---
title: "Complex Kinetics: Chains, Enzymes and Oscillations"
book: "University Chemistry — Year 3"
subject: chemistry
language: en
chapter: 13
exercises: 12
source: https://one-course.com/books/chemistry/4/en/chapter/13-complex-kinetics-chains-enzymes-and-oscillations
license: CC-BY-NC-SA-4.0
credit: "One Chemistry Book, One Course (one-course.com)"
---

# Chapter 13 — Complex Kinetics: Chains, Enzymes and Oscillations

A beaker of a stirred solution turns colourless, then amber, then blue, then colourless again, and keeps doing so for many minutes; poured into a dish, the same mixture draws spirals that rotate and travel. When Boris Belousov described such a reaction in 1951, his manuscript was rejected: a chemical system, the referee thought, cannot run back and forth, since it must go downhill towards equilibrium. It does go downhill; it simply does not go straight. This chapter treats mechanisms with many steps whose intermediates are regenerated: [chain reactions](#def-b3-complex-kinetics-chain) and explosions, molecules that need collisions to fall apart alone, [enzymes](#def-b3-complex-kinetics-enzyme), and the [oscillating reactions](#def-b3-complex-kinetics-oscillating), with the methods that follow reactions too fast to mix by hand.

**You already know.**

The Year 1 volume defined a reaction mechanism as a sequence of elementary steps, the rate-determining step, the pre-equilibrium and steady-state approximations, and catalysts. The Year 2 volume named the steps of a radical chain (initiation, propagation, termination), radical initiators and the kinetic chain length of a polymerisation, and fitted least-squares lines. [Chapter 12](https://one-course.com/books/chemistry/4/en/chapter/12-theories-of-reaction-rates#ch-b3-rate-theories) gave [transition-state theory](https://one-course.com/books/chemistry/4/en/chapter/12-theories-of-reaction-rates#def-b3-rate-theories-tst) and the diffusion limit of about $10^{10}\,\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{s}^{-1}$.

## 13.1 Chain reactions

**Definition 13.1 (Chain reaction, chain carrier).**

A *chain reaction* is a reaction in which reactive intermediates, the *chain carriers*, are regenerated by a cycle of propagation steps, so that one initiation event converts many reactant molecules.

Bodenstein measured in 1906 the rate of $\ce{H2 + Br2 -> 2 HBr}$ and found a law that no single step could produce: the rate grows as $[\ce{Br2}]^{1/2}$ and falls as HBr accumulates. The mechanism was written down in 1919:

| initiation | $\ce{Br2 -> 2 Br}$ | $k_1$ |
| --- | --- | --- |
| propagation | $\ce{Br + H2 -> HBr + H}$ | $k_2$ |
| propagation | $\ce{H + Br2 -> HBr + Br}$ | $k_3$ |
| inhibition | $\ce{H + HBr -> H2 + Br}$ | $k_4$ |
| termination | $\ce{2 Br -> Br2}$ | $k_{-1}$ |

(collisions with a third body M that carry energy in and out of the initiation and termination steps are left implicit).

**Theorem 13.2 (The hydrogen–bromine rate law).**

With the steady-state approximation for Br and H, the rate $v = \frac12\,\dd[\ce{HBr}]/\dd t$ of the mechanism is

$$
v = \frac{k[\ce{H2}][\ce{Br2}]^{1/2}}{1 + k'[\ce{HBr}]/[\ce{Br2}]},
  \qquad k = k_2\Bigl(\frac{k_1}{k_{-1}}\Bigr)^{1/2},\ k' = \frac{k_4}{k_3}.
$$

**Proof.** Steady state for H: $k_2[\ce{Br}][\ce{H2}] = k_3[\ce{H}][\ce{Br2}] + k_4[\ce{H}][\ce{HBr}]$. Steady state for Br: $2k_1[\ce{Br2}] - k_2[\ce{Br}][\ce{H2}] + k_3[\ce{H}][\ce{Br2}] + k_4[\ce{H}][\ce{HBr}] - 2k_{-1}[\ce{Br}]^2 = 0$. Adding the two equations, the propagation terms cancel: $k_1[\ce{Br2}] = k_{-1}[\ce{Br}]^2$, so $[\ce{Br}] =
(k_1/k_{-1})^{1/2}[\ce{Br2}]^{1/2}$. Then $[\ce{H}] = k_2[\ce{Br}][\ce{H2}]/(k_3[\ce{Br2}] + k_4[\ce{HBr}])$ and

$$
\frac{\dd[\ce{HBr}]}{\dd t} = k_2[\ce{Br}][\ce{H2}] + k_3[\ce{H}][\ce{Br2}] - k_4[\ce{H}][\ce{HBr}] = 2k_3[\ce{H}][\ce{Br2}]
  = \frac{2k_2[\ce{Br}][\ce{H2}]}{1 + (k_4/k_3)[\ce{HBr}]/[\ce{Br2}]},
$$

using the first equation; substituting $[\ce{Br}]$ gives the result. ∎

![The propagation cycle of the hydrogen–bromine chain: each turn consumes one H2 and one Br2, makes two HBr and gives back the bromine atom that started it. Initiation feeds the cycle with carriers, termination removes them; the inhibition step turns an H atom back into Br at the cost of an HBr already made.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-complex-kinetics/fig-968787372f3e.svg)

*The propagation cycle of the hydrogen–bromine chain: each turn consumes one $\ce{H2}$ and one $\ce{Br2}$, makes two HBr and gives back the bromine atom that started it. Initiation feeds the cycle with carriers, termination removes them; the inhibition step turns an H atom back into Br at the cost of an HBr already made.*

![The hydrogen–bromine mechanism integrated step by step (model rate constants, reduced units, k' = 0.1): conversion against time (left), and the rate 1/2 ( HBr)/ t of the full mechanism compared with the steady-state law (right). After an induction period, during which the bromine atoms build up, the two coincide to better than 1 %.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-complex-kinetics/fig-e0ad4d211012.svg)

*The hydrogen–bromine mechanism integrated step by step (model rate constants, reduced units, $k' = 0.1$): conversion against time (left), and the rate $\frac12\dd[\ce{HBr}]/\dd t$ of the full mechanism compared with the steady-state law (right). After an induction period, during which the bromine atoms build up, the two coincide to better than 1 %.*

**Method 13.3 (The rate law of a chain mechanism).**

1. Write a steady-state equation for each [chain carrier](#def-b3-complex-kinetics-chain) .
2. Add them: the propagation steps, which only exchange one carrier for another, cancel, leaving initiation equal to termination; this gives the carrier concentration.
3. Use one carrier’s equation to express the other carriers.
4. Write the rate of formation of a product from the propagation steps.

