during preincubation slows the loss. What is X most likely to be?

An active-site-directed irreversible inhibitor that modifies the enzyme covalently

A competitive reversible inhibitor, since the substrate protects the enzyme from it

An uncompetitive reversible inhibitor that binds only to the enzyme–substrate complex

An active-site-directed irreversible inhibitor that modifies the enzyme covalently

A non-specific denaturant whose action is unaffected by binding of the substrate

Which of the following statements about order and molecularity are correct?

The order of a reaction can be zero or fractional, whereas the molecularity of an elementary step is a positive whole number

For an elementary step, the order in each reactant equals that reactant's coefficient in the step

The overall order of a multistep reaction can always be read from the coefficients of its balanced equation

The order of a reaction can be zero or fractional, whereas the molecularity of an elementary step is a positive whole number

Termolecular elementary steps are common because three-body collisions are frequent in solution

For an elementary step, the order in each reactant equals that reactant's coefficient in the step

Which of the following are true of a reaction that is first order in its only reactant A?

A plot of ln[A] against time is a straight line of slope −k

Its half-life does not depend on the initial concentration of A

About ten half-lives are needed for 99.9% of A to react

A plot of ln[A] against time is a straight line of slope −k

Its half-life does not depend on the initial concentration of A

Its rate constant is expressed in units of mol L⁻¹ s⁻¹

About ten half-lives are needed for 99.9% of A to react

For an enzyme that follows the Michaelis–Menten scheme E + S ⇌ ES → E + P, which statements are correct?

At [S] ≫ Km the initial rate is nearly independent of [S], that is, zero order in substrate

kcat/Km has units of M⁻¹ s⁻¹ and cannot exceed the rate at which enzyme and substrate meet by diffusion

Doubling the total enzyme concentration at a fixed [S] doubles both Vmax and Km

At [S] ≫ Km the initial rate is nearly independent of [S], that is, zero order in substrate

Km equals the dissociation constant of ES whatever the relative sizes of k₋₁ and kcat

kcat/Km has units of M⁻¹ s⁻¹ and cannot exceed the rate at which enzyme and substrate meet by diffusion

Which of the following statements about irreversible enzyme inhibitors are correct?

Diisopropyl fluorophosphate inactivates serine proteases by phosphorylating the active-site serine

Iodoacetamide inactivates some enzymes by alkylating the thiol group of an essential cysteine

Diisopropyl fluorophosphate inactivates serine proteases by phosphorylating the active-site serine

Activity lost to an irreversible inhibitor returns once the inhibitor is removed by dialysis

Iodoacetamide inactivates some enzymes by alkylating the thiol group of an essential cysteine

An irreversible inhibitor raises the apparent Km while leaving the apparent Vmax unchanged

Reaction Kinetics: Rate Laws, Activation Energy, Catalysis and Enzyme Inhibition

This chapter answers Section XL0.7 of XL0 Chemistry, the part of the Life Sciences (XL) paper that every XL candidate answers: “Rate constant, order of reaction, molecularity, activation energy, zero, first and second order kinetics, catalysis and elementary enzyme reactions. Reversible and irreversible inhibition of enzymes.” It builds the rate law and separates order from molecularity, integrates the zero-, first- and second-order laws, links temperature to rate through the Arrhenius equation, explains how catalysts and enzymes lower the barrier, and ends with Michaelis–Menten kinetics and reversible and irreversible inhibition at the depth a chemistry section needs. The numericals are the units of k, orders from initial rates, half-lives and conversion times, activation energy from two temperatures, and enzyme rates with and without an inhibitor.

1. Rate, rate law, order and molecularity

The rate of a reaction aA + bB → cC + dD is the change of concentration per unit time divided by the stoichiometric coefficient, so that it is the same whichever species is followed: rate = −(1/a)d[A]/dt = (1/c)d[C]/dt. In 2N₂O₅ → 4NO₂ + O₂, NO₂ appears four times as fast as O₂ and twice as fast as N₂O₅ disappears. Experiment gives a rate law, rate = k[A]^m[B]^n, in which k is the rate constant (fixed at a given temperature, independent of concentration) and m and n are the orders with respect to A and B; the overall order is m + n. Orders are found by experiment, not read from the balanced equation, and may be zero, fractional or even negative. Since a rate always has units of mol L⁻¹ s⁻¹, the units of k reveal the overall order n: (mol L⁻¹)^(1−n) s⁻¹, that is mol L⁻¹ s⁻¹ for zero order, s⁻¹ for first order and L mol⁻¹ s⁻¹ for second order.

