Pollutants — NOₓ, soot, unburned hydrocarbons: Worked examples

This page is the "get your hands dirty" companion to Pollutants — NOₓ, soot, unburned hydrocarbons (index 5.3.9). We will not introduce new theory — we will stress-test the master result

$\frac{d [ NO ]}{d t} = 2 A [N_{2}] [O] e^{- E_{a} / R T}$

against every kind of number a problem can throw at you: high and low temperature, lean and rich mixtures, zero-input degenerate cases, and limiting behaviour. Before any example, we build a map of all the cases so you can see that nothing is missing.

Intuition Read this first — what all the symbols mean

Everything below is built from these plain ideas. Nothing else is assumed.

$T$ = temperature in kelvin (K). Absolute zero is $0$ K; a hot flame is around $2000$ K. Think of it as "how violently the molecules are jiggling."
$[X]$ = concentration of species X — literally "how many moles of X sit in one cubic metre," in units of $mol m^{- 3}$ . So $[N_{2}]$ is how crowded the nitrogen is, $[O]$ how crowded the oxygen atoms are, $[NO]$ how much nitric oxide has built up. The square brackets are just shorthand for "concentration of."
$\frac{d [ NO ]}{d t}$ = the rate at which $[NO]$ grows, in $mol m^{- 3} s^{- 1}$ — "how fast NO piles up each second."
$E_{a}$ = activation energy — the minimum energy a collision must carry to make a reaction happen, in $J mol^{- 1}$ . Big $E_{a}$ = a very picky, hard-to-trigger reaction. (Built in Arrhenius Equation and Activation Energy.)
$A$ = the pre-exponential (frequency) factor — the rate the reaction would run if every collision were energetic enough. It bundles how often molecules collide and how well aligned they must be. For a bimolecular gas reaction its units are $m^{3} mol^{- 1} s^{- 1}$ , chosen precisely so the whole rate law comes out in $mol m^{- 3} s^{- 1}$ . Think of $A$ as "the ceiling" and $e^{- E_{a} / R T}$ as "the fraction of that ceiling you actually reach at temperature $T$ ."
$R = 8.314 J mol^{- 1} K^{- 1}$ = the gas constant, the fixed number that converts "energy per mole" into "per degree of temperature."
$φ$ (phi) = equivalence ratio: fuel-to-air compared to the perfect (stoichiometric) ratio. $φ < 1$ = lean (extra air), $φ = 1$ = exact, $φ > 1$ = rich (extra fuel). (Built in Equivalence Ratio and Flammability Limits.)

The single tool doing all the heavy lifting is $e^{- E_{a} / R T}$ — the Arrhenius exponential. We use this function and not a straight line because the fraction of collisions energetic enough to react grows explosively as $T$ rises, and only an exponential captures "explosive."

The scenario matrix

Every problem this topic can ask lives in one of these cells. The examples below are labelled with the cell they cover, and together they fill the whole table.

#	Case class	What is special about it	Covered by
A	High-T NOₓ ratio	Two hot temperatures, find the fold-change	Ex 1
B	Modest-T-cut design	"Small" 150 K drop, big NOₓ effect	Ex 2
C	Degenerate: $T \to$ low	Limiting behaviour, exponent $\to$ huge negative	Ex 3
D	Zero input	$[O] = 0$ or $[N_{2}] = 0$ → rate is exactly zero	Ex 4
E	Lean side ( $φ < 1$ )	High T + plenty O → NOₓ peak; CO low	Ex 5
F	Rich side ( $φ > 1$ )	Soot window; NOₓ suppressed, CO high	Ex 6
G	Quench / cold wall	CO frozen, UHC escapes — a real-world word problem	Ex 7
H	Exam twist	"Doubling every 70 K" claim — derive the K needed	Ex 8

Two constants are reused throughout, so compute them once:

Cell A — high-temperature NOₓ fold-change

Worked example Ex 1 — hot combustor: 2100 K vs 1900 K

A gas-turbine hot streak runs at $2100$ K; a design fix lowers it to $1900$ K. By what factor does the thermal-NOₓ rate change? (Assume $[N_{2}], [O]$ roughly unchanged.)

