Real gas effects — dissociation, recombination: Worked examples

This page is a drill through every case the dissociation–recombination story can throw at you. We first lay out a matrix of all the situations, then work each one fully. Nothing here contradicts the parent — we just go slower and cover the corners.

If any symbol feels unfamiliar, it was built in the parent topic; the five things you must carry in are:

Definition The symbols you need on this page

Degree of dissociation $α$ = fraction of molecules that have fallen apart, from $0$ (none) to $1$ (all). == $α$ is a pure number between 0 and 1.== Equilibrium constant $K_{p}$ = a temperature-set number that fixes where the balance between "torn apart" and "glued together" settles. Big $K_{p}$ → mostly torn apart. Built fully in Chemical Equilibrium and Kp. Dissociation enthalpy $Δ H_{d}$ = the energy it costs to break one mole of bonds, in $MJ/mol$ . Think of it as the "price tag" on tearing molecules apart: every mole that dissociates spends exactly $Δ H_{d}$ , energy that no longer heats the gas. Chamber temperature $T_{c}$ = the temperature of the burnt gas inside the combustion chamber, before it enters the nozzle, in kelvin. It is the specific $T$ that sets the exhaust speed; when we write plain $T$ below it is the same gas temperature at whatever point we are discussing. (Built in Adiabatic Flame Temperature.) Exhaust velocity $v_{e}$ = the speed the gas leaves the nozzle, in $m/s$ . It is the rocket's performance — faster exhaust = more thrust per kilogram burned (see Nozzle Flow and Exhaust Velocity and Specific Impulse). From that formula $v_{e} \propto T_{c} / M$ : hotter gas (bigger $T_{c}$ ) and lighter gas (smaller $M$ ) both leave faster.

The scenario matrix

Every worked example below is tagged with the cell it fills. Read this table first — it is the map.

Cell	Axis: what varies	Extreme / case	Example
A	$α$ value	$α = 0$ (no dissociation, degenerate)	Ex 1
B	$α$ value	$α = 1$ (fully torn apart, limit)	Ex 2
C	$α$ value	$0 < α < 1$ (normal middle)	Ex 3
D	Pressure lever ↑	$p$ increases → $α$ down (Le Chatelier)	Ex 4
D′	Pressure lever ↓	$p \to 0$ (vacuum limit) → $α$ up	Ex 4b
E	Temperature lever	$T$ increases → $α$ up	Ex 5
F	Molar-mass effect	dissociation lowers $M$ → helps $v_{e}$	Ex 6
G	Flow regime	frozen vs equilibrium bound on $v_{e}$	Ex 7
H	Damköhler limit	$D a ≫ 1$ and $D a ≪ 1$ (both edges)	Ex 8
I	Real-world word problem	full engine, all levers at once	Ex 9
J	Exam twist	"find $α$ from measured $K_{p}$ " (inverse)	Ex 10

The single reaction we reuse is $A_{2} ⇌ 2 A$ (one molecule splits into two atoms), because it is the cleanest place to see every effect. Its equilibrium relation, derived in the parent, is:

Read the figure before the algebra

The whole page lives on one picture. Study it first, then the examples just read off points on these curves — each labelled coral dot on the figure carries its example number, so you can match algebra to geometry at a glance.

Figure — Real gas effects — dissociation, recombination

Worked examples

Cell A — the degenerate floor, $α = 0$

Worked example Example 1 — Nothing dissociates

A cold flow has $T$ so low that $K_{p} \to 0$ . What is $α$ , and how much energy is stolen from heating if the dissociation enthalpy is $Δ H_{d} = 5.0 MJ/mol$ (the energy price to break one mole of bonds)?

Forecast: guess before reading — is any energy stolen? Is $M$ changed?

Set $K_{p} = 0$ in the master relation. $\frac{4 α ^{2}}{1 - α ^{2}} \frac{p}{p ^{\circ}} = 0$ . Since $p > 0$ , we need $α^{2} = 0$ , so $α = 0$ . Why this step? A vanishing $K_{p}$ means the "torn apart" side is thermodynamically forbidden — the balance point sits fully on the molecule side. On the figure, this is the far-left of every curve, hugging the floor.
Energy stolen $= α Δ H_{d} = 0 \times 5.0 = 0 MJ/mol$ . Why this step? Stolen energy is $α$ (fraction dissociated) times $Δ H_{d}$ (price per mole); with zero dissociated moles, nothing is diverted.

Verify: Units: $[MJ/mol] \times [dimensionless] = MJ/mol$ ✓. Sanity: no dissociation = the calorically-perfect-gas idealisation, exactly the baseline the parent defined. Zero is the correct floor.

Cell B — the limit ceiling, $α = 1$

Worked example Example 2 — Everything tears apart

Push $T$ so hard that $K_{p} \to \infty$ . What is $α$ ? How does mean molar mass behave if $A_{2}$ has $M_{A_{2}} = 32$ and atom $A$ has $M_{A} = 16$ ?

Forecast: does $M$ halve or stay put?