Molecularity is the number of reactant particles that come together in one elementary step: one (unimolecular, as in a decay or isomerisation), two (bimolecular) or, rarely, three (termolecular), because a simultaneous three-body collision is improbable. It is a theoretical count, always a small whole number, and has meaning only for an elementary step. For an elementary step the order in each reactant equals its coefficient in that step, so order and molecularity coincide; for a multistep reaction the rate law is set by the slowest, rate-determining step and the steps before it. A reaction can be first order yet bimolecular: in the acid hydrolysis of an ester in water, or the inversion of sucrose, water is in such excess that its concentration hardly changes and is absorbed into an observed constant k′ = k[H₂O]. Such a reaction is pseudo-first order.

Worked example (initial rates). For A + B → P, doubling [A] at fixed [B] raises the initial rate from 1.5 × 10⁻³ to 6.0 × 10⁻³ mol L⁻¹ s⁻¹, while doubling [B] at fixed [A] leaves it unchanged. Then 4 = 2^m gives m = 2 and 1 = 2^n gives n = 0, so rate = k[A]², overall second order, with k in L mol⁻¹ s⁻¹. B's absence from the rate law means that B enters after the rate-determining step, not that B is unnecessary.

2. Zero-, first- and second-order kinetics

Integrating the rate law gives concentration as a function of time, and whichever plot comes out straight identifies the order from a single run. The half-life t½, the time for [A] to fall to half its value, is the quickest fingerprint: it grows with [A]₀ for zero order, does not depend on [A]₀ for first order, and shrinks as [A]₀ rises for second order.

Integrated rate laws for a single reactant A (rate written as −d[A]/dt)
Order	Rate law	Integrated form	Straight-line plot	Half-life	Units of k
Zero	k	[A] = [A]₀ − kt	[A] against t, slope −k	[A]₀/2k, proportional to [A]₀	mol L⁻¹ s⁻¹
First	k[A]	ln([A]₀/[A]) = kt	ln[A] against t, slope −k	0.693/k, independent of [A]₀	s⁻¹
Second	k[A]²	1/[A] = 1/[A]₀ + kt	1/[A] against t, slope +k	1/(k[A]₀), inversely proportional to [A]₀	L mol⁻¹ s⁻¹

First order matters most in biology: radioactive decay, most drug elimination and every pseudo-first-order reaction follow it. Since ln([A]₀/[A]) = kt, or t = (2.303/k) log([A]₀/[A]), the time to reach any fixed fraction does not depend on the starting amount. After n half-lives the fraction left is (1/2)ⁿ, so 87.5% conversion takes three half-lives and 99.9% about ten (ln 1000/ln 2 = 9.97). Zero order arises when something other than the reactant limits the rate: a saturated enzyme or catalyst surface, or a photochemical reaction limited by light. Ethanol elimination is the usual biological case, because liver alcohol dehydrogenase is saturated at ordinary blood levels, so a fixed amount is removed per hour. A zero-order reaction is complete at t = [A]₀/k, whereas first- and second-order reactions only approach completion.

Worked example. A first-order reaction with k = 0.0231 min⁻¹ has t½ = 0.693/0.0231 = 30.0 min, and needs t = (2.303/0.0231) log 10 = 99.7 min for 90% conversion, about 3.3 half-lives. A second-order reaction with k = 0.10 L mol⁻¹ s⁻¹ and [A]₀ = 0.50 mol L⁻¹ has t½ = 1/(0.10 × 0.50) = 20 s; the next half-life, starting from 0.25 mol L⁻¹, is 40 s. Successive half-lives that double are the signature of second order.