The number of propagation cycles per initiation event, here $v/k_1[\ce{Br2}]$, is the kinetic chain length of the Year 2 volume; for the hydrogen–bromine reaction it is of the order of $10^3$ to $10^6$ depending on the conditions. Inhibition by the product explains why the rate falls faster than the reactants are consumed.

## 13.2 Branching chains and explosions

**Definition 13.4 (Chain branching, explosion limits).**

*Chain branching* is a propagation step that produces more [chain carriers](#def-b3-complex-kinetics-chain) than it consumes, such as $\ce{H + O2 -> OH + O}$, which turns one carrier into two. The *explosion limits* of a mixture are the pressures, at a given temperature, at which its slow reaction turns into an explosion.

**Proposition 13.5 (Branching criterion).**

Let carriers be created at the rate $v_i$, multiply by branching with the rate constant $f$ and disappear by termination with the rate constant $g$: $\dd n/\dd t = v_i + (f - g)n$. If $f < g$, $n$ tends to the steady value $v_i/(g - f)$; if $f > g$, $n$ grows exponentially and the reaction explodes.

**Proof.** The linear equation has the solution $n(t) = \frac{v_i}{g - f}\bigl(1 - \eu^{-(g - f)t}\bigr)$, which tends to $v_i/(g - f)$ when $g > f$; when $f > g$ the exponent is positive and $n$ grows as $\eu^{(f - g)t}$ without bound (until the reactants run out). ∎

For hydrogen–oxygen mixtures at a few hundred degrees, $f$ grows with the pressure (it is proportional to $[\ce{O2}]$), termination at the walls of the vessel is fast at low pressure (the radicals diffuse there easily), and termination by three-body collisions ($\ce{H + O2 + M -> HO2 + M}$) grows as the square of the pressure. Branching wins between a first limit, set by the walls, and a second limit, set by three-body termination; at still higher pressure a third limit appears where the heat released can no longer escape. The last is a thermal explosion, driven by the exponential growth of the rate with temperature rather than by branching; most industrial explosions are of that kind.

## 13.3 Unimolecular reactions

A molecule such as cyclopropane isomerises alone, with first-order kinetics at ordinary pressures; yet at low pressure the rate constant falls. The energy needed comes from collisions.

**Definition 13.6 (Lindemann mechanism, fall-off region).**

The *Lindemann mechanism* of a unimolecular reaction is the sequence $\mathrm{A + M \rightleftharpoons A^* + M}$ ($k_1$, $k_{-1}$), activation and deactivation by collision with any molecule M, followed by $\mathrm{A^* \to P}$ ($k_2$). The *fall-off region* is the range of pressures where the observed first-order rate constant falls from its high-pressure value towards second-order behaviour.

**Theorem 13.7 (Lindemann rate constant).**

With the steady-state approximation for $\mathrm{A^*}$, $v = k_{\mathrm{uni}}[\mathrm A]$ with

$$
k_{\mathrm{uni}} = \frac{k_1k_2[\mathrm M]}{k_{-1}[\mathrm M] + k_2},
$$

which tends to $k_\infty = k_1k_2/k_{-1}$ at high pressure (first order) and to $k_1[\mathrm M]$ at low pressure (second order); it is $k_\infty/2$ at $[\mathrm M]_{1/2} = k_2/k_{-1}$.

**Proof.** $\dd[\mathrm{A^*}]/\dd t = k_1[\mathrm A][\mathrm M] - k_{-1}[\mathrm{A^*}][\mathrm M] - k_2[\mathrm{A^*}] = 0$ gives $[\mathrm{A^*}] =
k_1[\mathrm A][\mathrm M]/(k_{-1}[\mathrm M] + k_2)$ and $v = k_2[\mathrm{A^*}]$. If $k_{-1}[\mathrm M] \gg k_2$, deactivation outruns reaction and $\mathrm{A^*}$ is in pre-equilibrium: $k_{\mathrm{uni}} \to k_1k_2/k_{-1}$. If $k_{-1}[\mathrm M]
\ll k_2$, every activated molecule reacts and activation is rate-determining: $k_{\mathrm{uni}} \to k_1[\mathrm M]$. ∎

![The Lindemann fall-off curve (model constants): first order with the limiting constant k_∈fty at high pressure, second order (k_1( M)) at low pressure, and half of k_∈fty at ( M)_1/2 = k_2/k_-1. Real fall-off curves are broader: the rate constant k_2 of an energised molecule grows with its energy.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-complex-kinetics/fig-cd9fb7f5b919.svg)

*The Lindemann fall-off curve (model constants): first order with the limiting constant $k_\infty$ at high pressure, second order ($k_1[\mathrm M]$) at low pressure, and half of $k_\infty$ at $[\mathrm M]_{1/2} = k_2/k_{-1}$. Real fall-off curves are broader: the rate constant $k_2$ of an energised molecule grows with its energy.*

## 13.4 Enzyme kinetics

**Definition 13.8 (Enzyme, active site, enzyme–substrate complex).**

An *enzyme* is a biological catalyst, almost always a protein, that speeds up a specific reaction of its substrate. The *active site* is the pocket of the enzyme where the substrate binds and reacts; the bound species is the *enzyme–substrate complex* ES.

**Definition 13.9 (Michaelis constant, maximum rate, catalytic and specificity constants).**

For the mechanism $\mathrm{E + S \rightleftharpoons ES \to E + P}$ ($k_1$, $k_{-1}$, $k_2$) at total [enzyme](#def-b3-complex-kinetics-enzyme) concentration $[\mathrm E]_0$, the *catalytic constant* is $k_{\mathrm{cat}} = k_2$, the *maximum rate* is $V_{\max} = k_{\mathrm{cat}}[\mathrm E]_0$, the *Michaelis constant* is $K_M = (k_{-1} + k_2)/k_1$, and the *specificity constant* is $k_{\mathrm{cat}}/K_M$.