Forecast: guess the factor before reading on. Most people say "about 20% less." The exponential will surprise you — hold that guess.

Step 1. Since $[N_{2}], [O], A$ (and the factor 2) cancel in a ratio, only the exponentials survive: $ratio = \frac{e ^{- E_{a} / (R \cdot 1900)}}{e ^{- E_{a} / (R \cdot 2100)}} = exp [- \frac{E _{a}}{R} (\frac{1}{1900} - \frac{1}{2100})] .$ Why this step? Taking a ratio kills every messy constant — a standard trick so we only chase the part that actually matters, the temperature term.

Step 2. The bracket: $\frac{1}{1900} - \frac{1}{2100} = 5.263 \times 1 0^{- 4} - 4.762 \times 1 0^{- 4} = 5.01 \times 1 0^{- 5} K^{- 1}$ . Why this step? We are cooling, so $1/ T$ gets bigger (smaller $T$ → bigger $1/ T$ ). The bracket is positive.

Step 3. Multiply: $38369 \times 5.01 \times 1 0^{- 5} = 1.923$ . With the leading minus, exponent $= - 1.923$ . Why this step? Cooler = slower, so the factor must be less than 1; the minus sign enforces that.

Step 4. $e^{- 1.923} = 0.146$ . Why this step? The exponent by itself is just a number; raising $e$ to it converts "how far below the reference exponent we are" back into an actual ratio of rates — that final $0.146$ is the answer we can report.

Verify: the rate falls to $\approx 0.15$ of the original — a ~7× reduction from a 200 K cut. Sanity check: this exceeds the parent note's 5× for a 2200→2000 K cut, which makes sense because at lower absolute temperature the same 200 K is a larger fractional change in $1/ T$ . Your "20% less" forecast was off by an order of magnitude — that is the exponential's whole point.

Cell B — a "small" temperature cut with a big payoff

Worked example Ex 2 — is a 150 K reduction worth the engineering?

Lean-premixed staging (see Lean Premixed Combustion & Staging) can shave peak flame T from $2000$ K to $1850$ K. Management asks: "Only 150 K — is it worth it?" Quantify the NOₓ benefit.

Forecast: 150 K out of 2000 K is 7.5%. Do you expect a ~7.5% NOₓ cut, or much more?

Step 1. Ratio $= exp [- \frac{E _{a}}{R} (\frac{1}{1850} - \frac{1}{2000})]$ . Why this step? Same ratio trick as Ex 1 — only the exponential matters.

Step 2. Bracket $= 5.405 \times 1 0^{- 4} - 5.000 \times 1 0^{- 4} = 4.054 \times 1 0^{- 5} K^{- 1}$ . Why this step? Confirms sign: cooling raises $1/ T$ , bracket positive.

Step 3. $38369 \times 4.054 \times 1 0^{- 5} = 1.555$ ; exponent $= - 1.555$ ; $e^{- 1.555} = 0.211$ . Why this step? We multiply $E_{a} / R$ by the bracket to get the exponent, then raise $e$ to it to turn that exponent back into a physical ratio of rates — $0.211$ is the fraction of the original NOₓ rate that survives the cooling.

Verify: NOₓ drops to $\approx 0.21$ — nearly a 5× reduction from a "mere" 7.5% temperature cut. Units check: the bracket is $K^{- 1}$ , times $E_{a} / R$ in K, gives a pure number — good. Answer to management: absolutely worth it. This is the quantitative heart of Adiabatic Flame Temperature control for emissions.

Cell C — limiting / degenerate: temperature goes low

Worked example Ex 3 — what does the formula predict as the flame cools toward the walls?

As gas approaches a cold wall, $T$ drops from $2000$ K toward, say, $1000$ K, then $600$ K. Track the thermal-NOₓ rate factor relative to $2000$ K.

Forecast: does NOₓ formation slow gently, or shut off like a switch?