⚠️ Check how k is defined

The first-order half-life formula is not valid for other orders. For a reaction written 2A → P, check whether k is defined from −d[A]/dt or from −(1/2)d[A]/dt: the second definition turns the integrated law into 1/[A] = 1/[A]₀ + 2kt.

3. Temperature and activation energy

Rates rise steeply with temperature, often doubling or tripling for a 10 K rise near room temperature. Arrhenius expressed this as k = A e^(−Ea/RT), where Ea is the activation energy, the minimum energy a colliding pair needs to react, and A is the pre-exponential factor. In logarithmic form ln k = ln A − Ea/RT, so a plot of ln k against 1/T is a straight line of slope −Ea/R. With rate constants at two temperatures, ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂).

Collision theory explains the form: e^(−Ea/RT) is the fraction of collisions carrying at least Ea, and A gathers the collision frequency and a steric factor for correct orientation. Transition-state theory pictures the reactants climbing an energy hill to a short-lived activated complex (the transition state) at its top; Ea is the height of the hill above the reactants. In the energy profile of an exothermic step the products lie below the reactants by ΔH, and Ea(forward) − Ea(reverse) = ΔH. The barrier and the thermodynamics are therefore separate quantities: a strongly favourable reaction can still be immeasurably slow.

Worked example. If a rate constant doubles between 298 K and 308 K, Ea = R ln 2/(1/298 − 1/308) = (8.314 × 0.693)/(1.090 × 10⁻⁴) ≈ 52.9 kJ mol⁻¹. The larger Ea, the more sensitive k is to temperature: over the same interval, a reaction with twice the activation energy has its rate constant multiplied by the square of the factor.

4. Catalysis, from protons to enzymes

A catalyst increases the rate without being consumed, by opening an alternative pathway of lower activation energy. It lowers the barrier of the forward and the reverse reaction by the same amount, so it speeds both by the same factor, shortens the time to reach equilibrium, and leaves ΔG°, ΔH and the equilibrium constant unchanged; it cannot make a non-spontaneous reaction go. A homogeneous catalyst is in the same phase as the reactants (H⁺ in ester hydrolysis, acid–base catalysis in general); a heterogeneous catalyst acts at a surface on which reactants adsorb (nickel in the hydrogenation of oils), and once the surface is fully covered the rate becomes zero order in the gas.

Enzymes are protein (and a few RNA) catalysts, highly specific and commonly accelerating reactions by factors of 10⁶ to 10¹² or more. They bind the transition state more tightly than the substrate, and combine proximity and orientation of the reacting groups, general acid–base catalysis by side chains such as His, Asp and Glu, covalent catalysis through a transient enzyme–substrate bond (the active-site Ser of chymotrypsin forms an acyl-enzyme), and metal-ion catalysis (the Zn²⁺ of carbonic anhydrase activates a water molecule). Because k depends exponentially on the barrier, a modest drop in Ea gives a huge gain: at 298 K, lowering Ea by 34 kJ mol⁻¹ with A unchanged multiplies k by e^(34000/(8.314 × 298)) ≈ 9 × 10⁵. Enzymes also show an optimum temperature, above which thermal denaturation overtakes the Arrhenius rise.

🧠 How fast, never how far

A catalyst or enzyme changes how fast equilibrium is reached, never where it lies. Any option claiming that a catalyst shifts K, makes ΔG° more negative or raises the yield at equilibrium is wrong.

5. Elementary enzyme kinetics: the Michaelis–Menten model

The simplest scheme is E + S ⇌ ES → E + P, with k₁ and k₋₁ for binding and release and k₂ (= kcat) for the catalytic step. Assuming that [ES] stays constant after a brief initial burst (the steady-state assumption of Briggs and Haldane) and that [S] ≫ [E]ₜ, the initial rate is v₀ = Vmax[S]/(Km + [S]), with Vmax = kcat[E]ₜ and Km = (k₋₁ + kcat)/k₁.

