Result

The steady-state approximation

T-043Home CU-202Threads kinetics

Statement

In a mechanism of elementary steps obeying mass action, an intermediate whose chemical lifetime is far shorter than the timescale on which the reactants change settles, after a brief induction period, onto a quasi-stationary concentration equal to its total rate of formation divided by its total first-order coefficient of loss. Formally: let the intermediate be created and destroyed only by steps of the mechanism, let its lifetime be \(\tau_{\text{I}}\), the reciprocal of the total pseudo-first-order coefficient for its loss, and let the reactants change on the timescale \(\tau_{\text{R}}\). If \(\varepsilon=\tau_{\text{I}}/\tau_{\text{R}}\ll1\), then for \(t\gg\tau_{\text{I}}\) the condition \(d[\text{I}]/dt\approx0\) determines \([\text{I}]_{ss}\) algebraically, and substituting it into the product-forming step yields an observed rate law whose fractional error is of order \(\varepsilon\).

Why it matters

Experiment delivers a rate law; a mechanism is a hypothesis about elementary steps. The two can only be compared if the mechanism can be made to predict a rate law, and for any mechanism containing a reactive intermediate that means solving a coupled, nonlinear system of differential equations for which no closed form generally exists. The steady-state approximation (SSA, also called the Bodenstein quasi-stationary-state hypothesis) replaces one differential equation per intermediate by one algebraic equation, and the algebra is elementary. Every predicted rate law in gas-phase kinetics, combustion, atmospheric chemistry, polymerisation and enzymology is obtained this way.

The approximation also has a precise mathematical status, which is what makes it more than a convenient fudge: it is the leading term of a singular-perturbation expansion in the small parameter \(\varepsilon\), the ratio of the intermediate’s lifetime to the reaction’s. That framing tells you not only what the SSA predicts but how wrong it is, and exactly when it fails — during the induction period, and whenever the intermediate is not fleeting. Both statements are quantified below and checked against the one case that can be solved exactly.

Hypotheses

Every step of the mechanism is elementary, so its rate law follows from its molecularity.Only for an elementary step may the rate be written as the product of the concentrations of the species that collide in it. Applied to an overall equation — writing rate \(=k[\text{H}_2][\text{Br}_2]\) for \(\text{H}_2+\text{Br}_2\to2\,\text{HBr}\) — the whole construction is meaningless, since that reaction is in fact a radical chain whose observed order in \(\text{Br}_2\) is one half.

The species treated by the SSA is a genuine intermediate: it is absent initially and is consumed by first-order or pseudo-first-order steps.The loss terms must be linear in \([\text{I}]\) for \([\text{I}]_{ss}\) to be a ratio of the form stated. A second-order loss such as radical recombination, \(-2k_t[\text{I}]^2\), still gives an algebraic equation, but a quadratic one; the steady-state concentration then scales as the square root of the initiation rate, which is where the half-order rate laws of chain kinetics come from.

Timescale separation: \(\varepsilon=\tau_{\text{I}}/\tau_{\text{R}}\ll1\).With \(\tau_{\text{I}}=(k_{-1}+k_2)^{-1}\) and \(\tau_{\text{R}}=k_1^{-1}\) for the two-step mechanism below, \(\varepsilon=k_1/(k_{-1}+k_2)\). If \(\varepsilon\sim1\) the intermediate accumulates to a concentration comparable with the reactant and the SSA is wrong by a factor of order unity, not by a small correction (see Fails without).

The observation time lies outside the initial boundary layer: \(t\gg\tau_{\text{I}}\).The SSA cannot satisfy the initial condition \([\text{I}](0)=0\); it asserts a non-zero concentration at \(t=0\). The true solution needs a time of order \(\tau_{\text{I}}\) to climb onto the quasi-stationary branch, and during that induction period the SSA over-predicts \([\text{I}]\), and hence the rate, by up to 100 percent.

The fast subsystem is stable: the eigenvalues of \(\partial(d[\text{I}]/dt)/\partial[\text{I}]\) have negative real parts bounded away from zero.This is the hypothesis of Tikhonov’s theorem, and it is what guarantees that the algebraic solution is an attractor rather than a repeller. Chain branching (an intermediate that produces more of itself) flips the sign, and the steady state then exists on paper but is never approached — the physical outcome is an explosion.

Proof

The argument is carried out on the smallest mechanism that contains the whole structure — one reactant, one intermediate, one product — and is then compared with the exact solution of the same mechanism, which is available in closed form. The generalisation to many intermediates is Step 10.

\[ \text{A}\ \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}}\ \text{I}\ \overset{k_2}{\longrightarrow}\ \text{P} \]

The mechanism: reversible formation of the intermediate, then irreversible conversion to product. Each step is elementary, so its rate is the rate constant times the concentration of the single species entering it (Hypotheses). A

\[ \frac{d[\text{A}]}{dt}=-k_1[\text{A}]+k_{-1}[\text{I}],\qquad \frac{d[\text{I}]}{dt}=k_1[\text{A}]-(k_{-1}+k_2)[\text{I}],\qquad \frac{d[\text{P}]}{dt}=k_2[\text{I}] \]

