Combustion of hypergolics — N₂O₄ + UDMH - MMH; ignition delay: Worked examples

Definition The symbols we reuse (all from the parent) plus two book-keeping symbols

$τ_{i g n}$ (read "tau", the Greek t) = ignition delay, a time in seconds. Picture a stopwatch that starts the instant fuel and oxidiser touch and stops when the flame appears.
$T$ = absolute temperature in kelvin (K). Kelvin never goes negative; $0 K$ is the coldest possible. Room temperature $\approx 300 K$ .
$E_{a}$ = activation energy in J/mol — the energy "hill" the pre-ignition reaction must climb before it runs away. See Arrhenius equation.
$R = 8.314 J mol^{- 1} K^{- 1}$ = the gas constant, a fixed conversion number.
$C$ = a constant (units of time) that soaks up everything not depending on $T$ .
$O / F$ = oxidiser-to-fuel mass ratio — kilograms of oxidiser burned per kilogram of fuel. See Stoichiometry and limiting reagent.
$M$ = molar mass, the mass of one mole of a substance, in grams per mole (g/mol). Example: one mole of NTO weighs $92$ g, so $M (N_{2} O_{4}) = 92$ g/mol. Think of it as the "price tag" that converts a count of moles into a mass.
$n$ = amount in moles (mol), i.e. how many moles of a substance you have. It links to mass by $n = \frac{mass}{M}$ — divide grams by the price-tag $M$ to get the count.

The one master law we test everywhere: $τ_{i g n} = C e^{E_{a} / R T}, ln τ_{i g n} = ln C + \frac{E _{a}}{R} \cdot \frac{1}{T} .$ That second form is a straight line if we plot $ln τ_{i g n}$ (vertical) against $1/ T$ (horizontal). We will lean on that line all page.

The scenario matrix

Every problem this topic can throw is one of these cells. Each worked example below is tagged with the cell(s) it covers.

#	Cell class	What makes it distinct	Covered by
A	Two-point $E_{a}$ extract (normal)	Two $(τ, T)$ pairs, both finite, $T$ rising ⇒ $τ$ falling	Ex 1
B	Predict $τ$ at a new $T$	Know $E_{a}, C$ ; ask for delay at unseen temperature	Ex 2
C	Cold-start / low- $T$ limit	$T$ small ⇒ exponent huge ⇒ $τ$ blows up (the "hard start" danger)	Ex 3
D	High- $T$ / $T \to \infty$ limit	Exponent $\to 0$ ⇒ $τ \to C$ (floor set by mixing, not chemistry)	Ex 3
E	Degenerate: $E_{a} \to 0$	Delay stops depending on temperature — the line goes flat	Ex 4
F	Sign check of the slope	Is the $ln τ$ –vs– $1/ T$ slope $+$ or $-$ ? (a common flip mistake)	Ex 5 (figure)
G	Stoichiometry: mass from moles	Balanced equation → $O / F$ by mass, both fuels	Ex 6
H	Limiting reagent / off-ratio	Feed the "wrong" ratio; who runs out, how much leftover	Ex 7
I	Word problem (real engine)	Tank sizing / propellant split for a mission mass	Ex 8
J	Exam twist: change of base / units	Given data in $lo g_{10}$ or ms; watch the conversions	Ex 9

The Arrhenius line — the picture behind cells A–F

Figure — Combustion of hypergolics — N₂O₄ + UDMH - MMH; ignition delay

Look at the figure. The horizontal axis is $1/ T$ (so cold is on the right, hot on the left). The vertical axis is $ln τ_{i g n}$ . The data fall on a straight line. Its slope is $E_{a} / R$ — a positive number, so the line rises to the right (toward cold). That single geometric fact powers Examples 1–5.

Ex 1 — Cell A: extract $E_{a}$ from two points

Worked example Two operating points

A thruster shows $τ_{1} = 8 ms$ at $T_{1} = 250 K$ and $τ_{2} = 2 ms$ at $T_{2} = 300 K$ . Find $E_{a}$ .

Forecast: Will $E_{a}$ come out large (like a bond energy, hundreds of kJ/mol) or small (tens of kJ/mol)? Guess before reading — hypergolics are famous for igniting even when frigid.

Step 1. Write the line at both points and subtract. $ln τ_{1} - ln τ_{2} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}}) .$ Why this step? The unknown $ln C$ appears in both equations identically; subtracting cancels it, leaving only $E_{a}$ . This is exactly the "two points determine a line" idea from the figure.

Step 2. Numbers in. $ln \frac{8}{2} = ln 4 = 1.3863, \frac{1}{250} - \frac{1}{300} = 6.667 \times 1 0^{- 4} K^{- 1} .$ Why this step? Note the ratio $τ_{1} / τ_{2}$ is all that matters, so the "ms" units cancel — we never had to convert to seconds.