Km is the substrate concentration at which v₀ = Vmax/2. It equals the dissociation constant of ES only when kcat ≪ k₋₁; only then does a low Km mean tight binding. kcat, the turnover number, is the number of substrate molecules converted per active site per second at saturation. kcat/Km, the specificity constant (M⁻¹ s⁻¹), is the second-order rate constant for E + S at low [S]; its ceiling of about 10⁸ to 10⁹ M⁻¹ s⁻¹ is set by diffusion. The hyperbola links back to order: at [S] ≪ Km, v₀ ≈ (Vmax/Km)[S], first order in S; at [S] ≫ Km, v₀ ≈ Vmax, zero order.

Worked example. For Km = 2.0 mM and [S] = 6.0 mM, v₀/Vmax = 6.0/(2.0 + 6.0) = 0.75. If Vmax = 0.40 µmol L⁻¹ s⁻¹ with [E]ₜ = 2.0 nmol L⁻¹, kcat = (0.40 × 10⁻⁶)/(2.0 × 10⁻⁹) = 200 s⁻¹. A useful ladder: [S] = Km gives 50%, 3Km gives 75% and 9Km gives 90% of Vmax.

Because Vmax is approached only asymptotically, the double-reciprocal Lineweaver–Burk plot is used to read the constants: 1/v₀ = (Km/Vmax)(1/[S]) + 1/Vmax, a straight line with y-intercept 1/Vmax, x-intercept −1/Km and slope Km/Vmax. Its weakness is that the lowest-[S] points, the least precise, lie farthest out and dominate the fit.

6. Reversible and irreversible inhibition

A reversible inhibitor binds non-covalently and leaves when diluted or dialysed away. Write α = 1 + [I]/Kᵢ for its binding to free E and α′ = 1 + [I]/Kᵢ′ for its binding to ES; the pattern depends on which form it binds.

Reversible inhibition at a glance
Type	Binds	Apparent Vmax	Apparent Km	Lineweaver–Burk lines
Competitive	Free E only, usually at the active site	Unchanged	αKm (raised)	Meet on the y-axis
Uncompetitive	ES only	Vmax/α′	Km/α′ (lowered)	Parallel
Pure noncompetitive (Kᵢ = Kᵢ′)	E and ES equally	Vmax/α	Unchanged	Meet on the x-axis
Mixed (Kᵢ ≠ Kᵢ′)	E and ES unequally	Vmax/α′	Kmα/α′ (raised or lowered)	Meet left of the y-axis, off the x-axis

A competitive inhibitor is overcome by raising [S], since substrate outcompetes it for the same site; malonate acting on succinate dehydrogenase is the classic case. Worked example: with [I] = 2Kᵢ, α = 3. At [S] = Km the uninhibited rate is Vmax/2, but the inhibited rate is Vmax·Km/(3Km + Km) = Vmax/4; [S] must be raised to 3Km to restore half-maximal rate.

An irreversible inhibitor forms a covalent bond, or binds so tightly that it practically never leaves, and so removes active enzyme from the pool. Diisopropyl fluorophosphate (DIFP) phosphorylates the active-site serine of serine proteases and acetylcholinesterase; iodoacetamide alkylates essential cysteine thiols; aspirin acetylates a serine of cyclooxygenase; penicillin acylates the bacterial transpeptidase. Mechanism-based (suicide) inhibitors are unreactive until the target enzyme begins its own catalytic cycle on them and turns them into a reactive group that attacks the active site. The enzyme that survives behaves normally, so an irreversible inhibitor lowers the apparent Vmax and leaves Km unchanged.

⚠️ Irreversible or noncompetitive?

Both lower Vmax and leave Km unchanged. Remove the free inhibitor by dialysis or gel filtration: a reversible inhibitor's effect disappears, an irreversible one's stays. Irreversible inactivation also deepens with the time of exposure, and if substrate present during the exposure slows it, the inhibitor is attacking the active site.