Summing every elementary step’s contribution to each species gives the exact kinetic system. No approximation has been made yet; note that the three right-hand sides sum to zero, which is conservation of matter. A

\[ \tau_{\text{I}}=\frac{1}{k_{-1}+k_2},\qquad \tau_{\text{R}}=\frac{1}{k_1},\qquad \varepsilon=\frac{\tau_{\text{I}}}{\tau_{\text{R}}}=\frac{k_1}{k_{-1}+k_2} \]

Identify the two timescales in the problem before approximating anything: \(\tau_{\text{I}}\) is the mean lifetime of a molecule of \(\text{I}\) once formed, \(\tau_{\text{R}}\) the timescale on which \(\text{A}\) is drained. Their ratio is the only dimensionless group the mechanism contains, so it is the only thing an approximation can be small in. B

\[ a=\frac{[\text{A}]}{[\text{A}]_0},\qquad u=\frac{[\text{I}]}{\varepsilon\,[\text{A}]_0},\qquad \tau=\frac{t}{\tau_{\text{R}}}\ \ \Longrightarrow\ \ \varepsilon\frac{du}{d\tau}=a-u,\qquad \frac{da}{d\tau}=-a+\phi\,u \]

Rescaling concentrations by \([\text{A}]_0\), the intermediate additionally by \(\varepsilon\), and time by \(\tau_{\text{R}}\) puts the system in standard singular-perturbation form, with \(\phi=k_{-1}/(k_{-1}+k_2)\in[0,1]\). The scaling of \(u\) is the content of the step: it states in advance that the intermediate is smaller than the reactant by the factor \(\varepsilon\), which is exactly what the SSA will confirm. B

\[ \varepsilon\to0:\qquad 0=a-u\ \ \Longrightarrow\ \ u=a\ \ \Longleftrightarrow\ \ [\text{I}]_{ss}=\frac{k_1[\text{A}]}{k_{-1}+k_2} \]

Setting \(\varepsilon=0\) demotes the differential equation for \(u\) to an algebraic one — this is the steady-state approximation, and in these variables it is visibly a statement about a small parameter rather than about a concentration being constant. Undoing the scaling returns the familiar form: formation rate over total loss coefficient. B

\[ \frac{d[\text{P}]}{dt}=k_2[\text{I}]_{ss}=\frac{k_1k_2}{k_{-1}+k_2}\,[\text{A}]\ \equiv\ k_{\text{obs}}[\text{A}] \]

Substituting the algebraic result into the product-forming step of Step 2 eliminates the unobservable \([\text{I}]\) and leaves a rate law in measurable quantities alone. The mechanism therefore predicts first-order kinetics with the composite constant \(k_{\text{obs}}=k_1k_2/(k_{-1}+k_2)\) — the prediction that experiment can falsify. A

\[ \left|\frac{d[\text{I}]_{ss}}{dt}\right| = \frac{k_1}{k_{-1}+k_2}\left|\frac{d[\text{A}]}{dt}\right| \approx \varepsilon\,k_{\text{obs}}[\text{A}] \qquad\text{versus}\qquad k_1[\text{A}] \]

Self-consistency check: differentiate the quantity that was assumed constant and compare it with the terms that were kept. The neglected derivative is smaller than either individual term of Step 2 by the factor \(\varepsilon\), so the approximation does not contradict itself — \(d[\text{I}]/dt\) is negligible relative to the formation and loss rates, not absolutely zero. C

\[ k_{-1}=0:\qquad \frac{d[\text{I}]}{dt}+k_2[\text{I}]=k_1[\text{A}]_0e^{-k_1t} \ \ \Longrightarrow\ \ [\text{I}](t)=\frac{k_1[\text{A}]_0}{k_2-k_1}\left(e^{-k_1t}-e^{-k_2t}\right) \]

For the irreversible sequential case the system is linear and solvable exactly: \([\text{A}]=[\text{A}]_0e^{-k_1t}\), then the integrating factor \(e^{k_2t}\) and the initial condition \([\text{I}](0)=0\) give the closed form. This is the benchmark against which the approximation can be measured rather than asserted. B

\[ \frac{[\text{I}]_{\text{exact}}}{[\text{I}]_{ss}}=\frac{1}{1-\varepsilon}\left(1-e^{-(k_2-k_1)t}\right),\qquad \varepsilon=\frac{k_1}{k_2} \]

Dividing Step 8 by \([\text{I}]_{ss}=(k_1/k_2)[\text{A}]\) isolates the two and only two error terms: a uniform relative error \(\varepsilon/(1-\varepsilon)\) that persists for all time, and a transient \(e^{-(k_2-k_1)t}\) that dies within a few multiples of \(\tau_{\text{I}}\). The SSA is therefore accurate to \(O(\varepsilon)\) once \(t\gg\tau_{\text{I}}\), and to nothing at all before that. C

\[ \frac{d\mathbf{x}}{dt}=\mathbf{f}(\mathbf{x},\mathbf{y}),\qquad \varepsilon\frac{d\mathbf{y}}{dt}=\mathbf{g}(\mathbf{x},\mathbf{y}),\qquad \mathbf{g}(\mathbf{x},\mathbf{y}_{ss})=0 \]

