Chemistry · Lesson

Reaction networks and kinetic modelling

Write balances for a declared reaction network before eliminating intermediates. Use conservation and positivity checks while keeping equilibrium, a driven steady state and mechanism identifiability distinct.

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By the end, you can

Calculate an accumulating intermediate in a two-step first-order network.
Separate reaction-event rate from species depletion.
Compare equilibrium composition, relaxation and driven steady throughput.

A network supplies several linked species equations

For an invented irreversible A→I→P network with first-order coefficients k₁ and k₂, write d[A]/dt=−k₁[A], d[I]/dt=k₁[A]−k₂[I] and d[P]/dt=k₂[I]. Assume a well-mixed uniform fixed-volume system, fixed conditions and the same conserved constituent amount in A,I and P. Then [A]+[I]+[P] is constant.

Overall stoichiometry alone does not determine every empirical rate exponent. Elementary-step assumptions and measured information are separate inputs. A model should preserve its conserved constituents and nonnegative concentrations. A fitted concentration curve can agree with more than one mechanism, so fit quality alone does not prove a unique sequence.

An intermediate can first accumulate and then disappear

For [A]₀=1.00 mol L⁻¹, [I]₀=[P]₀=0 and stipulated k₁=0.200 s⁻¹, k₂=0.500 s⁻¹, the exact model gives [A]=[A]₀exp(−k₁t), [I]=[A]₀k₁[exp(−k₁t)−exp(−k₂t)]/(k₂−k₁), and [P]=[A]₀−[A]−[I]. No quasi-steady intermediate approximation is used.

The intermediate peaks at t=ln(k₂/k₁)/(k₂−k₁)≈ 3.0543 s, when its formation and removal rates match. At 5.00 s, [A]≈ 0.36788, [I]≈ 0.19053 and [P]≈ 0.44159 mol L⁻¹. Equal coefficients require the appropriate limit [I]=[A]₀kt exp(−kt), rather than division by zero.

An intermediate accumulates without breaking the conserved balance

Concentration / mol L⁻¹

Elapsed time / s

Starting species A
Intermediate I
Final species P
— Exact intermediate concentration

Underlying data

Elapsed time / s	Concentration / mol L⁻¹
0	1 · Starting species A
1	0.8187307530779818 · Starting species A
2	0.6703200460356393 · Starting species A
5	0.36787944117144233 · Starting species A
10	0.1353352832366127 · Starting species A
20	0.01831563888873418 · Starting species A
0	0 · Intermediate I
1	0.1414667289102323 · Intermediate I
2	0.2016270699094647 · Intermediate I
5	0.19052962836502904 · Intermediate I
10	0.08573155749168483 · Intermediate I
20	0.01218015930598113 · Intermediate I
0	0 · Final species P
1	0.03980251801178589 · Final species P
2	0.12805288405489598 · Final species P
5	0.4415909304635286 · Final species P
10	0.7789331592717025 · Final species P
20	0.9695042018052846 · Final species P

Original uniform fixed-volume A→I→P model, [A]₀=1.00 mol L⁻¹, other initial amounts 0, k₁=0.200 s⁻¹ and k₂=0.500 s⁻¹. Selected markers show all three concentrations; the dashed curve draws only the intermediate I. One conserved constituent unit is retained in each species, so [A]+[I]+[P]=1.00 mol L⁻¹. No real reaction data or quasi-steady elimination is supplied.

Stoichiometry turns one event rate into several species rates

For a different stipulated elementary step 2A→B, define event rate v=k[A]². Then d[A]/dt=−2v and d[B]/dt=v. The conserved unit balance is [A]+2[B], rather than [A]+[B]. A coefficient defined directly for A depletion would differ by the factor 2.

With k=3.00 L mol⁻¹ s⁻¹ and current [A]=0.100 mol L⁻¹, v=0.0300 mol L⁻¹ s⁻¹ while A disappears at 0.0600 mol L⁻¹ s⁻¹. A deterministic concentration law is also not automatically a finite-particle stochastic propensity; identical-pair counting needs its own convention.

Stationarity can arise for different reasons

For a closed ideal elementary A⇌B model with compatible forward 0.300 s⁻¹ and reverse 0.100 s⁻¹ coefficients and total 1.00 mol L⁻¹, equilibrium has [B]=0.750 and [A]=0.250 mol L⁻¹. The gross forward and reverse rates both equal 0.0750 mol L⁻¹ s⁻¹. Scaling both coefficients by 10 preserves that equilibrium ratio but makes relaxation 10 times faster.

A separate open compartment has d[I]/dt=F−k_out[I], with maintained feed F=0.100 mol L⁻¹ s⁻¹ and removal coefficient 0.200 s⁻¹. Its steady concentration is 0.500 mol L⁻¹ with continuing throughput 0.100 mol L⁻¹ s⁻¹. Local perturbations decay here because the derivative is−k_out. Zero accumulation alone establishes neither equilibrium nor stability for every network.

Worked examples

Check the exact three-species model at one time

Use the specified A→I→P model and t=5.00 s. Find all three concentrations and check their sum.

[A]=exp(−0.200× 5.00)≈ 0.367879 mol L⁻¹.
[I]=(0.200/0.300)[exp(−1)−exp(−2.5)]≈ 0.190530 mol L⁻¹.
[P]=1.00−[A]−[I]≈ 0.441591 mol L⁻¹.
Use the unrounded values to check [A]+[I]+[P]=1.00 mol L⁻¹.

At 5.00 s, approximately 0.368 M A,0.191 M I and 0.442 M P. Their independently rounded sum can differ slightly from the exact conserved value.

Find throughput without calling it equilibrium

For the separate open compartment, F=0.100 mol L⁻¹ s⁻¹ and k_out=0.200 s⁻¹. Find [I]_steady and removal throughput.

Set accumulation to zero: F−k_out[I]=0.
[I]_steady=F/k_out=0.500 mol L⁻¹.
Removal throughput=k_out[I]_steady=0.100 mol L⁻¹ s⁻¹.
A small concentration perturbation obeys d(δ[I])/dt=−k_outδ[I], so it decays in this model.

The concentration is stationary at 0.500 mol L⁻¹, with feed and removal both 0.100 mol L⁻¹ s⁻¹. This driven throughput is not thermodynamic equilibrium.

Try it yourself

For event rate v=0.0300 mol L⁻¹ s⁻¹ in 2A→B, what is A’s positive disappearance rate?
Show the answer and reasoning
0.0600 mol L⁻¹ s⁻¹.
Each event consumes two A units.
Which combination is conserved by the stated 2A→B step at fixed volume?
Show the answer and reasoning
[A]+2[B].
Each B contains the two conserved A units.
If both compatible A⇌B coefficients are multiplied by 10, does the equilibrium ratio change?
Show the answer and reasoning
No.
Their ratio stays fixed while the relaxation rate scales.
Does one matching concentration fit prove a unique mechanism?
Show the answer and reasoning
No.
Other networks can produce compatible observables; additional evidence is needed.

What to carry forward

Write species balances and conserved quantities first.
Use an explicit approximation before eliminating an intermediate.
Keep event rate distinct from each species rate.
Stationary concentration can coexist with continuing driven flux.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.