Key takeaways

Orders are the exponents of the experimental rate law; molecularity is the number of particles in one elementary step, and the two coincide only for an elementary step. The units of k, (mol L⁻¹)^(1−n) s⁻¹, give the overall order n.
Zero, first and second order: [A] = [A]₀ − kt, ln[A] = ln[A]₀ − kt and 1/[A] = 1/[A]₀ + kt, with half-lives [A]₀/2k, 0.693/k and 1/(k[A]₀). Only the first-order half-life is independent of [A]₀.
Arrhenius: k = A e^(−Ea/RT) and ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂); a larger Ea means a steeper temperature dependence.
A catalyst lowers Ea for the forward and reverse reactions alike and never changes K or ΔG°. Enzymes add transition-state binding, proximity, acid–base, covalent and metal-ion catalysis.
Michaelis–Menten: v₀ = Vmax[S]/(Km + [S]); Km is [S] at Vmax/2, kcat = Vmax/[E]ₜ, and kcat/Km measures specificity. The Lineweaver–Burk intercepts are 1/Vmax and −1/Km.
Competitive inhibition raises Km, uncompetitive lowers Km and Vmax by the same factor, and pure noncompetitive lowers only Vmax. Irreversible inhibition also lowers Vmax, but it deepens with time and does not reverse on dialysis.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The rate constant of a reaction is reported in L mol⁻¹ s⁻¹. What is the overall order of the reaction?
1. Order 0
2. Order 1
3. Order 2
4. Order 3
Show answer
Answer: C — Order 2
The units of k are (mol L⁻¹)^(1−n) s⁻¹. For n = 2 this is (mol L⁻¹)⁻¹ s⁻¹ = L mol⁻¹ s⁻¹. First order would give s⁻¹ (the tempting slip, because seconds appear in both), zero order mol L⁻¹ s⁻¹ and third order L² mol⁻² s⁻¹.
For the reaction A + B → products at a fixed temperature, three initial-rate runs gave: run 1, [A] = 0.10 M, [B] = 0.10 M, rate = 2.0 × 10⁻⁴ M s⁻¹; run 2, [A] = 0.20 M, [B] = 0.10 M, rate = 8.0 × 10⁻⁴ M s⁻¹; run 3, [A] = 0.20 M, [B] = 0.30 M, rate = 2.4 × 10⁻³ M s⁻¹. Which rate law fits these data?
1. rate = k[A]²[B]
2. rate = k[A][B]²
3. rate = k[A][B]
4. rate = k[A]²[B]²
Show answer
Answer: A — rate = k[A]²[B]
Runs 1 and 2: [A] doubles at fixed [B] and the rate rises 4-fold, so 2^m = 4 and m = 2. Runs 2 and 3: [B] triples at fixed [A] and the rate rises 3-fold, so n = 1. Hence rate = k[A]²[B]. The tempting k[A][B]² swaps the orders by comparing the wrong pair of runs; k[A][B] would give only a 2-fold rise in runs 1 to 2.
In a series of runs of a reaction with a single reactant A, halving the initial concentration [A]₀ doubles the half-life. What is the order of the reaction?
1. Zero order, whose half-life is [A]₀/2k
2. First order, whose half-life is 0.693/k
3. Third order, whose half-life is 3/(2k[A]₀²)
4. Second order, whose half-life is 1/(k[A]₀)
Show answer
Answer: D — Second order, whose half-life is 1/(k[A]₀)
A half-life inversely proportional to [A]₀ is the second-order signature: t½ = 1/(k[A]₀), so halving [A]₀ doubles t½. A zero-order half-life would halve, a first-order one would not change, and a third-order one (proportional to 1/[A]₀²) would become four times longer.
The acid-catalysed hydrolysis of an ester in dilute aqueous solution follows first-order kinetics in the ester, although water is consumed in the reaction. What is the best explanation?
1. Water acts only as a catalyst, so it is regenerated and does not enter the rate law
2. Water is in such large excess that its concentration stays constant and is absorbed into k
3. The rate-determining step is unimolecular, so its molecularity fixes the order at one
4. The H⁺ catalyst is consumed at the same rate as the ester, which cancels its effect
Show answer
Answer: B — Water is in such large excess that its concentration stays constant and is absorbed into k
This is a pseudo-first-order reaction: about 55 mol L⁻¹ of water barely changes as a little ester reacts, so rate = k[ester][H₂O] behaves as k′[ester] with k′ = k[H₂O]. Water is a reactant, not a catalyst, so the option calling water a catalyst misreads its role; H⁺ is the catalyst and is not consumed.