**Theorem 13.10 (Michaelis–Menten equation).**

When $[\mathrm S] \gg [\mathrm E]_0$ and ES is in a steady state, the initial rate of formation of the product is

$$
v = \frac{V_{\max}[\mathrm S]}{K_M + [\mathrm S]} .
$$

This statement is the *Michaelis–Menten equation*.

**Proof.** Steady state: $k_1[\mathrm E][\mathrm S] = (k_{-1} + k_2)[\mathrm{ES}]$, and $[\mathrm E] = [\mathrm E]_0 - [\mathrm{ES}]$ (the free substrate is practically all the substrate since $[\mathrm S] \gg [\mathrm E]_0$). Hence $[\mathrm{ES}]
= [\mathrm E]_0[\mathrm S]/(K_M + [\mathrm S])$ and $v = k_2[\mathrm{ES}]$. ∎

**Remark 13.11.**

Michaelis and Menten (1913) assumed instead a rapid pre-equilibrium of E, S and ES; the same equation follows with $K_M$ replaced by the dissociation constant $k_{-1}/k_1$, the limit of $K_M$ when $k_2 \ll k_{-1}$. The [catalytic constant](#def-b3-complex-kinetics-michaelis) is what biochemists call the turnover number of the [enzyme](#def-b3-complex-kinetics-enzyme): the number of substrate molecules one [active site](#def-b3-complex-kinetics-enzyme) converts per second when saturated.

At $[\mathrm S] = K_M$ the rate is $V_{\max}/2$; at $[\mathrm S] \ll K_M$ it is $(k_{\mathrm{cat}}/K_M)[\mathrm E]_0[\mathrm S]$: the [specificity constant](#def-b3-complex-kinetics-michaelis) is the second-order rate constant of the free [enzyme](#def-b3-complex-kinetics-enzyme) with its substrate, and it cannot exceed the rate at which they meet, the diffusion limit of [Chapter 12](https://one-course.com/books/chemistry/4/en/chapter/12-theories-of-reaction-rates#ch-b3-rate-theories). A few [enzymes](#def-b3-complex-kinetics-enzyme) come close; the “average” [enzyme](#def-b3-complex-kinetics-enzyme), in a survey of several thousand, has $k_{\mathrm{cat}}/K_M$ of about $10^5$ $\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{s}^{-1}$, some four orders of magnitude below.

**Proposition 13.12 (Lineweaver–Burk form).**

The [Michaelis–Menten equation](#thm-b3-complex-kinetics-michaelis-menten) is equivalent to

$$
\frac1v = \frac{1}{V_{\max}} + \frac{K_M}{V_{\max}}\,\frac{1}{[\mathrm S]},
$$

a straight line of $1/v$ against $1/[\mathrm S]$ with intercept $1/V_{\max}$, slope $K_M/V_{\max}$, and intercept $-1/K_M$ on the $1/[\mathrm S]$ axis.

**Proof.** Invert: $1/v = (K_M + [\mathrm S])/(V_{\max}[\mathrm S])$. At $1/v = 0$, $1/[\mathrm S] = -1/K_M$. ∎

**Method 13.13 (KMK_MKM​ and Vmax⁡V_{\max}Vmax​ by nonlinear least squares).**

1. Measure initial rates at substrate concentrations spread from about $K_M/5$ to $10K_M$ .
2. Fit $v = V_{\max}[\mathrm S]/(K_M + [\mathrm S])$ directly, by minimising $\sum(v_i - v(\mathrm S_i))^2$ (iteratively, starting from the values of a Lineweaver–Burk line); the standard errors come from the residuals and the sensitivity of $v$ to each parameter.
3. Use the Lineweaver–Burk plot only to see the pattern: inverting small rates magnifies their errors, so the least-squares line through $1/v$ is dominated by the least precise points, and its parameters are biased.

**Definition 13.14 (Inhibition).**

An inhibitor I binds the [enzyme](#def-b3-complex-kinetics-enzyme) reversibly. In *competitive inhibition* it binds only the free [enzyme](#def-b3-complex-kinetics-enzyme), at the [active site](#def-b3-complex-kinetics-enzyme), with the *inhibition constant* $K_i$ (dissociation constant of EI); in *uncompetitive inhibition* it binds only ES, with the constant $K_i'$; in *mixed inhibition* it binds both.

**Proposition 13.15 (Inhibited rate laws).**

With $\alpha = 1 + [\mathrm I]/K_i$ and $\alpha' = 1 + [\mathrm I]/K_i'$, the rate keeps the Michaelis–Menten form $v = V^{\mathrm{app}}[\mathrm S]/(K_M^{\mathrm{app}} + [\mathrm S])$ with: competitive, $K_M^{\mathrm{app}} = \alpha K_M$, $V^{\mathrm{app}} = V_{\max}$; uncompetitive, $K_M^{\mathrm{app}} = K_M/\alpha'$, $V^{\mathrm{app}} =
V_{\max}/\alpha'$; mixed, $K_M^{\mathrm{app}} = \alpha K_M/\alpha'$, $V^{\mathrm{app}} = V_{\max}/\alpha'$.

**Proof.** Mixed case (the others are its limits $K_i' \to \infty$ and $K_i \to \infty$): the [enzyme](#def-b3-complex-kinetics-enzyme) is shared among E, EI, ES and ESI with $[\mathrm{EI}] = [\mathrm E][\mathrm I]/K_i$, $[\mathrm{ESI}] = [\mathrm{ES}][\mathrm I]/K_i'$ and the steady state $[\mathrm E][\mathrm S] = K_M[\mathrm{ES}]$. So $[\mathrm E]_0 = [\mathrm E]\alpha + [\mathrm{ES}]\alpha' =
[\mathrm{ES}](\alpha K_M/[\mathrm S] + \alpha')$ and $v = k_2[\mathrm{ES}] = V_{\max}[\mathrm S]/(\alpha K_M + \alpha'[\mathrm S])$; dividing numerator and denominator by $\alpha'$ gives the stated form. ∎

**Method 13.16 (Recognising the type of inhibition).**

1. Measure $v([\mathrm S])$ at several inhibitor concentrations and draw the Lineweaver–Burk lines.
2. Lines meeting on the $1/v$ axis (same $V_{\max}$ ): competitive; a large excess of substrate overcomes the inhibitor.
3. Parallel lines (slope $K_M/V_{\max}$ unchanged): uncompetitive.
4. Lines meeting to the left of the $1/v$ axis: mixed.
5. Obtain $K_i$ by fitting the apparent constants against $[\mathrm I]$ : for a competitive inhibitor $K_M^{\mathrm{app}} = K_M + (K_M/K_i)[\mathrm I]$ , a straight line.