With \(n\) intermediates \(\mathbf{y}\) the same scaling gives one algebraic equation per intermediate, to be solved simultaneously. Tikhonov’s theorem states that the solution of the full system converges to that of the reduced system as \(\varepsilon\to0\), uniformly outside the initial layer, provided the root \(\mathbf{y}_{ss}\) is isolated and asymptotically stable — every eigenvalue of \(\partial\mathbf{g}/\partial\mathbf{y}\) having negative real part. For a linear mechanism the algebraic system is linear, so \(\mathbf{y}_{ss}\) is unique whenever the loss matrix is non-singular. C

Result

\[ [\text{I}]_{ss}=\frac{\sum(\text{formation rates of I})}{\sum(\text{first-order loss coefficients of I})},\qquad \frac{d[\text{P}]}{dt}=\frac{k_1k_2}{k_{-1}+k_2}[\text{A}],\qquad \text{error}=O\!\left(\varepsilon\right),\ \ \varepsilon=\frac{k_1}{k_{-1}+k_2} \]

Reading. One algebraic equation per intermediate replaces one differential equation per intermediate. The intermediate is not assumed to be zero and is not assumed constant: what is assumed is that its net rate of change is small compared with the individual formation and loss rates, which is true to relative order \(\varepsilon\) once the induction period is over.

Scope. Valid for \(t\gg\tau_{\text{I}}\) and \(\varepsilon\ll1\), with the fast subsystem stable. It predicts the functional form of the rate law and the composite constant \(k_{\text{obs}}\); it cannot resolve the individual constants \(k_{-1}\) and \(k_2\) separately, only their ratio, and it does not conserve matter exactly — the mass balance is short by \([\text{I}]_{ss}\), itself an \(O(\varepsilon)\) quantity.

Corollaries & converses

Rate-determining-step limit. If \(k_2\gg k_{-1}\), then \(k_{\text{obs}}\to k_1\): every intermediate formed goes on to product, and the first step alone controls the rate.
Pre-equilibrium limit. If \(k_{-1}\gg k_2\), then \(k_{\text{obs}}\to (k_1/k_{-1})k_2=Kk_2\), the pre-equilibrium result. The pre-equilibrium approximation is thus a special case of the SSA, not an alternative to it, and it overestimates \(k_{\text{obs}}\) by the factor \(1+k_2/k_{-1}\).
Lindemann–Hinshelwood unimolecular kinetics. With collisional activation \(\text{A}+\text{M}\rightleftharpoons\text{A}^{*}+\text{M}\) followed by \(\text{A}^{*}\to\text{P}\), the SSA gives \(k_{\text{uni}}=k_1k_2[\text{M}]/(k_{-1}[\text{M}]+k_2)\): second order at low pressure, first order at high pressure, with the reciprocal form \(1/k_{\text{uni}}=1/k_{\infty}+1/(k_1[\text{M}])\) linear in \(1/[\text{M}]\).
Michaelis–Menten kinetics. The same algebra applied to an enzyme–substrate complex, with the enzyme mass balance \([\text{E}]_T=[\text{E}]+[\text{ES}]\) closing the system, gives \(K_M=(k_{-1}+k_2)/k_1\) and the hyperbolic rate law (Michaelis-Menten kinetics).
Chain reactions. Applying the SSA to every radical and adding the resulting equations makes all propagation terms cancel in pairs, leaving rate of initiation = rate of termination. With second-order termination this produces the characteristic half-order dependence on the initiator.
Converse fails. Agreement between an SSA-predicted rate law and the measured one supports a mechanism but does not establish it: distinct mechanisms can share a rate law, which is why isotopic labelling, trapping and direct detection of intermediates remain necessary.

Fails without

Timescale separation dropped (\(\varepsilon\sim1\)): take \(k_1=k_2=k\) in the sequential mechanism. The exact solution degenerates to \([\text{I}]=k t[\text{A}]_0e^{-kt}\), peaking at \([\text{I}]=[\text{A}]_0/e=0.368[\text{A}]_0\) at \(t=1/k\) — the “intermediate” is a major component. The SSA predicts \([\text{I}]_{ss}=(k_1/k_2)[\text{A}]=[\text{A}]\), and at \(t=0.2/k\) it gives \(0.819[\text{A}]_0\) against the true \(0.164[\text{A}]_0\): too large by a factor of \(5.0\).
Induction period included (\(t\lesssim\tau_{\text{I}}\)): the SSA assigns \([\text{I}]\) a finite value at \(t=0\) although the true value is zero, so it violates the initial condition outright. With \(k_1=1.00\times10^{-3}\,\text{s}^{-1}\) and \(k_2=0.500\,\text{s}^{-1}\), and \([\text{A}]_0=0.100\,\text{mol}\,\text{dm}^{-3}\), the exact \([\text{I}]\) at \(t=1.0\,\text{s}\) is \(7.87\times10^{-5}\,\text{mol}\,\text{dm}^{-3}\), while the SSA gives \(2.00\times10^{-4}\,\text{mol}\,\text{dm}^{-3}\), high by a factor of \(2.5\). Only after \(t\approx4.6/k_2\approx9\,\text{s}\) does the transient fall below one percent.
Stability of the fast subsystem dropped (branching): in a branched chain such as \(\text{H}\cdot+\text{O}_2\to\text{OH}\cdot+\text{O}\), one chain carrier produces two, so the loss coefficient of the radical pool becomes \(k_{\text{term}}-k_{\text{branch}}\). The SSA answer \([\text{R}]_{ss}\propto1/(k_{\text{term}}-k_{\text{branch}})\) diverges and then turns negative as branching overtakes termination at the explosion limit; the true behaviour is exponential growth, and no stationary state is ever reached.
Uniqueness and stability of the algebraic root dropped (oscillations): in autocatalytic schemes of Belousov–Zhabotinsky type the steady state exists but is unstable, and concentrations circulate around it in a limit cycle. A stationary rate law then describes no part of the observed behaviour, however small the intermediates’ concentrations are.