Which statement correctly describes the effect of a catalyst on a reversible reaction at a fixed temperature?
1. It lowers the activation energy of only the forward reaction, so the equilibrium shifts towards products
2. It makes ΔG° of the reaction more negative, and that is why the reaction becomes faster
3. It lowers the barrier for the forward and reverse reactions equally, so K is unchanged
4. It raises the average kinetic energy of the molecules, so more collisions exceed Ea
Show answer
Answer: C — It lowers the barrier for the forward and reverse reactions equally, so K is unchanged
A catalyst provides a pathway whose transition state is lower by the same amount when approached from either side, so kf and kr rise by the same factor and K = kf/kr is unchanged. It does not alter ΔG°, which depends only on the initial and final states, and it does not change the temperature or the energy distribution; raising the temperature is what increases the average kinetic energy.
Reactions P and Q have the same pre-exponential factor; their activation energies are 40 kJ mol⁻¹ (P) and 80 kJ mol⁻¹ (Q), both independent of temperature. The temperature is raised from 300 K to 310 K. Which statement is correct? (R = 8.314 J mol⁻¹ K⁻¹)
1. k of Q rises by the larger factor, although k of Q remains far smaller than k of P
2. k of P rises by the larger factor, because a lower barrier is crossed more easily
3. k of P and k of Q rise by the same factor, because the temperature rise is the same
4. k of Q rises by the larger factor and overtakes k of P at 310 K
Show answer
Answer: A — k of Q rises by the larger factor, although k of Q remains far smaller than k of P
k₂/k₁ = exp[(Ea/R)(1/300 − 1/310)] with 1/300 − 1/310 = 1.075 × 10⁻⁴ K⁻¹. For P: exp(0.517) = 1.68; for Q: exp(1.035) = 2.81, the square of P's factor. But with equal A, kQ/kP = exp(−40000/RT) ≈ 10⁻⁷ at these temperatures, so Q stays far slower. A lower barrier gives a larger k, not a larger relative increase.
For an enzyme obeying Michaelis–Menten kinetics, what does the Michaelis constant Km represent operationally?
1. The dissociation constant of the ES complex, whatever the relative sizes of k₋₁ and kcat
2. The maximum number of substrate molecules converted per active site per second
3. The substrate concentration needed to saturate every active site of the enzyme
4. The substrate concentration at which the initial rate is exactly half of Vmax
Show answer
Answer: D — The substrate concentration at which the initial rate is exactly half of Vmax
Putting v₀ = Vmax/2 into v₀ = Vmax[S]/(Km + [S]) gives [S] = Km. Km = (k₋₁ + kcat)/k₁ equals the ES dissociation constant only when kcat ≪ k₋₁, so that option is not general. The maximum conversions per site per second is kcat, and saturation is approached only as [S] ≫ Km, never at Km.
Lineweaver–Burk plots of an enzyme measured with and without a fixed concentration of a reversible inhibitor are parallel straight lines, the inhibited line lying above the other. What does this indicate?
1. Competitive inhibition: the apparent Km rises while Vmax is unchanged
2. Uncompetitive inhibition: apparent Vmax and Km both fall by the same factor
3. Pure noncompetitive inhibition: apparent Vmax falls while Km is unchanged
4. Mixed inhibition: the inhibitor binds E and ES with different affinities
Show answer
Answer: B — Uncompetitive inhibition: apparent Vmax and Km both fall by the same factor
An inhibitor that binds only ES gives 1/v₀ = (Km/Vmax)(1/[S]) + α′/Vmax: the slope Km/Vmax is unchanged and only the intercept rises, so the lines are parallel, with Vmax/α′ and Km/α′. Competitive inhibition changes the slope and keeps the y-intercept (lines meet on the y-axis), pure noncompetitive changes both slope and y-intercept by α (lines meet on the x-axis), and mixed lines cross left of the y-axis.
An enzyme is preincubated with compound X. Samples taken at intervals are passed through a gel-filtration desalting column to remove free X and then assayed at saturating substrate. Activity falls steadily with preincubation time and is not recovered after X is removed; including a high substrate concentration during preincubation slows the loss. What is X most likely to be?