![Michaelis–Menten kinetics with K_M = 0.80\, mM and V_ = 0.50\, µ M\, s-1 (model): no inhibitor (black), a competitive inhibitor at ( I) = 2K_i (blue, K_M tripled, same V_) and an uncompetitive one at ( I) = K_i' (red, K_M and V_ halved). Right: the Lineweaver–Burk lines; the competitive line meets the uninhibited one on the 1/v axis, the uncompetitive one is parallel to it (1/( S) in mM-1, 1/v in s\, µ M-1).](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-complex-kinetics/fig-a013ce583d1b.svg)

*Michaelis–Menten kinetics with $K_M = 0.80\,\mathrm{mM}$ and $V_{\max} = 0.50\,\text{µ}\mathrm{M}\,\mathrm{s}^{-1}$ (model): no inhibitor (black), a competitive inhibitor at $[\mathrm I] = 2K_i$ (blue, $K_M$ tripled, same $V_{\max}$) and an uncompetitive one at $[\mathrm I] = K_i'$ (red, $K_M$ and $V_{\max}$ halved). Right: the Lineweaver–Burk lines; the competitive line meets the uninhibited one on the $1/v$ axis, the uncompetitive one is parallel to it ($1/[\mathrm S]$ in $\mathrm{mM}^{-1}$, $1/v$ in $\mathrm{s}\,\text{µ}\mathrm{M}^{-1}$).*

## 13.5 Autocatalysis, oscillations and fast reactions

**Definition 13.17 (Autocatalytic reaction).**

An *autocatalytic reaction* is one in which a product catalyses its own formation, as in $\mathrm{A + X \to 2\,X}$.

**Proposition 13.18 (Logistic law).**

For $\mathrm{A + X \to 2\,X}$ with rate constant $k$ and initial concentrations $a_0$ and $x_0$, with $N = a_0 + x_0$,

$$
x(t) = \frac{N}{1 + (N/x_0 - 1)\eu^{-kNt}},
$$

an S-shaped curve that reaches half its final value at $t_{1/2} = \ln(N/x_0 - 1)/kN$.

**Proof.** $\dd x/\dd t = kx(N - x)$, since $a = N - x$. Separating, $\frac{\dd x}{x(N - x)} = \frac1N\bigl(\frac1x + \frac{1}{N - x}\bigr)\dd x
= k\,\dd t$, so $\ln\frac{x}{N - x} = kNt + \ln\frac{x_0}{N - x_0}$, which rearranges into the stated form; $x = N/2$ when $(N/x_0 - 1)\eu^{-kNt} = 1$. ∎

**Definition 13.19 (Oscillating reaction, limit cycle).**

An *oscillating reaction* is one in which the concentrations of intermediates rise and fall periodically while the overall reaction proceeds towards equilibrium. A *limit cycle* is a closed trajectory in the space of concentrations towards which neighbouring trajectories converge, so that the oscillation has an amplitude independent of the starting point.

**Theorem 13.20 (Lotka–Volterra model).**

For the autocatalytic scheme $\mathrm{A + X \to 2\,X}$, $\mathrm{X + Y \to 2\,Y}$, $\mathrm{Y \to B}$ with A held constant, the concentrations obey $\dot x = x(a - by)$, $\dot y = y(dx - c)$ with positive constants, and the quantity

$$
V = dx - c\ln x + by - a\ln y
$$

is constant along each trajectory: the trajectories are closed curves around the steady state $(c/d, a/b)$, and the concentrations oscillate with an amplitude fixed by the starting point.

**Proof.** $\dot V = (d - c/x)\dot x + (b - a/y)\dot y = (dx - c)(a - by) + (by - a)(dx - c) = 0$. $V$ is a sum of two convex functions with a single minimum at $(c/d, a/b)$, so its level curves are closed curves around that point; a trajectory stays on one of them and, the velocity never vanishing elsewhere, goes round it periodically. ∎

The Lotka–Volterra oscillations are fragile: every starting point gives its own orbit, and any perturbation moves the system to another. Real chemical oscillators have a [limit cycle](#def-b3-complex-kinetics-oscillating), which needs a nonlinearity that destabilises the steady state.

**Proposition 13.21 (Instability of the Brusselator).**

The model $\dot X = A - (B + 1)X + X^2Y$, $\dot Y = BX - X^2Y$ (reduced concentrations, $A$ and $B$ held constant) has the single steady state $(A, B/A)$, which is stable if $B < 1 + A^2$ and unstable if $B > 1 + A^2$; the system then settles on a [limit cycle](#def-b3-complex-kinetics-oscillating).

**Proof.** Setting both derivatives to zero: $BX = X^2Y$ gives $XY = B$, and then $A - X = 0$. The Jacobian matrix at $(A, B/A)$ is

$$
\begin{pmatrix} -(B + 1) + 2XY & X^2 \\ B - 2XY & -X^2 \end{pmatrix}
  = \begin{pmatrix} B - 1 & A^2 \\ -B & -A^2 \end{pmatrix},
$$

with trace $B - 1 - A^2$ and determinant $A^2 > 0$. Small deviations evolve as $\eu^{\lambda t}$ with $\lambda$ the [eigenvalues](https://one-course.com/books/chemistry/4/en/chapter/1-quantum-mechanics-for-chemists-model-systems#def-b3-quantum-model-systems-operator), whose product is the determinant (positive) and whose sum is the trace: both have negative real parts when the trace is negative, positive real parts when it is positive. The existence of the [limit cycle](#def-b3-complex-kinetics-oscillating) for $B > 1 + A^2$ (the trajectories being confined to a bounded region) is admitted. ∎

**Method 13.22 (The linear stability of a steady state).**

1. Find the steady state by setting all time derivatives to zero.
2. Compute the Jacobian matrix of partial derivatives of the rates there.
3. For two variables: stable if the trace is negative and the determinant positive; an unstable state with complex [eigenvalues](https://one-course.com/books/chemistry/4/en/chapter/1-quantum-mechanics-for-chemists-model-systems#def-b3-quantum-model-systems-operator) ( $\mathrm{tr}^2 < 4\det$ ) gives growing oscillations, a candidate [limit cycle](#def-b3-complex-kinetics-oscillating) .