Common errors

“\(d[\text{I}]/dt=0\) means \([\text{I}]=0\).” It means the net rate of change is negligible against the individual rates. The steady-state concentration is small but finite, and it is exactly the quantity substituted back into the rate law.
“\([\text{I}]_{ss}\) is a constant, so it may be treated as one throughout.” \([\text{I}]_{ss}=(k_1/(k_{-1}+k_2))[\text{A}]\) decays as \([\text{A}]\) decays. It is quasi-stationary: slaved to the reactant, not fixed. Integrating the rate law while holding \([\text{I}]_{ss}\) fixed gives a zero-order decay that is simply wrong.
“Steady state means equilibrium.” Equilibrium requires every step to be balanced by its own reverse (detailed balance) and carries no net flux; the steady state here carries the full reaction flux and is maintained precisely because the system is far from equilibrium.
“The SSA and the pre-equilibrium approximation are two names for the same thing.” Pre-equilibrium is the \(k_{-1}\gg k_2\) corner of the SSA. Using it when \(k_2\sim k_{-1}\) inflates \(k_{\text{obs}}\) by \(1+k_2/k_{-1}\), which at \(k_2=k_{-1}\) is a factor of two.
Omitting a loss channel. Every step that consumes the intermediate belongs in the denominator. Dropping one of two comparable channels changes \(k_{\text{obs}}\) by a factor of order one and can change the predicted order in a reagent.
Applying the SSA to a reactant or product. Only species that are formed and destroyed within the mechanism, and never accumulate, qualify. Setting \(d[\text{A}]/dt\approx0\) for a reactant asserts that nothing happens.

Discussion

Max Bodenstein introduced the quasi-stationary-state hypothesis in 1913 while analysing the photochemical hydrogen–chlorine and hydrogen–bromine reactions, where the intermediates are atoms present at concentrations far below detection and the alternative was no theory at all. The device spread quickly: to unimolecular gas-phase reactions through Lindemann and Hinshelwood in the 1920s, and to enzymes through Briggs and Haldane in 1925, whose steady-state treatment replaced Michaelis and Menten’s more restrictive rapid-equilibrium assumption while leaving the algebraic form of the rate law unchanged — only the interpretation of \(K_M\) changes.

The practical payoff is that the SSA turns an intractable inverse problem into an algebraic one. A measured rate law of the saturating form \(v=a[\text{S}]/(b+[\text{S}])\), or one showing inverse dependence on a product, is immediately recognisable as the signature of a mechanism with a reversibly formed intermediate; the fitted composite constants then constrain ratios of elementary rate constants even when the individual constants are unmeasurable. What the SSA cannot do is separate \(k_{-1}\) from \(k_2\): the rate law depends on them only through \(k_1k_2/(k_{-1}+k_2)\), so no amount of concentration-dependence data will resolve them, and a fast spectroscopic experiment that observes the intermediate directly is required.

Mathematically the SSA is the leading-order outer solution of a singularly perturbed system, and the induction period is its inner (boundary-layer) solution; matched asymptotics supplies the uniformly valid composite approximation, and Tikhonov’s theorem supplies the conditions under which the outer solution is legitimate at all. This is more than pedantry in modern practice: chemical mechanisms with thousands of species are numerically stiff precisely because their radical intermediates have lifetimes many orders of magnitude shorter than the simulation timestep, and the standard remedy — quasi-steady-state model reduction — is this theorem applied automatically, with the species eligible for elimination selected by exactly the eigenvalue criterion of Step 10.

Common misconceptions. That the SSA is an assumption about the mechanism rather than about a regime: the same mechanism satisfies it at one temperature or pressure and violates it at another, because \(\varepsilon\) depends on rate constants that depend on both. And that its accuracy improves as the reaction proceeds: for the sequential mechanism the relative error \(\varepsilon/(1-\varepsilon)\) is independent of time once the transient has decayed, so a long experiment does not make a poor approximation good.