1. A competitive reversible inhibitor, since the substrate protects the enzyme from it
2. An uncompetitive reversible inhibitor that binds only to the enzyme–substrate complex
3. An active-site-directed irreversible inhibitor that modifies the enzyme covalently
4. A non-specific denaturant whose action is unaffected by binding of the substrate
Show answer
Answer: C — An active-site-directed irreversible inhibitor that modifies the enzyme covalently
Loss of activity that grows with time and survives removal of free X shows covalent, irreversible inactivation; protection by substrate shows that X reacts at the active site. A reversible competitive inhibitor would be stripped off by gel filtration and activity would return. An uncompetitive inhibitor needs ES, so substrate would increase, not decrease, its effect, and a non-specific denaturant would not be slowed by substrate binding.
Which of the following statements about order and molecularity are correct?
1. The overall order of a multistep reaction can always be read from the coefficients of its balanced equation
2. The order of a reaction can be zero or fractional, whereas the molecularity of an elementary step is a positive whole number
3. Termolecular elementary steps are common because three-body collisions are frequent in solution
4. For an elementary step, the order in each reactant equals that reactant's coefficient in the step
Show answer
Answer: B — The order of a reaction can be zero or fractional, whereas the molecularity of an elementary step is a positive whole number; D — For an elementary step, the order in each reactant equals that reactant's coefficient in the step
Order is an experimental exponent and can be zero or fractional; molecularity counts the particles in an elementary step and is 1, 2 or rarely 3. For an elementary step the rate law follows its stoichiometry. The overall equation of a multistep reaction says nothing reliable about order, and termolecular steps are rare because simultaneous three-body collisions are improbable.
Which of the following are true of a reaction that is first order in its only reactant A?
1. A plot of ln[A] against time is a straight line of slope −k
2. Its half-life does not depend on the initial concentration of A
3. Its rate constant is expressed in units of mol L⁻¹ s⁻¹
4. About ten half-lives are needed for 99.9% of A to react
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Answer: A — A plot of ln[A] against time is a straight line of slope −k; B — Its half-life does not depend on the initial concentration of A; D — About ten half-lives are needed for 99.9% of A to react
ln[A] = ln[A]₀ − kt is linear with slope −k; t½ = 0.693/k has no [A]₀ in it; and 99.9% conversion needs t = ln(1000)/k = 9.97 × (0.693/k), about ten half-lives. A first-order rate constant has units of s⁻¹ (time⁻¹); mol L⁻¹ s⁻¹ belongs to zero order.
For an enzyme that follows the Michaelis–Menten scheme E + S ⇌ ES → E + P, which statements are correct?
1. Doubling the total enzyme concentration at a fixed [S] doubles both Vmax and Km
2. At [S] ≫ Km the initial rate is nearly independent of [S], that is, zero order in substrate
3. Km equals the dissociation constant of ES whatever the relative sizes of k₋₁ and kcat
4. kcat/Km has units of M⁻¹ s⁻¹ and cannot exceed the rate at which enzyme and substrate meet by diffusion
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Answer: B — At [S] ≫ Km the initial rate is nearly independent of [S], that is, zero order in substrate; D — kcat/Km has units of M⁻¹ s⁻¹ and cannot exceed the rate at which enzyme and substrate meet by diffusion
When [S] ≫ Km, v₀ → Vmax and is zero order in S. kcat/Km = k₁kcat/(k₋₁ + kcat) is at most k₁, the encounter rate constant, which diffusion limits to about 10⁸ to 10⁹ M⁻¹ s⁻¹. Vmax = kcat[E]ₜ doubles with enzyme but Km, a ratio of rate constants, does not change; and Km equals the ES dissociation constant only when kcat ≪ k₋₁.