Step 3. Solve. $E_{a} = \frac{R l n 4}{6.667 \times 1 0 ^{- 4}} = \frac{8.314 \times 1.3863}{6.667 \times 1 0 ^{- 4}} \approx 1.73 \times 1 0^{4} J/mol = 17.3 kJ/mol .$ Why this step? We isolate $E_{a}$ by dividing both sides of Step 1 by the bracket $(1/ T_{1} - 1/ T_{2})$ and multiplying by $R$ — pure algebra to get the single unknown alone on one side, then substitute the two numbers from Step 2.

Verify: $17.3$ kJ/mol is small (a C–H bond is ~410 kJ/mol) — matching the forecast that hypergolics ignite easily. Units: $\frac{J mol ^{- 1} K ^{- 1}}{K ^{- 1}} = J mol^{- 1}$ . ✓

Ex 2 — Cell B: predict the delay at a new temperature

Worked example Delay at 275 K

Using $E_{a} = 17.3 kJ/mol$ and the point $τ = 2 ms$ at $300 K$ , predict $τ$ at $T = 275 K$ .

Forecast: $275 K$ is colder than $300 K$ . Should the delay go up or down? (Cold = sluggish, remember the "damp paper" analogy.)

Step 1. Anchor to the known point. $ln \frac{τ}{τ _{0}} = \frac{E _{a}}{R} (\frac{1}{T} - \frac{1}{T _{0}}), T_{0} = 300, τ_{0} = 2 ms .$ Why this step? We already know one point on the line; the line's slope $E_{a} / R$ tells us how far the new point sits.

Step 2. Plug in. $\frac{1}{275} - \frac{1}{300} = 3.0303 \times 1 0^{- 4} K^{- 1}, \frac{E _{a}}{R} = \frac{17300}{8.314} = 2081 K .$ $ln \frac{τ}{2} = 2081 \times 3.0303 \times 1 0^{- 4} = 0.6306.$ Why this step? $E_{a} / R$ is the slope in kelvin; multiplying by the horizontal shift $Δ (1/ T)$ gives the vertical rise in $ln τ$ .

Step 3. Undo the log. $τ = 2 e^{0.6306} = 2 \times 1.879 = 3.76 ms .$ Why this step? Step 2 gave us $ln (τ /2)$ , but we want $τ$ itself, not its logarithm. The exponential $e^{(\cdot)}$ is the exact inverse of $ln$ , so applying it to both sides "unwraps" the log and hands back the raw delay.

Verify: $3.76 > 2$ ms — the delay grew when we cooled down, exactly as forecast. Colder is on the right of the figure, higher up the line. ✓

Ex 3 — Cells C & D: the two temperature limits

Worked example Hard-start floor and ceiling

With the same $C$ and $E_{a}$ , describe $τ$ as (i) $T \to$ very small (cold-soaked in shadow, say 150 K) and (ii) $T \to \infty$ . First get $C$ from Ex 2's data.

Forecast: One limit gives a dangerous long delay; one gives a floor the delay can never beat. Which is which?

Step 1. Find $C$ . From $τ = C e^{E_{a} / R T}$ at $300 K$ : $C = τ e^{- E_{a} / R T} = 2 ms \times e^{- 2081/300} = 2 \times e^{- 6.937} = 2 \times 9.71 \times 1 0^{- 4} = 1.94 \times 1 0^{- 3} ms .$ Why this step? $C$ is the intercept-type constant; once known, we can evaluate any limit.

Step 2. Cold case (Cell C), $T = 150 K$ . $τ = 1.94 \times 1 0^{- 3} e^{2081/150} = 1.94 \times 1 0^{- 3} e^{13.87} = 1.94 \times 1 0^{- 3} \times 1.058 \times 1 0^{6} \approx 2053 ms \approx 2.05 s .$ Why this step? Small $T$ makes the exponent huge; $τ$ explodes. This is the physical "hard start" — propellant pools for ~2 seconds before igniting, then detonates. This is why engineers heat the chamber before firing.

Step 3. Hot limit (Cell D), $T \to \infty$ . $\frac{E _{a}}{R T} \to 0 \Rightarrow e^{E_{a} / R T} \to e^{0} = 1 \Rightarrow τ \to C = 1.94 \times 1 0^{- 3} ms .$ Why this step? Chemistry becomes instant when hot; what's left is the mixing time floor $C$ — you can never ignite faster than you can mix.

Verify: Cold delay (2050 ms) $≫$ warm delay (2 ms) $≫$ floor ( $0.0019$ ms), a monotone ladder — consistent with the single rising line in the figure (cold = far right = highest). ✓

Ex 4 — Cell E (degenerate): what if $E_{a} = 0$ ?