![Top: the Brusselator (A = 1, B = 3 > 1 + A2) settles into sustained oscillations of period 7.2 (reduced time). Bottom left: two trajectories, one starting near the unstable steady state (dot) and one far outside, wind onto the same limit cycle. Bottom right: Lotka–Volterra orbits, each fixed by its starting point, around the steady state (dot).](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-complex-kinetics/fig-9426116e5591.svg)

*Top: the Brusselator ($A = 1$, $B = 3 > 1 + A^2$) settles into sustained oscillations of period 7.2 (reduced time). Bottom left: two trajectories, one starting near the unstable steady state (dot) and one far outside, wind onto the same [limit cycle](#def-b3-complex-kinetics-oscillating). Bottom right: Lotka–Volterra orbits, each fixed by its starting point, around the steady state (dot).*

The Belousov–Zhabotinsky reaction, the oxidation of malonic acid by bromate catalysed by cerium or by a ferroin indicator, runs through such a cycle: an autocatalytic production of $\ce{HBrO2}$ switches on when bromide falls below a threshold, oxidises the catalyst (the colour change), and is switched off when the oxidised catalyst regenerates bromide. Unstirred, in a thin layer, the oscillation propagates as waves.

![Spiral waves of the Belousov–Zhabotinsky reaction in a thin layer of solution: each blue front is a wave of oxidation of the catalyst, which travels into the reduced (red) medium and cannot re-enter the region it just left.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-complex-kinetics/img-a59a0c24bc84.jpg)

*Spiral waves of the Belousov–Zhabotinsky reaction in a thin layer of solution: each blue front is a wave of oxidation of the catalyst, which travels into the reduced (red) medium and cannot re-enter the region it just left.*

**Definition 13.23 (Relaxation method, chemical relaxation time).**

A *relaxation method* perturbs a system at equilibrium suddenly (by a jump of temperature, pressure or electric field) and follows its return to the new equilibrium. For a small perturbation the deviation decays exponentially, with the *chemical relaxation time* $\tau$.

**Proposition 13.24 (Relaxation time of an association).**

For $\mathrm{A + B \rightleftharpoons C}$ ($k_1$, $k_{-1}$), after a small perturbation,

$$
\frac1\tau = k_1([\mathrm A]_e + [\mathrm B]_e) + k_{-1} .
$$

**Proof.** Write $[\mathrm C] = [\mathrm C]_e + x$, $[\mathrm A] = [\mathrm A]_e - x$, $[\mathrm B] = [\mathrm B]_e - x$. Then $\dot x =
k_1([\mathrm A]_e - x)([\mathrm B]_e - x) - k_{-1}([\mathrm C]_e + x)$; the terms without $x$ cancel at equilibrium, and dropping $x^2$: $\dot x = -[k_1([\mathrm A]_e + [\mathrm B]_e) + k_{-1}]x$. ∎

Measuring $\tau$ at several concentrations gives a line of $1/\tau$ against $[\mathrm A]_e +
[\mathrm B]_e$ whose slope is $k_1$ and intercept $k_{-1}$: both constants from experiments on a system that never leaves equilibrium by more than a few per cent. Manfred Eigen measured in this way the fastest reactions in solution, among them the neutralisation of $\ce{H3O+}$ by $\ce{OH-}$.

![Relaxation of A + B C after a temperature jump that lowered its equilibrium constant by 5 % (model: k_1 = 1.0 × 108\, L\, mol-1\, s-1, k_-1 = 1.0 × 103\, s-1, 1.0 × 10-4\, mol/ L of A and B in all): the full rate equation and the exponential with 1/ = k_1(( A)_e + ( B)_e) + k_-1 coincide.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-complex-kinetics/fig-0642cf0098d9.svg)

*Relaxation of $\mathrm{A + B \rightleftharpoons C}$ after a temperature jump that lowered its equilibrium constant by 5 % (model: $k_1 = 1.0 \times 10^{8}\,\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{s}^{-1}$, $k_{-1} = 1.0 \times 10^{3}\,\mathrm{s}^{-1}$, $1.0 \times 10^{-4}\,\mathrm{mol}/\mathrm{L}$ of A and B in all): the full rate equation and the exponential with $1/\tau = k_1([\mathrm A]_e + [\mathrm B]_e) + k_{-1}$ coincide.*

**Definition 13.25 (Stopped flow, flash photolysis).**

The *stopped-flow method* mixes two reactant solutions in about a millisecond by driving them through a mixer into an observation cell, then stops the flow and records the absorbance or [fluorescence](https://one-course.com/books/chemistry/4/en/chapter/7-electronic-spectroscopy-and-photophysics#def-b3-electronic-spectroscopy-luminescence). *Flash photolysis* creates a reactive species by a short intense light pulse and follows its reactions with a second, delayed light beam.

![A stopped-flow apparatus. The drive pushes two syringes; the solutions meet in the mixer and fill the observation cell within about a millisecond; the plunger of the stop syringe hits a block, the flow stops, and the detector records the reaction of the freshly mixed solution.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-complex-kinetics/fig-93f5084c8273.svg)

*A stopped-flow apparatus. The drive pushes two syringes; the solutions meet in the mixer and fill the observation cell within about a millisecond; the plunger of the stop syringe hits a block, the flow stops, and the detector records the reaction of the freshly mixed solution.*

![A stopped-flow burst trace (data of the weekend problem): the first product of a two-step enzyme appears in a fast burst followed by a straight line; the dashed line extrapolates the steady state back to t = 0.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-complex-kinetics/fig-50be528c8795.svg)

*A stopped-flow burst trace (data of the weekend problem): the first product of a two-step [enzyme](#def-b3-complex-kinetics-enzyme) appears in a fast burst followed by a straight line; the dashed line extrapolates the steady state back to $t = 0$.*

**In the lab — A stopped-flow measurement.**

The two syringes are filled with [enzyme](#def-b3-complex-kinetics-enzyme) and substrate solutions in the same buffer, thermostated, and fired several times to flush the cell before the recorded shots; each trace is averaged over five to ten shots. The dead time, the age of the mixture when observation starts, is measured once with a reaction of known rate.