Worked examples

Example 1. For the sequential mechanism \(\text{A}\to\text{I}\to\text{P}\) take \(k_1=1.00\times10^{-3}\,\text{s}^{-1}\), \(k_2=0.500\,\text{s}^{-1}\) and \([\text{A}]_0=0.100\,\text{mol}\,\text{dm}^{-3}\). Find \([\text{I}]\) at \(t=600\,\text{s}\) from the SSA and exactly, quantify the error, and state when the SSA becomes usable.

\[ \varepsilon=\frac{k_1}{k_2}=\frac{1.00\times10^{-3}\,\text{s}^{-1}}{0.500\,\text{s}^{-1}}=2.00\times10^{-3} \]

Test the hypothesis before using the result. The intermediate’s lifetime \(\tau_{\text{I}}=1/k_2=2.00\,\text{s}\) is \(500\) times shorter than \(\tau_{\text{R}}=1/k_1=1000\,\text{s}\), so an \(O(\varepsilon)\) error of about \(0.2\) percent is expected. A

\[ [\text{A}](600\,\text{s})=[\text{A}]_0e^{-k_1t}=0.100\,e^{-0.600}=5.488\times10^{-2}\,\text{mol}\,\text{dm}^{-3} \]

The reactant decays with its own first-order constant because, with \(k_{-1}=0\), nothing returns to \(\text{A}\). Symbols first, numbers last: \(k_1t=(1.00\times10^{-3})(600)=0.600\) is dimensionless as it must be. A

\[ [\text{I}]_{ss}=\frac{k_1}{k_2}[\text{A}]=\left(2.00\times10^{-3}\right)\left(5.488\times10^{-2}\right)=1.098\times10^{-4}\,\text{mol}\,\text{dm}^{-3} \]

The steady-state result of Step 5 of the Proof with \(k_{-1}=0\). The intermediate sits three orders of magnitude below the reactant, consistent with calling it an intermediate. A

\[ [\text{I}]_{\text{exact}}=\frac{k_1[\text{A}]_0}{k_2-k_1}\left(e^{-k_1t}-e^{-k_2t}\right)=\frac{1.00\times10^{-4}}{0.499}\left(0.5488-e^{-300}\right)=1.100\times10^{-4}\,\text{mol}\,\text{dm}^{-3} \]

Step 8 of the Proof evaluated at the same time; \(e^{-300}\) is smaller than \(10^{-130}\) and is discarded. The prefactor \((1.00\times10^{-3})(0.100)/(0.500-0.001)=2.004\times10^{-4}\,\text{mol}\,\text{dm}^{-3}\). B

\[ \frac{[\text{I}]_{\text{exact}}-[\text{I}]_{ss}}{[\text{I}]_{\text{exact}}}=\frac{1.0998-1.0976}{1.0998}=2.0\times10^{-3}=\varepsilon \]

The measured discrepancy reproduces the predicted \(\varepsilon/(1-\varepsilon)\approx\varepsilon\) of Step 9 — the SSA is low by \(0.20\) percent, and the error estimate is not merely an order of magnitude but the actual number. B

\[ e^{-k_2t}\le0.01\ \Longrightarrow\ t\ge\frac{\ln100}{k_2}=\frac{4.605}{0.500\,\text{s}^{-1}}=9.2\,\text{s} \]

The transient term sets the induction period: after \(9.2\,\text{s}\) — \(1.5\) percent of the \(600\,\text{s}\) observation and \(0.9\) percent of \(\tau_{\text{R}}\) — the boundary-layer error has fallen below one percent, so the whole of a practical measurement lies in the valid region. A

\[ [\text{I}]_{ss}=1.098\times10^{-4}\,\text{mol}\,\text{dm}^{-3},\quad [\text{I}]_{\text{exact}}=1.100\times10^{-4}\,\text{mol}\,\text{dm}^{-3},\quad \text{error}=0.20\ \text{percent},\quad t_{\text{ind}}\approx9\,\text{s} \]

Reading. With a five-hundred-fold separation of timescales the SSA is accurate to two parts in a thousand for all but the first nine seconds of a ten-minute experiment.

Scope. The rate of product formation inherits the same accuracy: \(k_2[\text{I}]_{ss}=5.488\times10^{-5}\) against the exact \(5.499\times10^{-5}\,\text{mol}\,\text{dm}^{-3}\,\text{s}^{-1}\).

Example 2. A gas-phase isomerisation \(\text{A}\to\text{P}\) is first order in \(\text{A}\) but its first-order constant depends on the total pressure — the Lindemann–Hinshelwood fall-off. Two runs at \(500\,\text{K}\) give \(k_{\text{uni}}=2.00\times10^{-4}\,\text{s}^{-1}\) at \([\text{M}]=1.00\times10^{-2}\,\text{mol}\,\text{dm}^{-3}\) and \(k_{\text{uni}}=5.00\times10^{-5}\,\text{s}^{-1}\) at \([\text{M}]=1.00\times10^{-3}\,\text{mol}\,\text{dm}^{-3}\). Extract the activation rate constant \(k_1\) and the high-pressure limit \(k_{\infty}\), and find the pressure at which the reaction has fallen to half of \(k_{\infty}\).

\[ \text{A}+\text{M}\ \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}}\ \text{A}^{*}+\text{M},\qquad \text{A}^{*}\overset{k_2}{\longrightarrow}\text{P} \]

A unimolecular reaction still needs a bimolecular collision to supply the energy; \(\text{A}^{*}\) is the energised molecule, and \(\text{M}\) any collision partner (see Kinetic theory and gas pressure). A

\[ [\text{A}^{*}]_{ss}=\frac{k_1[\text{A}][\text{M}]}{k_{-1}[\text{M}]+k_2}\ \Longrightarrow\ k_{\text{uni}}=\frac{k_2[\text{A}^{*}]_{ss}}{[\text{A}]}=\frac{k_1k_2[\text{M}]}{k_{-1}[\text{M}]+k_2} \]