Step 1. Factor at $T$ vs $2000$ K: $exp [- \frac{E _{a}}{R} (\frac{1}{T} - \frac{1}{2000})]$ . Why this step? Reusing the ratio form lets us just plug several $T$ values.

Step 2 — $T = 1000$ K. Bracket $= 1.000 \times 1 0^{- 3} - 5.000 \times 1 0^{- 4} = 5.000 \times 1 0^{- 4}$ . Product $= 38369 \times 5.0 \times 1 0^{- 4} = 19.18$ . Factor $= e^{- 19.18} = 4.7 \times 1 0^{- 9}$ . Why this step? Halving $T$ doubles $1/ T$ , which throws the exponent to nearly $- 20$ — the exponential collapses.

Step 3 — $T = 600$ K. Bracket $= 1.6667 \times 1 0^{- 3} - 5.0 \times 1 0^{- 4} = 1.1667 \times 1 0^{- 3}$ . Product $= 44.76$ . Factor $= e^{- 44.76} \approx 3.3 \times 1 0^{- 20}$ . Why this step? Confirms the trend: colder still → utterly negligible.

Verify: as $T \to$ low, $1/ T \to \infty$ , so $- E_{a} / (R T) \to - \infty$ , so $e^{- E_{a} / R T} \to 0$ . Thermal NOₓ switches off below ~1500 K — it does not fade linearly, it vanishes. This is why the Quenching and Wall Heat Transfer region makes essentially zero thermal NOₓ (but, as Ex 7 shows, it does trap CO/UHC).

Cell D — zero input (multiplicative degeneracy)

Worked example Ex 4 — pure fuel jet with no air entrained yet, and a pure inert stream

(a) In the very first instant of a fuel jet before any air mixes in, $[O] = 0$ . (b) In a stream of pure argon at $2200$ K, $[N_{2}] = 0$ . What is the thermal-NOₓ rate in each?

Forecast: temperature is high in both — does that guarantee NOₓ?

Step 1 — case (a). The rate is $2 A [N_{2}] [O] e^{- E_{a} / R T}$ . With $[O] = 0$ , the whole product is $0 \times (anything) = 0$ . Why this step? The rate is a product; a single zero factor forces the entire rate to zero, no matter how hot.

Step 2 — case (b). With $[N_{2}] = 0$ , again a zero factor → rate $= 0$ . Why this step? Same logic — no nitrogen means there is nothing to oxidise.

Verify: heat alone is necessary but not sufficient. You need all three ingredients — hot and oxygen atoms and nitrogen — simultaneously. This is exactly why NOₓ peaks near stoichiometric (both O and N₂ available) rather than in the hottest pure-fuel core.

Cell E — lean side, $φ < 1$ (the NOₓ danger zone)

The figure below plots the three pollutant levels against the equivalence ratio $φ$ . Horizontal axis: $φ$ from lean ( $0.5$ ) to rich ( $2.0$ ), with the stoichiometric line $φ = 1$ dashed. Vertical axis: relative pollutant level (0 to ~1.2, arbitrary units, each curve scaled to its own peak). Three curves: NOₓ (teal) peaks just left of $φ = 1$ ; CO/UHC (orange) is U-shaped, high at both lean and rich extremes, minimum near $φ \approx 0.95$ ; soot (plum) is zero until $φ > 1$ then climbs. Dotted markers flag $φ = 0.9$ (Ex 5) and $φ = 1.6$ (Ex 6).

Figure — Pollutants — NOₓ, soot, unburned hydrocarbons

Worked example Ex 5 — reading the trade-off curve at

$φ = 0.9$ A premixed flame runs slightly lean, $φ = 0.9$ . Using the master trade-off figure above, rank NOₓ, CO, and soot from highest to lowest and justify each with the chemistry.

Forecast: which single pollutant is worst here?

Step 1. Locate $φ = 0.9$ on the figure — just left of the dashed stoichiometric line. Read the three curves: NOₓ (teal) is near its peak; CO (orange) is near its minimum; soot (plum) is essentially zero. Why this step? The whole topic reduces to one picture; the curves are the answer, we just read them.