**Safety.**

![](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-complex-kinetics/fig-aa048b319e7f.svg)

![](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-complex-kinetics/fig-ed6bb6b332c4.svg)

![](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-complex-kinetics/fig-d16e3fb60326.svg)

Bromine: a volatile, dense red-brown liquid; fatal if inhaled, causes severe burns, very toxic to aquatic life. Handled only in a fume hood, with gloves and face protection; a solution of sodium thiosulfate is kept at hand to reduce spills.

**History — Bodenstein and Belousov.**

Max Bodenstein and Samuel Lind measured the hydrogen–bromine rate law in 1906; Jens Christiansen, Karl Herzfeld and Michael Polanyi explained it in 1919 with the chain mechanism of this chapter. Boris Belousov, a biochemist in Moscow, found his [oscillating reaction](#def-b3-complex-kinetics-oscillating) around 1951 while looking for a chemical model of a metabolic cycle; his manuscript was rejected, and only a short abstract appeared in 1959. Anatol Zhabotinsky took up the reaction in the 1960s and explained it; the full mechanism was worked out in the early 1970s.

## 13.6 Exercises

**Exercise 13.1 ★.**

For the chain $\ce{Cl2 -> 2 Cl}$, $\ce{Cl + H2 -> HCl + H}$, $\ce{H + Cl2 -> HCl + Cl}$, $\ce{2 Cl -> Cl2}$, name each step and the [chain carriers](#def-b3-complex-kinetics-chain), and write the overall reaction.

**Solution of Exercise 13.1.**

$\ce{Cl2 -> 2 Cl}$: initiation; $\ce{Cl + H2 -> HCl + H}$ and $\ce{H + Cl2 -> HCl + Cl}$: propagation; $\ce{2 Cl -> Cl2}$: termination. Carriers: Cl and H. Overall (sum of the two propagation steps): $\ce{H2 + Cl2 -> 2 HCl}$.

**Exercise 13.2 ★.**

A Lindemann system has $k_2 = 1.0 \times 10^{7}\,\mathrm{s}^{-1}$ and $k_{-1} = 1.0 \times 10^{10}\,\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{s}^{-1}$ (data of the exercise). Compute $[\mathrm M]_{1/2}$ and the corresponding pressure at $750\,\mathrm{K}$.

**Solution of Exercise 13.2.**

$[\mathrm M]_{1/2} = k_2/k_{-1} = 1.0 \times 10^{-3}\,\mathrm{mol}/\mathrm{L} = 1.0\,\mathrm{mol}/\mathrm{m}^{3}$; $p = cRT = 1.0 \times 8.314 \times 750 =
6.2\,\mathrm{kPa}$.

**Exercise 13.3 ★.**

Express $v/V_{\max}$ at $[\mathrm S] = K_M$, $3K_M$ and $9K_M$. What concentration gives 90 % of $V_{\max}$?

**Solution of Exercise 13.3.**

$v/V_{\max} = [\mathrm S]/(K_M + [\mathrm S])$: $\frac12$, $\frac34$, $\frac{9}{10}$. 90 % of $V_{\max}$ needs $[\mathrm S] = 9K_M$.

**Exercise 13.4 ★.**

An [enzyme](#def-b3-complex-kinetics-enzyme) has $k_{\mathrm{cat}} = 100\,\mathrm{s}^{-1}$ and $K_M = 0.80\,\mathrm{mM}$. Compute its [specificity constant](#def-b3-complex-kinetics-michaelis) and compare it with the diffusion limit in water, $7.4 \times 10^{9}\,\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{s}^{-1}$, and with the typical $10^5$ $\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{s}^{-1}$.

**Solution of Exercise 13.4.**

$k_{\mathrm{cat}}/K_M = 100/0.80 \times 10^{-3} = 1.25 \times 10^{5}\,\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{s}^{-1}$: some 60 000 times below the diffusion limit, and close to the typical value.

**Exercise 13.5 ★★.**

The decomposition $\ce{CH3CHO -> CH4 + CO}$ is proposed to run by: $\ce{CH3CHO -> CH3 +
CHO}$ ($k_1$); $\ce{CH3 + CH3CHO -> CH4 + CH3CO}$ ($k_2$); $\ce{CH3CO -> CH3 + CO}$ ($k_3$); $\ce{2 CH3 -> C2H6}$ ($k_4$). Show that the rate of formation of methane is $k_2(k_1/2k_4)^{1/2}[\ce{CH3CHO}]^{3/2}$.

**Solution of Exercise 13.5.**

Steady state for $\ce{CH3CO}$: $k_2[\ce{CH3}][\ce{CH3CHO}] = k_3[\ce{CH3CO}]$. Steady state for $\ce{CH3}$: $k_1[\ce{CH3CHO}] - k_2[\ce{CH3}][\ce{CH3CHO}] + k_3[\ce{CH3CO}] - 2k_4[\ce{CH3}]^2 = 0$. Adding: $k_1[\ce{CH3CHO}] =
2k_4[\ce{CH3}]^2$, so $[\ce{CH3}] = (k_1/2k_4)^{1/2}[\ce{CH3CHO}]^{1/2}$, and $\dd[\ce{CH4}]/\dd t = k_2[\ce{CH3}][\ce{CH3CHO}] =
k_2(k_1/2k_4)^{1/2}[\ce{CH3CHO}]^{3/2}$.

**Exercise 13.6 ★★.**

Starting from equal concentrations of $\ce{H2}$ and $\ce{Br2}$, with $k' = 0.10$, by what factor has the rate fallen when half the bromine is consumed? How much of the fall is due to inhibition?

**Solution of Exercise 13.6.**

At half conversion $[\ce{H2}] = [\ce{Br2}] = \frac12$, $[\ce{HBr}] = 1$ (relative units): $v/v_0 = \frac12 \times (\frac12)^{1/2}/(1 + 0.10 \times 1/0.5) = 0.354/1.2 = 0.29$. Without inhibition it would be 0.354: inhibition divides the rate by a further 1.2.

**Exercise 13.7 ★★.**

In a model hydrogen–oxygen mixture (data of the exercise), branching has the rate constant $f = 1000\,p$, wall termination $6000/p$ and three-body termination $10\,p^2$, all in $\mathrm{s}^{-1}$ with $p$ in kPa. Find the pressure range in which the mixture explodes.

**Solution of Exercise 13.7.**

Explosion when $1000p > 6000/p + 10p^2$, i.e. $10p^3 - 1000p^2 + 6000 < 0$, whose positive roots are 2.48 and $99.9\,\mathrm{kPa}$: the mixture explodes between about 2.5 and $100\,\mathrm{kPa}$ (first and second limits of the model).