Formation rate over total loss coefficient, exactly as in the Result; the energised molecule is lost both by deactivating collisions (pseudo-first-order coefficient \(k_{-1}[\text{M}]\)) and by reaction. A

\[ \frac{1}{k_{\text{uni}}}=\frac{k_{-1}[\text{M}]+k_2}{k_1k_2[\text{M}]}=\frac{k_{-1}}{k_1k_2}+\frac{1}{k_1[\text{M}]}=\frac{1}{k_{\infty}}+\frac{1}{k_1[\text{M}]} \]

Inverting linearises the fall-off: a plot of \(1/k_{\text{uni}}\) against \(1/[\text{M}]\) has intercept \(1/k_{\infty}\) with \(k_{\infty}=k_1k_2/k_{-1}\), and slope \(1/k_1\). Rearranged before any number is inserted. B

\[ \frac{1}{k_{\infty}}+\frac{100}{k_1}=5.00\times10^{3}\,\text{s},\qquad \frac{1}{k_{\infty}}+\frac{1000}{k_1}=2.00\times10^{4}\,\text{s} \]

The two runs give two equations; the reciprocal concentrations are \(1/(1.00\times10^{-2})=100\) and \(1/(1.00\times10^{-3})=1000\,\text{dm}^{3}\,\text{mol}^{-1}\). A

\[ \frac{900}{k_1}=1.50\times10^{4}\,\text{s}\ \Longrightarrow\ k_1=6.00\times10^{-2}\,\text{dm}^{3}\,\text{mol}^{-1}\,\text{s}^{-1} \]

Subtracting eliminates the intercept — the standard two-point determination of a slope. The units are those of a bimolecular rate constant, as the activation step requires. A

\[ \frac{1}{k_{\infty}}=5.00\times10^{3}-\frac{100}{6.00\times10^{-2}}=5.00\times10^{3}-1.667\times10^{3}=3.333\times10^{3}\,\text{s}\ \Longrightarrow\ k_{\infty}=3.00\times10^{-4}\,\text{s}^{-1} \]

Back-substitution into the first equation. Sanity check: the higher-pressure run already sits at \(2.00/3.00=67\) percent of the limiting value, consistent with it lying above the fall-off midpoint. A

\[ k_{\text{uni}}=\tfrac12k_{\infty}\ \Longleftrightarrow\ \frac{1}{k_1[\text{M}]}=\frac{1}{k_{\infty}}\ \Longrightarrow\ [\text{M}]_{1/2}=\frac{k_{\infty}}{k_1}=\frac{3.00\times10^{-4}}{6.00\times10^{-2}}=5.00\times10^{-3}\,\text{mol}\,\text{dm}^{-3} \]

Half of the high-pressure limit is reached when the two reciprocal terms are equal — the definition of the fall-off midpoint. B

\[ p_{1/2}=cRT=\left(5.00\,\text{mol}\,\text{m}^{-3}\right)\left(8.314\,\text{J}\,\text{K}^{-1}\text{mol}^{-1}\right)\left(500\,\text{K}\right)=2.08\times10^{4}\,\text{Pa}=20.8\,\text{kPa} \]

Converting to pressure with the ideal-gas equation of state (The ideal gas law) and \(R=8.314\,\text{J}\,\text{K}^{-1}\text{mol}^{-1}\); note \(5.00\times10^{-3}\,\text{mol}\,\text{dm}^{-3}=5.00\,\text{mol}\,\text{m}^{-3}\). A

\[ k_1=6.00\times10^{-2}\,\text{dm}^{3}\,\text{mol}^{-1}\,\text{s}^{-1},\qquad k_{\infty}=3.00\times10^{-4}\,\text{s}^{-1},\qquad p_{1/2}=20.8\,\text{kPa} \]

Reading. The SSA converts a pressure-dependent “first-order” constant into two genuine constants: the collisional activation rate and the high-pressure limit. Below about \(21\,\text{kPa}\) activation cannot keep up and the reaction drifts toward second order overall.

Scope. Only the composite \(k_{\infty}=k_1k_2/k_{-1}\) is accessible from fall-off data; \(k_2\) and \(k_{-1}\) cannot be separated this way. The strict Lindemann form also predicts a straight reciprocal plot, whereas real fall-off curves bend, because \(k_2\) depends on how much energy the molecule actually carries — the correction supplied by RRKM theory.