Ex 5 — Cell F: which way does the slope go? (figure)

Worked example Slope sign check

A student plots $ln τ$ against $1/ T$ and gets a line falling to the right. Is their $E_{a}$ positive or negative — and did they make a mistake?

Forecast: The physics says cold ⇒ long delay. What must the correct slope's sign be?

Step 1. From $ln τ = ln C + \frac{E _{a}}{R} \cdot \frac{1}{T}$ , the slope equals $E_{a} / R$ . Why this step? This is literally the "rise over run" of the line; the coefficient of $1/ T$ is the slope.

Step 2. Physically $E_{a} > 0$ (there is always a real energy barrier), so $E_{a} / R > 0$ — the line must rise to the right (toward small $T$ , large $1/ T$ ), the green line in the figure. Why this step? Cold = large $1/ T$ = far right; delay there is long = high up. A rising-right line is the only one consistent with "cold is slow."

Step 3. A line falling to the right (red, dashed) would mean $E_{a} < 0$ : colder ⇒ faster ignition. That contradicts every experiment.

Verify: Recompute the slope from Ex 1: $E_{a} / R = 17300/8.314 = 2081 K > 0$ . Positive slope ⇒ rising-right line. The student flipped an axis (likely plotted $T$ not $1/ T$ ). ✓

Ex 6 — Cell G: stoichiometric $O / F$ by mass, both fuels

Worked example Oxidiser mass ratio

Using the balanced equations from the parent note, find the stoichiometric $O / F$ (mass) for UDMH and for MMH.

Forecast: NTO is heavy ( $M = 92$ g/mol). Do you expect $O / F$ near 1, near 2.5, or near 5?

Step 1. UDMH: $C_{2} H_{8} N_{2} + 2 N_{2} O_{4} \to \dots$ $M (UDMH) = 2 (12) + 8 (1) + 2 (14) = 60 g/mol, M (N_{2} O_{4}) = 2 (14) + 4 (16) = 92.$ Why this step? Coefficients give moles $n$ ; molar masses $M$ convert moles to mass via mass $= n M$ . See Bond enthalpy and reaction energetics for where these atoms come from.

Step 2. $(\frac{O}{F})_{UDMH} = \frac{2 \times 92}{1 \times 60} = \frac{184}{60} = 3.07.$ Why this step? $O / F$ by mass is goal = (mass of oxidiser)/(mass of fuel). We take the balanced coefficients (2 mol NTO per 1 mol UDMH), multiply each by its molar mass $M$ from Step 1 to turn moles into grams, and divide oxidiser-grams by fuel-grams — that quotient is exactly the ratio the engine's plumbing must deliver.

Step 3. MMH: $4 CH_{6} N_{2} + 5 N_{2} O_{4} \to \dots$ , $M (MMH) = 12 + 6 + 28 = 46$ . $(\frac{O}{F})_{MMH} = \frac{5 \times 92}{4 \times 46} = \frac{460}{184} = 2.5.$ Why this step? Same recipe; different balanced numbers. This reproduces the parent's $2.5$ for MMH.

Verify: Both land between 2 and 3.1 (forecast: ~2.5 range ✓). UDMH needs more oxidiser per kg because it carries more C and H to burn per mole. ✓

Ex 7 — Cell H: limiting reagent when the ratio is off

Worked example Feeding too little oxidiser

An engine is fed $1.0 kg MMH$ with only $2.0 kg NTO$ (below the stoichiometric $2.5$ ). Which runs out, and how much fuel is left unburnt?

Forecast: With less oxidiser than stoichiometry, the engine is fuel-rich. Which reagent limits the burn?

Step 1. Amounts (moles $n$ ) fed, using $n = mass / M$ : $n (MMH) = \frac{1000}{46} = 21.7 mol, n (NTO) = \frac{2000}{92} = 21.7 mol .$ Why this step? Stoichiometry and limiting reagent compares moles against the recipe, not raw masses; dividing each mass by its molar mass $M$ gives the mole count $n$ .

Step 2. Recipe ratio $NTO:MMH = 5 : 4 = 1.25$ . We have $21.7 : 21.7 = 1.00 < 1.25$ NTO per MMH. $NTO needed to burn all MMH = 21.7 \times \frac{5}{4} = 27.2 mol > 21.7 available .$ Why this step? Compare "needed" vs "available" — NTO falls short, so NTO is the limiting reagent.

Step 3. MMH consumed $= 21.7 \times \frac{4}{5} = 17.4 mol$ ; leftover MMH $= 21.7 - 17.4 = 4.34 mol$ . $leftover mass = n M = 4.34 \times 46 = 200 g = 0.20 kg .$ Why this step? Because NTO is the limiting reagent (Step 2), it dictates how much MMH can react: the recipe burns 4 mol MMH for every 5 mol NTO, so the $21.7$ mol NTO consumes only $21.7 \times \frac{4}{5}$ mol MMH. Whatever MMH the NTO could not pair with simply survives, so we subtract consumed from fed and convert those leftover moles $n$ back to grams with $M (MMH) = 46$ .