Step 2 — why NOₓ peaks here. Slightly lean means temperature is still very high (near-complete burn) and there is spare $O_{2}$ → plenty of O atoms feeding $[O]$ in the rate law. Both factors up → NOₓ up. Why this step? Ties the curve back to $2 A [N_{2}] [O] e^{- E_{a} / R T}$ : both $[O]$ and $T$ are large.

Step 3 — why CO is low. Extra oxygen and high T let $CO + OH \to CO_{2} + H$ run to completion. See CO Oxidation and Chemical Kinetics. Why this step? Lean + hot = the ideal condition for CO burnout.

Step 4 — why soot is zero. Soot needs rich, oxygen-starved carbon. Lean means excess O everywhere → no place for carbon to clump. Why this step? Confirms soot lives only on the rich side.

Verify: ranking is NOₓ ≫ CO > soot ≈ 0. This matches the figure and is the classic "lean-and-hot makes NOₓ" lesson.

Cell F — rich side, $φ > 1$ (the soot window)

Worked example Ex 6 — a rich diffusion flame at

$φ = 1.6$ A diffusion flame has a fuel-rich pocket at $φ = 1.6$ , hot but oxygen-starved. Which pollutants dominate, and which is suppressed?

Forecast: does high temperature here mean high NOₓ, like Ex 5?

Step 1. Read the same figure above at $φ = 1.6$ (right of stoichiometric, plum dotted marker): soot (plum) is rising steeply, CO (orange) is high again, NOₓ (teal) has dropped. Why this step? Same one-picture logic — read across.

Step 2 — why soot appears. Rich + hot = pyrolysis makes $C_{2} H_{2}$ ; carbon polymerises via HACA into PAHs and particles before oxygen can reach it. This is the "hot rich pocket" soot window. Why this step? Directly the soot pathway from the parent note.

Step 3 — why CO is high. Not enough O to finish $CO \to CO_{2}$ ; carbon oxidises only partway. Why this step? Oxygen shortage stalls CO burnout.

Step 4 — why NOₓ drops despite heat. Two reasons: (i) O atoms are scarce → $[O]$ is small in the rate law $2 A [N_{2}] [O] e^{- E_{a} / R T}$ , and a small factor pulls the whole product down even though $e^{- E_{a} / R T}$ stays large; (ii) fuel radicals compete for and scavenge the few O atoms present. So high $T$ alone cannot make NOₓ without O — the exact Ex 4 lesson, now on the rich side. Why this step? Reinforces that NOₓ needs the product of "hot" and "oxygen present"; kill either factor and NOₓ falls, which is why the hottest pocket is not the dirtiest for NOₓ.

Verify: ranking is soot ≈ CO ≫ NOₓ. Complements Ex 5: lean and rich sides trade NOₓ for soot/CO — the master dilemma.

Cell G — quench / cold-wall word problem

Worked example Ex 7 — engine crevice releases CO and UHC (real-world)

A spark-ignition engine emits high CO and unburned hydrocarbons even though the core burns clean and hot. Explain mechanistically, and estimate the CO-burnout slowdown if a gas parcel quenches from $1800$ K to $1200$ K.

Forecast: is the CO from the hot core or the cold edges?

Step 1. Mechanism: near cold walls and inside ring-gap crevices, heat is pulled out; a thin quench layer freezes the chemistry (see Quenching and Wall Heat Transfer). Fuel hides in crevices too narrow for the flame → UHC. CO cannot finish because $CO + OH \to CO_{2} + H$ needs hot gas and OH radicals. Why this step? Identifies the cold edges, not the hot core, as the source.

Step 2. Quantify the CO-oxidation slowdown. The $CO + OH$ step has a modest activation energy; take $E_{a}^{CO} \approx 70 kJ/mol$ (a representative kinetic value). Then $E_{a} / R = 70000/8.314 = 8420 K$ . Why this step? We reuse the Arrhenius ratio, but with the CO reaction's own smaller $E_{a}$ — different reaction, different pickiness.