Problems

The reaction \(\text{F}_2+2\,\text{NO}_2\to2\,\text{FNO}_2\) is proposed to proceed by \(\text{F}_2+\text{NO}_2\overset{k_1}{\to}\text{FNO}_2+\text{F}\) followed by \(\text{F}+\text{NO}_2\overset{k_2}{\to}\text{FNO}_2\). Derive the SSA rate law for \(\text{FNO}_2\) formation, and evaluate it and \([\text{F}]_{ss}\) for \(k_1=4.0\times10^{-2}\,\text{dm}^{3}\,\text{mol}^{-1}\text{s}^{-1}\), \(k_2=1.0\times10^{8}\,\text{dm}^{3}\,\text{mol}^{-1}\text{s}^{-1}\), \([\text{F}_2]=2.0\times10^{-2}\) and \([\text{NO}_2]=1.0\times10^{-2}\,\text{mol}\,\text{dm}^{-3}\).
Solution
Fluorine atoms are formed only in step 1 and consumed only in step 2, so \(d[\text{F}]/dt=k_1[\text{F}_2][\text{NO}_2]-k_2[\text{F}][\text{NO}_2]\approx0\), giving \([\text{F}]_{ss}=k_1[\text{F}_2]/k_2\) — the \([\text{NO}_2]\) cancels. Product appears in both steps, so \(d[\text{FNO}_2]/dt=k_1[\text{F}_2][\text{NO}_2]+k_2[\text{F}]_{ss}[\text{NO}_2]=2k_1[\text{F}_2][\text{NO}_2]\): second order overall, first order in each reactant, with the observed constant equal to twice that of the slow step.
Numerically \([\text{F}]_{ss}=(4.0\times10^{-2})(2.0\times10^{-2})/(1.0\times10^{8})=8.0\times10^{-12}\,\text{mol}\,\text{dm}^{-3}\), i.e. \(8\times10^{-10}\) of \([\text{NO}_2]\) — amply justifying the approximation — and the rate is \(2(4.0\times10^{-2})(2.0\times10^{-2})(1.0\times10^{-2})=1.6\times10^{-5}\,\text{mol}\,\text{dm}^{-3}\,\text{s}^{-1}\).
For \(\text{A}\rightleftharpoons\text{I}\to\text{P}\) with \(k_1=2.0\times10^{-2}\,\text{s}^{-1}\), \(k_{-1}=4.0\times10^{3}\,\text{s}^{-1}\) and \(k_2=1.0\times10^{2}\,\text{s}^{-1}\), compute \(k_{\text{obs}}\) from the SSA and from the pre-equilibrium approximation, and express the discrepancy in closed form.
Solution
SSA: \(k_{\text{obs}}=k_1k_2/(k_{-1}+k_2)=(2.0\times10^{-2})(1.0\times10^{2})/(4.1\times10^{3}\,\text{s}^{-1})=2.0/(4.1\times10^{3})=4.878\times10^{-4}\,\text{s}^{-1}\). Pre-equilibrium: \(k_1k_2/k_{-1}=2.0/(4.0\times10^{3})=5.000\times10^{-4}\,\text{s}^{-1}\).
The ratio is \(\dfrac{k_1k_2/k_{-1}}{k_1k_2/(k_{-1}+k_2)}=1+\dfrac{k_2}{k_{-1}}=1+0.025\), so pre-equilibrium overestimates by exactly \(2.5\) percent. The approximation is good here because \(k_2/k_{-1}\) is small; at \(k_2=k_{-1}\) it would be wrong by a factor of two. Note also \(\varepsilon=k_1/(k_{-1}+k_2)=4.9\times10^{-6}\), so the SSA itself is excellent.
For the sequential mechanism with \(k_1=1.0\times10^{-2}\,\text{s}^{-1}\) and \(k_2=1.0\,\text{s}^{-1}\), find the earliest time at which the SSA reproduces the exact \([\text{I}]\) to within one percent, and compare it with the reactant timescale.
Solution
Measure the error of the approximation against the truth. Inverting Step 9, \([\text{I}]_{ss}/[\text{I}]_{\text{exact}}=(1-\varepsilon)/(1-e^{-(k_2-k_1)t})=0.99/(1-e^{-0.99t})\) with \(\varepsilon=k_1/k_2=0.010\). Because \(1-e^{-0.99t}<1\), this ratio never falls below \(1-\varepsilon=0.99\): the SSA is never low by more than one percent, so only the upper edge of the band can be breached. The condition is \(0.99/(1-e^{-0.99t})\le1.01\), i.e. \(e^{-0.99t}\le1-0.99/1.01=0.0198\).
Hence \(0.99\,t\ge\ln(1/0.0198)=3.92\) and \(t\ge3.96\,\text{s}\), about four intermediate lifetimes \(1/k_2=1.0\,\text{s}\). The agreement then holds ever after: the ratio passes through unity at \(t=\ln(100)/0.99=4.65\,\text{s}\) and settles at the residual \(0.99\), the persistent \(O(\varepsilon)\) error of Step 9, so beyond \(3.96\,\text{s}\) the SSA is never worse than one percent — nor, at any time, better. Against the reactant timescale \(1/k_1=100\,\text{s}\) this induction period is \(4\) percent of the run: negligible for a rate law fitted over several half-lives, but fatal for a stopped-flow experiment resolving the first second.