Verify: $0.20 kg$ MMH unburnt out of $1.0 kg$ fed = $20%$ leftover. Cross-check: burnt MMH $= 0.80 kg$ needs $0.80 \times 2.5 = 2.0 kg$ NTO — exactly what we supplied. ✓ Fuel-rich, as forecast.

Ex 8 — Cell I: mission word problem

Worked example Sizing a lander's tanks

A Mars-lander stage must burn a total propellant of 700 kg using MMH/NTO at the stoichiometric $O / F = 2.5$ . How many kg of each go in the tanks?

Forecast: Since oxidiser is 2.5× the fuel, will the tanks be roughly equal, or the oxidiser tank much bigger?

Step 1. Let fuel mass $= f$ , oxidiser mass $= 2.5 f$ . Total: $f + 2.5 f = 700 \Rightarrow 3.5 f = 700 \Rightarrow f = 200 kg .$ Why this step? The $O / F$ ratio pins oxidiser mass to fuel mass; one unknown, one equation.

Step 2. Oxidiser $= 2.5 \times 200 = 500 kg$ . Why this step? We defined oxidiser as $2.5 f$ in Step 1 precisely because $O / F = 2.5$ means oxidiser mass is $2.5$ times fuel mass; now that Step 1 has fixed $f = 200$ kg, we just substitute it back into that ratio expression to read off the oxidiser mass.

Verify: $200 + 500 = 700$ ✓ and $500/200 = 2.5$ ✓. The NTO tank is much bigger — a real geometric constraint on lander design (relevant to Specific impulse and rocket performance via total impulse). Forecast confirmed: not equal tanks.

Ex 9 — Cell J: exam twist (log base and units)

Worked example Data given in

$lo g_{10}$ and milliseconds An exam gives the correlation $lo g_{10} τ = 0.20 + \frac{910}{T}$ with $τ$ in milliseconds (ms). Find $E_{a}$ (in kJ/mol) and $τ$ at $T = 300 K$ .

Forecast: The "910" is a slope, but in $lo g_{10}$ , not $ln$ . Do you use it directly for $E_{a} / R$ , or must you multiply by something?

Step 1. Convert base. Since $ln x = 2.303 lo g_{10} x$ : $ln τ = 2.303 (0.20 + \frac{910}{T}) = 0.4606 + \frac{2095.7}{T} .$ Why this step? Our master law $ln τ = ln C + (E_{a} / R) (1/ T)$ uses natural log; the exam used base 10. The slope must be rescaled by $ln 10 = 2.303$ before it means $E_{a} / R$ .

Step 2. Match to $ln τ = ln C + \frac{E _{a}}{R} \cdot \frac{1}{T}$ : $\frac{E _{a}}{R} = 2095.7 K \Rightarrow E_{a} = 2095.7 \times 8.314 = 17420 J/mol \approx 17.4 kJ/mol .$ Why this step? The coefficient of $1/ T$ is $E_{a} / R$ ; solve for $E_{a}$ by multiplying that slope by $R$ .

Step 3. Delay at 300 K, straight from the given equation: $lo g_{10} τ = 0.20 + \frac{910}{300} = 0.20 + 3.033 = 3.233 \Rightarrow τ = 1 0^{3.233} = 1710 ms \approx 1.71 s .$ Why this step? For $τ$ itself no base conversion is needed — the original equation is already in $lo g_{10}$ , so we plug in $T = 300$ , then invert the base-10 log by raising 10 to the power $3.233$ . Keeping the given ms unit, we finally convert to seconds.

Verify: $E_{a} \approx 17.4$ kJ/mol matches the ~17 kJ/mol scale of Ex 1 (same chemistry family) ✓. And $1710 ms = 1.71$ s is a long, cold-start-like delay — this fictitious data set describes a chilly engine. Units of $τ$ stayed ms → s throughout. ✓

Recall

Recall Cover the answers

Cold engine: does $τ$ grow or shrink, and which cell? ::: Grows (Cell C); low $T$ ⇒ huge exponent ⇒ long delay (hard start).
As $T \to \infty$ , $τ \to$ ? ::: $τ \to C$ , the mixing-time floor (Cell D).
Slope of $ln τ$ vs $1/ T$ ? ::: $E_{a} / R$ , positive ⇒ line rises to the right (Cell F).
How do you turn a $lo g_{10}$ slope into $E_{a} / R$ ? ::: Multiply the slope by $ln 10 = 2.303$ (Cell J).
MMH stoichiometric $O / F$ by mass? ::: $2.5$ (Cell G).

Parent: main topic note · see also Arrhenius equation, Redox in combustion, Adiabatic flame temperature.