**Exercise 13.8 ★★.**

Initial rates are $0.20\,\text{µ}\mathrm{M}\,\mathrm{s}^{-1}$ at $[\mathrm S] = 0.50\,\mathrm{mM}$ and $0.40\,\text{µ}\mathrm{M}\,\mathrm{s}^{-1}$ at $2.0\,\mathrm{mM}$. Find $K_M$ and $V_{\max}$.

**Solution of Exercise 13.8.**

$1/v = 1/V_{\max} + (K_M/V_{\max})/[\mathrm S]$: $5.0 = 1/V_{\max} + 2.0K_M/V_{\max}$ and $2.5 = 1/V_{\max} + 0.5K_M/V_{\max}$. Subtracting: $K_M/V_{\max} = 1.667\,\mathrm{s}\,\mathrm{mM}\,\text{µ}\mathrm{M}^{-1}$, then $1/V_{\max} = 1.667$: $V_{\max} = 0.60\,\text{µ}\mathrm{M}\,\mathrm{s}^{-1}$, $K_M = 1.0\,\mathrm{mM}$.

**Exercise 13.9 ★★.**

With $50\,\text{µ}\mathrm{M}$ of an inhibitor, the Lineweaver–Burk line of an [enzyme](#def-b3-complex-kinetics-enzyme) ($K_M = 0.80\,\mathrm{mM}$) meets the uninhibited one on the $1/v$ axis and crosses the $1/[\mathrm S]$ axis at $-0.417\,\mathrm{mM}^{-1}$. Identify the type of inhibition and compute $K_i$.

**Solution of Exercise 13.9.**

Same intercept on the $1/v$ axis: competitive. $K_M^{\mathrm{app}} = 1/0.417 = 2.40\,\mathrm{mM} = 3K_M$, so $1 + 50/K_i = 3$ and $K_i = 25\,\text{µ}\mathrm{M}$.

**Exercise 13.10 ★★★.**

For $\mathrm{A + X \to 2\,X}$ with $k = 0.50\,\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{s}^{-1}$, $a_0 = 1.00\,\mathrm{mol}/\mathrm{L}$ and $x_0 = 0.010\,\mathrm{mol}/\mathrm{L}$, compute the time at which half the final amount of X is present, and the time of [maximum rate](#def-b3-complex-kinetics-michaelis).

**Solution of Exercise 13.10.**

$N = 1.01\,\mathrm{mol}/\mathrm{L}$: $t_{1/2} = \ln(1.01/0.010 - 1)/(0.50 \times 1.01) = \ln100/0.505 = 9.1\,\mathrm{s}$. The rate $kx(N - x)$ is largest at $x = N/2$: the same instant, the inflexion of the S curve.

**Exercise 13.11 ★★★.**

For the Brusselator with $A = 1$, compute the [eigenvalues](https://one-course.com/books/chemistry/4/en/chapter/1-quantum-mechanics-for-chemists-model-systems#def-b3-quantum-model-systems-operator) of the Jacobian at the steady state for $B = 1.5$ and $B = 2.5$, and describe the behaviour near the steady state in each case.

**Solution of Exercise 13.11.**

Trace $B - 2$, determinant 1. $B = 1.5$: $\lambda = -0.25 \pm 0.968\iu$, a stable focus (damped oscillations towards the steady state). $B = 2.5$: $\lambda = 0.25 \pm 0.968\iu$, an unstable focus: oscillations grow until they reach the [limit cycle](#def-b3-complex-kinetics-oscillating).

**Exercise 13.12 ★★★.**

For $\mathrm{A + B \rightleftharpoons C}$ with $k_1 = 1.0 \times 10^{8}\,\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{s}^{-1}$ and $k_{-1} = 1.0 \times 10^{3}\,\mathrm{s}^{-1}$, and $[\mathrm A]_e = [\mathrm B]_e = 2.70 \times 10^{-5}\,\mathrm{mol}/\mathrm{L}$, compute $\tau$. How would you obtain both rate constants from relaxation measurements alone?

**Solution of Exercise 13.12.**

$1/\tau = 1.0 \times 10^{8} \times 5.40 \times 10^{-5} + 1.0 \times 10^{3} = 6.4 \times 10^{3}\,\mathrm{s}^{-1}$, $\tau = 156\,\text{µ}\mathrm{s}$. Measure $\tau$ at several total concentrations: $1/\tau$ against $[\mathrm A]_e + [\mathrm B]_e$ is a straight line of slope $k_1$ and intercept $k_{-1}$.

## 13.7 Problem: An Enzyme, an Inhibitor and a Dose

**Problem 13.1.**

Weekend problem — an enzyme, an inhibitor and a dose: the Michaelis–Menten parameters by least squares, the inhibition constant of a competitive inhibitor, the activity left at a given dose, and a stopped-flow burst

An [enzyme](#def-b3-complex-kinetics-enzyme) ($[\mathrm E]_0 = 5.0\,\mathrm{nM}$ in the assays) hydrolyses its substrate; initial rates (data of the problem):

| $[\mathrm S]$ / mM | 0.10 | 0.20 | 0.40 | 0.80 | 1.60 | 3.20 | 6.40 |
| --- | --- | --- | --- | --- | --- | --- | --- |
| $v$ / $\text{µ}\mathrm{M}\,\mathrm{s}^{-1}$ | 0.0566 | 0.0981 | 0.169 | 0.247 | 0.340 | 0.397 | 0.447 |

In the presence of an inhibitor, fits of the same kind give $V_{\max}$ unchanged and the apparent $K_M$: 0.79, 1.47, 2.38 and $4.02\,\mathrm{mM}$ at $[\mathrm I] = 0$, 20, 50 and $100\,\text{µ}\mathrm{M}$. A stopped-flow experiment with $[\mathrm E]_0 = 2.0\,\text{µ}\mathrm{M}$ and saturating substrate gives the burst trace of the figure of section 5 (a burst of $1.28\,\text{µ}\mathrm{M}$ with rate constant $625\,\mathrm{s}^{-1}$, then $200\,\text{µ}\mathrm{M}\,\mathrm{s}^{-1}$).