Which of the following statements about irreversible enzyme inhibitors are correct?
1. Diisopropyl fluorophosphate inactivates serine proteases by phosphorylating the active-site serine
2. Activity lost to an irreversible inhibitor returns once the inhibitor is removed by dialysis
3. Iodoacetamide inactivates some enzymes by alkylating the thiol group of an essential cysteine
4. An irreversible inhibitor raises the apparent Km while leaving the apparent Vmax unchanged
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Answer: A — Diisopropyl fluorophosphate inactivates serine proteases by phosphorylating the active-site serine; C — Iodoacetamide inactivates some enzymes by alkylating the thiol group of an essential cysteine
DIFP phosphorylates the reactive serine of serine proteases (and acetylcholinesterase), and iodoacetamide alkylates cysteine SH groups; both are covalent. Covalent modification is not undone by dialysis. Because an irreversible inhibitor removes active enzyme while the surviving enzyme keeps its normal Km, it lowers apparent Vmax with Km unchanged, the opposite of the claim that it raises Km and spares Vmax.
A reaction that is first order in its only reactant has a rate constant of 1.155 × 10⁻² min⁻¹. Taking ln 2 = 0.693, the time required for 87.5% of the reactant to be consumed is ____ min (to the nearest integer).
Numerical answer — type the value.
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Answer: 180
t½ = 0.693/(1.155 × 10⁻²) = 60.0 min. When 87.5% is consumed, 12.5% = (1/2)³ remains, so three half-lives have passed: 3 × 60.0 = 180 min. Directly, t = ln 8/k = 2.079/0.01155 = 180 min.
The rate constant of a reaction becomes four times larger when the temperature is raised from 298 K to 318 K. Assuming the activation energy is independent of temperature and taking R = 8.314 J mol⁻¹ K⁻¹, the activation energy is ____ kJ mol⁻¹ (to one decimal place).
Numerical answer — type the value.
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Answer: 54.6
ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂). Here ln 4 = 1.3863 and 1/298 − 1/318 = 0.0033557 − 0.0031447 = 2.1105 × 10⁻⁴ K⁻¹. Ea = 8.314 × 1.3863/(2.1105 × 10⁻⁴) = 54,610 J mol⁻¹ = 54.6 kJ mol⁻¹. Using log₁₀ 4 without the factor 2.303 is the common slip.
An enzyme has Km = 4.0 mM and Vmax = 120 µmol min⁻¹ for its substrate. A competitive inhibitor with Kᵢ = 0.50 mM is present at 1.0 mM. Assuming Michaelis–Menten kinetics, the initial rate at [S] = 4.0 mM is ____ µmol min⁻¹ (to the nearest integer).
Numerical answer — type the value.
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Answer: 30
α = 1 + [I]/Kᵢ = 1 + 1.0/0.50 = 3, so the apparent Km = 3 × 4.0 = 12 mM while Vmax is unchanged. v₀ = 120 × 4.0/(12 + 4.0) = 480/16 = 30 µmol min⁻¹. Without the inhibitor the rate would be 60 (half of Vmax, since [S] = Km); lowering Vmax instead of raising Km would also give a wrong value.
A reaction 2A → products obeys the rate law −d[A]/dt = k[A]² with k = 0.50 L mol⁻¹ min⁻¹. Starting from [A]₀ = 0.25 mol L⁻¹, the time taken for [A] to fall to 0.050 mol L⁻¹ is ____ min (to the nearest integer).
Numerical answer — type the value.
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Answer: 32
With the rate defined as −d[A]/dt = k[A]², the integrated law is 1/[A] = 1/[A]₀ + kt. So t = (1/0.050 − 1/0.25)/0.50 = (20 − 4.0)/0.50 = 32 min. Applying a first-order formula, t = ln(0.25/0.050)/k = 3.2, is the trap.
An enzyme obeying Michaelis–Menten kinetics has Km = 2.0 mM. The substrate concentration at which the initial rate equals 80% of Vmax is ____ mM (to the nearest integer).
Numerical answer — type the value.
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Answer: 8
0.80 = [S]/(Km + [S]) gives 0.80Km = 0.20[S], so [S] = 4Km = 4 × 2.0 = 8 mM. A quick check: [S] = Km gives 50% and [S] = 3Km gives 75%, so 80% needs a little more than 3Km.