Ozone decomposes as \(2\,\text{O}_3\to3\,\text{O}_2\) by \(\text{O}_3+\text{M}\rightleftharpoons\text{O}_2+\text{O}+\text{M}\) (\(k_1,k_{-1}\)) and \(\text{O}+\text{O}_3\overset{k_2}{\to}2\,\text{O}_2\). Derive the SSA rate law, take the limit in which deactivation dominates, and evaluate the rate for \([\text{O}_3]=1.0\times10^{-6}\), \([\text{O}_2]=1.0\times10^{-3}\,\text{mol}\,\text{dm}^{-3}\) and a fitted \(2k_1k_2/k_{-1}=1.0\times10^{2}\,\text{s}^{-1}\).
Solution
Oxygen atoms are formed by the forward first step and lost by its reverse and by the second step: \(d[\text{O}]/dt=k_1[\text{O}_3][\text{M}]-k_{-1}[\text{O}_2][\text{O}][\text{M}]-k_2[\text{O}][\text{O}_3]\approx0\), so \([\text{O}]_{ss}=k_1[\text{O}_3][\text{M}]/(k_{-1}[\text{O}_2][\text{M}]+k_2[\text{O}_3])\). Ozone is consumed by the forward first step and the second, and regenerated by the reverse; using the steady-state condition to replace the first two terms, \(-d[\text{O}_3]/dt=2k_2[\text{O}]_{ss}[\text{O}_3]=\dfrac{2k_1k_2[\text{O}_3]^2[\text{M}]}{k_{-1}[\text{O}_2][\text{M}]+k_2[\text{O}_3]}\).
When \(k_{-1}[\text{O}_2][\text{M}]\gg k_2[\text{O}_3]\) the collision partner cancels top and bottom, leaving \(-d[\text{O}_3]/dt=(2k_1k_2/k_{-1})[\text{O}_3]^2/[\text{O}_2]\): second order in ozone, inhibited by its own product, and independent of total pressure — a rate law no one would guess from the stoichiometry, and the form actually observed. The composite constant has units of \(\text{s}^{-1}\), since \(k_1\) and \(k_2\) are bimolecular and \(k_{-1}\) is termolecular: \((\text{dm}^{3}\,\text{mol}^{-1}\text{s}^{-1})^{2}/(\text{dm}^{6}\,\text{mol}^{-2}\text{s}^{-1})=\text{s}^{-1}\), matching \([\text{O}_3]^2/[\text{O}_2]\) being a concentration. Numerically \((1.0\times10^{2})(1.0\times10^{-6})^2/(1.0\times10^{-3})=(1.0\times10^{2})(1.0\times10^{-12})/(1.0\times10^{-3})=1.0\times10^{-7}\,\text{mol}\,\text{dm}^{-3}\,\text{s}^{-1}\). Doubling \([\text{O}_2]\) halves the rate; halving \([\text{O}_3]\) quarters it.
For the chain \(\text{Br}_2+\text{M}\overset{k_1}{\to}2\,\text{Br}\cdot+\text{M}\), \(\text{Br}\cdot+\text{H}_2\overset{k_2}{\to}\text{HBr}+\text{H}\cdot\), \(\text{H}\cdot+\text{Br}_2\overset{k_3}{\to}\text{HBr}+\text{Br}\cdot\), \(\text{H}\cdot+\text{HBr}\overset{k_4}{\to}\text{H}_2+\text{Br}\cdot\), \(2\,\text{Br}\cdot+\text{M}\overset{k_5}{\to}\text{Br}_2+\text{M}\), derive the rate of formation of \(\text{HBr}\). Then, taking the fitted ratio \(k_4/k_3=0.12\), find the fractional reduction in rate once \([\text{HBr}]=[\text{Br}_2]\).
Solution
Apply the SSA to both radicals. Adding the two steady-state equations makes the propagation terms cancel in pairs and leaves initiation equal to termination: \(2k_1[\text{Br}_2][\text{M}]=2k_5[\text{Br}\cdot]^2[\text{M}]\), so \([\text{Br}\cdot]_{ss}=(k_1/k_5)^{1/2}[\text{Br}_2]^{1/2}\) — the collision partner cancels, and the square root comes from the second-order termination. From the hydrogen-atom equation, \([\text{H}\cdot]_{ss}=k_2[\text{Br}\cdot]_{ss}[\text{H}_2]/(k_3[\text{Br}_2]+k_4[\text{HBr}])\).
HBr is made in steps 2 and 3 and destroyed in step 4: \(d[\text{HBr}]/dt=k_2[\text{Br}\cdot][\text{H}_2]+[\text{H}\cdot](k_3[\text{Br}_2]-k_4[\text{HBr}])\). Substituting \([\text{H}\cdot]_{ss}\) and combining over the common denominator gives \(k_2[\text{Br}\cdot][\text{H}_2]\left(1+\dfrac{k_3[\text{Br}_2]-k_4[\text{HBr}]}{k_3[\text{Br}_2]+k_4[\text{HBr}]}\right)=\dfrac{2k_2[\text{Br}\cdot][\text{H}_2]k_3[\text{Br}_2]}{k_3[\text{Br}_2]+k_4[\text{HBr}]}\), that is
\[ \frac{d[\text{HBr}]}{dt}=\frac{2k_2(k_1/k_5)^{1/2}[\text{H}_2][\text{Br}_2]^{1/2}}{1+(k_4/k_3)[\text{HBr}]/[\text{Br}_2]} \]
which is the empirical Bodenstein–Lind form: first order in \(\text{H}_2\), half order in \(\text{Br}_2\) initially, and retarded by the product. With \(k_4/k_3=0.12\) and \([\text{HBr}]/[\text{Br}_2]=1\) the denominator is \(1.12\), so the rate is \(1/1.12=0.893\) of its unretarded value — a reduction of \(10.7\) percent. The retardation is competitive: HBr and \(\text{Br}_2\) compete for the same hydrogen atoms.