**Part I — $K_M$ and $V_{\max}$.**

1. Why are initial rates used?
2. Draw the Lineweaver–Burk plot; its least-squares line gives $K_M =  0.773\,\mathrm{mM}$ and $V_{\max} = 0.491\,\text{µ}\mathrm{M}\,\mathrm{s}^{-1}$ . Which points dominate it?
3. Why is a direct nonlinear fit preferred?
4. The nonlinear fit gives $K_M = 0.801 \pm 0.023\,\mathrm{mM}$ and $V_{\max} = 0.502 \pm  0.005\,\text{µ}\mathrm{M}\,\mathrm{s}^{-1}$ . Are the Lineweaver–Burk values compatible with them?
5. Compute $k_{\mathrm{cat}}$ .
6. Compute the [specificity constant](#def-b3-complex-kinetics-michaelis) .
7. Compare it with the diffusion limit and with a typical [enzyme](#def-b3-complex-kinetics-enzyme) .

**Part II — The inhibitor.**

8. What type of inhibition do the data show?
9. Write $K_M^{\mathrm{app}}$ as a function of $[\mathrm I]$ .
10. The least-squares line of $K_M^{\mathrm{app}}$ against $[\mathrm I]$ has intercept $0.799\,\mathrm{mM}$ and slope $0.0321\,\mathrm{mM}\,\text{µ}\mathrm{M}^{-1}$ . Deduce $K_i$ .
11. Its standard error is $0.9\,\text{µ}\mathrm{M}$ . Give a 95 % interval ( $t = 4.30$ for two degrees of freedom).
12. What does $K_i$ measure?

**Part III — Activity left.**

13. Express the fraction of activity left, $v_i/v_0$ , as a function of $[\mathrm S]$ and $[\mathrm I]$ .
14. Evaluate it at $[\mathrm S] = K_M$ and $[\mathrm I] = K_i$ .
15. At $[\mathrm S] = K_M$ and $[\mathrm I] = 10K_i$ .
16. At $[\mathrm S] = 10K_M$ and $[\mathrm I] = 10K_i$ . Comment.
17. Show that 90 % inhibition at $[\mathrm S] = K_M$ needs $[\mathrm I] = 18K_i$ .
18. Compute that concentration with the fitted $K_i$ .

**Part IV — The burst.**

19. The [enzyme](#def-b3-complex-kinetics-enzyme) works in two steps, $\mathrm{E + S \to E{-}acyl + P_1}$ ( $k_2$ ) then $\mathrm{E{-}acyl \to E +  P_2}$ ( $k_3$ ). Why does $\mathrm P_1$ appear in a burst?
20. The burst amplitude is $[\mathrm E]_0\bigl(k_2/(k_2 + k_3)\bigr)^2$ . Deduce $k_2/(k_2 + k_3)$ .
21. From the steady slope, compute $k_{\mathrm{cat}}$ and compare with Part I.
22. The burst rate constant is $k_2 + k_3$ . Deduce $k_2$ and $k_3$ .
23. Check that $k_{\mathrm{cat}} = k_2k_3/(k_2 + k_3)$ .
24. State the result: the inhibitor concentration that gives 90 % inhibition at $[\mathrm S] = K_M$ .

**Solution of Problem 13.1.**

**1.** At the start $[\mathrm S]$ is known and the product, which could inhibit or react back, is absent. **2.** The low-$[\mathrm S]$ points, at large $1/[\mathrm S]$ and $1/v$: they set the slope, and the inversion magnifies their errors. **3.** It weights each measured rate as measured; the Lineweaver–Burk line gives biased parameters. **4.** $K_M$: $0.773$ is 1.2 standard errors below 0.801; $V_{\max}$: $0.491$ is 2.2 standard errors below 0.502. Both are low, as expected from the bias. **5.** $k_{\mathrm{cat}} = 0.502/0.0050 = 100\,\mathrm{s}^{-1}$. **6.** $100/0.801 \times 10^{-3} = 1.25 \times 10^{5}\,\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{s}^{-1}$. **7.** Some 60 000 times below the diffusion limit; a typical [enzyme](#def-b3-complex-kinetics-enzyme). **8.** Competitive: $V_{\max}$ unchanged, apparent $K_M$ growing with $[\mathrm I]$. **9.** $K_M^{\mathrm{app}} = K_M(1 + [\mathrm I]/K_i) = K_M + (K_M/K_i)[\mathrm I]$. **10.** $K_i = 0.799/0.0321 = 24.9\,\text{µ}\mathrm{M}$. **11.** $24.9 \pm 4.30 \times 0.9 = 24.9 \pm 3.9\,\text{µ}\mathrm{M}$. **12.** The dissociation constant of the enzyme–inhibitor complex: the inhibitor concentration that occupies half the free [enzyme](#def-b3-complex-kinetics-enzyme). **13.** $v_i/v_0 = (K_M + [\mathrm S])/(K_M(1 + [\mathrm I]/K_i) + [\mathrm S])$. **14.** $2/3 = 0.67$. **15.** $2/12 = 0.17$. **16.** $11/21 = 0.52$: at high substrate concentration the substrate outcompetes the inhibitor. **17.** At $[\mathrm S] = K_M$, $v_i/v_0 = 2/(2 + [\mathrm I]/K_i) = 0.10$ gives $[\mathrm I]/K_i = 18$. **18.** $18 \times 24.9 = 448\,\text{µ}\mathrm{M}$. **19.** The first step is fast: every [enzyme](#def-b3-complex-kinetics-enzyme) molecule quickly releases one $\mathrm P_1$ and is trapped as the [acyl-enzyme](#def-b3-complex-kinetics-enzyme); afterwards $\mathrm P_1$ appears only as fast as the slow second step frees the [enzyme](#def-b3-complex-kinetics-enzyme). **20.** $1.28/2.0 = 0.64 = (k_2/(k_2 + k_3))^2$, so $k_2/(k_2 + k_3) = 0.80$. **21.** $200/2.0 = 100\,\mathrm{s}^{-1}$, the same as in Part I. **22.** $k_2 = 0.80 \times 625 = 500\,\mathrm{s}^{-1}$, $k_3 = 125\,\mathrm{s}^{-1}$. **23.** $500 \times 125/625 = 100\,\mathrm{s}^{-1}$. **24.** **$[\mathrm I]_{90\,\%} = 18K_i \approx 0.45\,\mathrm{mM}$ at $[\mathrm S] = K_M$.**