Step 3. Ratio $= exp [- 8420 (\frac{1}{1200} - \frac{1}{1800})]$ . Bracket $= 8.333 \times 1 0^{- 4} - 5.556 \times 1 0^{- 4} = 2.778 \times 1 0^{- 4}$ . Product $= 2.339$ ; factor $= e^{- 2.339} = 0.0964$ . Why this step? Quantifies "frozen": cooling to 1200 K slashes CO burnout to ~10% of its 1800 K rate.

Verify: CO oxidation runs ~10× slower after quench → CO is "frozen in." Note the smaller $E_{a}$ (70 vs 319 kJ/mol) makes CO less temperature-sensitive than NOₓ — a genuinely different qualitative behaviour, which is why cooling helps NOₓ far more than it hurts CO. Real-world fix: keep exhaust hot long enough (or catalyst) to finish CO.

Cell H — exam twist: derive the "70 K doubling" claim

Worked example Ex 8 — how many kelvin doubles thermal NOₓ near 2000 K?

The parent note claims NOₓ "roughly doubles every ~70 K" near $2000$ K. Derive the exact $Δ T$ that doubles the rate at $2000$ K, from first principles.

Forecast: will the answer come out near 70 K, confirming the rule of thumb?

Step 1. Set the ratio equal to 2, going from $T = 2000$ to $T + Δ T$ (hotter → faster). In a ratio the constants $2 A [N_{2}] [O]$ cancel, leaving only the exponentials: $2 = exp [- \frac{E _{a}}{R} (\frac{1}{T + Δ T} - \frac{1}{T})] .$ Why this step? "Doubles" literally means ratio $= 2$ ; we solve for the unknown $Δ T$ .

Step 2. Take $ln$ of both sides: $ln 2 = - \frac{E _{a}}{R} (\frac{1}{T + Δ T} - \frac{1}{T}) = \frac{E _{a}}{R} \cdot \frac{Δ T}{T ( T + Δ T )}$ . Why this step? $ln$ is the tool that undoes the exponential — the only way to free $Δ T$ from the exponent.

Step 3. For small $Δ T$ , approximate $T (T + Δ T) \approx T^{2}$ : $Δ T \approx \frac{R T ^{2} l n 2}{E _{a}} = \frac{8.314 \times 200 0 ^{2} \times 0.6931}{319000} .$ Why this step? Near a point, the reaction rate looks locally exponential in $T$ ; this linearised form gives a clean formula.

Step 4. Compute: numerator $= 8.314 \times 4000000 \times 0.6931 = 2.305 \times 1 0^{7}$ ; divide by $319000$ → $Δ T \approx 72.3 K$ .

Verify: $Δ T \approx 72$ K — the rule of thumb "~70 K" is confirmed. Note $Δ T \propto T^{2}$ : at $2200$ K the doubling gap would be $72.3 \times (2200/2000)^{2} \approx 87$ K, so the rule loosens as flames get hotter — a subtle point that trips up exam candidates.

Recall Self-test before you move on

A 200 K cut from 2100→1900 K reduces NOₓ by roughly what factor? ::: About 7× (to $\sim 0.15$ ). With $[O] = 0$ but $T = 2200$ K, what is the thermal-NOₓ rate? ::: Exactly zero — a product with a zero factor is zero. On the rich side ( $φ > 1$ ), which pollutant dominates and which is suppressed? ::: Soot (and CO) dominate; NOₓ is suppressed (little O). Why is CO less temperature-sensitive than NOₓ? ::: Its $E_{a}$ (~70 kJ/mol) is far smaller than NOₓ's (~319 kJ/mol). Roughly what $Δ T$ doubles NOₓ at 2000 K, and how does it scale? ::: ~72 K, and it scales as $T^{2}$ . What does the factor of 2 in the rate law represent? ::: Each slow O+N₂ event makes two NO — one directly, one from the leftover N atom. What do the square brackets $[X]$ mean, and in what units? ::: Concentration of species X, in mol m⁻³.