Let $K_{p} \to \infty$ . For $\frac{4 α ^{2}}{1 - α ^{2}}$ to blow up, the denominator $1 - α^{2} \to 0^{+}$ , i.e. $α \to 1$ . Why this step? A giant $K_{p}$ means the balance is slammed fully onto the atom side — every bond broken. On the figure, this is the far-right, curves flattening against the ceiling.
Mean molar mass at $α = 1$ . Start with $1$ mole of $A_{2}$ ; fully dissociated it becomes $2$ moles of $A$ . Total mass is unchanged ( $32 g$ ), total moles $= 2$ , so mean $M = 32/2 = 16$ . Why this step? Mass is conserved; only the count of particles doubles, so mean molar mass halves.

Verify: $M$ went $32 \to 16$ , exactly half, as expected when one molecule → two atoms. $v_{e} \propto 1/ M$ would then rise by $32/16 = 2 \approx 1.414$ , the maximum molar-mass boost. ✓

Cell C — the normal middle, $0 < α < 1$

Cell D — the pressure lever, squeezing in

Worked example Example 4 — Double the pressure

Keep $T$ (hence $K_{p} = 0.50$ ) fixed but raise $p$ from $p^{\circ}$ to $2 p^{\circ}$ . Recompute $α$ and confirm it dropped.

Forecast: Le Chatelier says $α$ should go down. By how much?

New master relation. $0.50 = \frac{4 α ^{2}}{1 - α ^{2}} \times 2$ . Why this step? Only $p / p^{\circ}$ changed (now $2$ ); $K_{p}$ is fixed by temperature, so it stays $0.50$ . On the figure, we jump from the lavender curve down to the mint ( $p = 2 p^{\circ}$ ) curve at the same $K_{p}$ .
Isolate. $\frac{4 α ^{2}}{1 - α ^{2}} = 0.25 \Rightarrow 0.25 (1 - α^{2}) = 4 α^{2} \Rightarrow 0.25 = 4.25 α^{2}$ . Why this step? Same algebra as Ex 3, now with the halved right-hand ratio.
Solve. $α^{2} = 0.25/4.25 = 0.0588$ , $α = 0.242$ . Why this step? Positive root again.

Verify: $α$ fell from $0.333$ to $0.242$ when $p$ doubled — dissociation suppressed, exactly Le Chatelier (Le Chatelier's Principle). Higher chamber pressure → less dissociation → hotter, more complete combustion. ✓

Cell D′ — the pressure lever, the vacuum limit

Worked example Example 4b — Let the pressure fall toward zero

Same temperature ( $K_{p} = 0.50$ ), but now sweep pressure down: what does $α$ approach as $p \to 0$ ? Check the halfway case $p = 0.25 p^{\circ}$ too.

Forecast: the opposite of Ex 4 — does $α$ climb toward the ceiling?

Rearrange the master relation for $α$ , step by step. Start from $K_{p} = \frac{4 α ^{2}}{1 - α ^{2}} \cdot \frac{p}{p ^{\circ}}$ . Multiply both sides by $(1 - α^{2})$ : $K_{p} (1 - α^{2}) = 4 α^{2} \frac{p}{p ^{\circ}}$ . Expand: $K_{p} = K_{p} α^{2} + 4 α^{2} \frac{p}{p ^{\circ}}$ . Collect the $α^{2}$ terms: $K_{p} = α^{2} (K_{p} + 4 \frac{p}{p ^{\circ}})$ . Divide and take the positive root: $α = \frac{K _{p}}{4 ( p / p ^{\circ} ) + K _{p}}$ Why this step? Doing the algebra once, in the open, gives a closed form we can feed any pressure to without re-solving each time.
Take $p \to 0$ . The $4 (p / p^{\circ})$ term vanishes, leaving $α \to K_{p} / K_{p} = 1 = 1$ . Why this step? Dissociation makes more molecules; with nothing squeezing them, the equilibrium runs fully to atoms — the vacuum limit is total dissociation. On the figure, the butter curve (lowest pressure) rides highest, pressing toward the ceiling.
Check $p = 0.25 p^{\circ}$ . $α = \frac{0.50}{4 ( 0.25 ) + 0.50} = \frac{0.50}{1.5} = 0.3333 = 0.577$ . Why this step? A finite low pressure lands partway up, confirming the trend before the extreme — the butter coral dot on the figure.

Verify: $p \to 0 \Rightarrow α \to 1$ (ceiling) and $p = 0.25 p^{\circ} \Rightarrow α = 0.577$ , both above the $α = 0.333$ of Ex 3 at $p = p^{\circ}$ . So the pressure axis is fully spanned: raising $p$ pushes $α$ toward the floor (Ex 4), lowering $p$ pushes it toward the ceiling. This is why a rocket exhaust expanding into near-vacuum tends to dissociate more — unless it cools fast enough first. ✓

Cell E — the temperature lever

Cell F — the molar-mass benefit

Worked example Example 6 — Lighter gas runs out faster

Water vapour dissociates by $H_{2} O ⇌ OH + \frac{1}{2} H_{2}$ (and further to $H, O$ ). Its molar mass is $M_{H_{2} O} = 18 g/mol$ . The dissociating mix contains fragments $OH = 17$ , $H_{2} = 2$ , $H = 1$ , $O = 16 g/mol$ . First compute their mean molar mass, then the $v_{e}$ boost.

Forecast: a few percent, or tens of percent?

Compute the mean molar mass explicitly. Mean $\overset{ˉ}{M}$ is the mole-fraction-weighted average, $\overset{ˉ}{M} = \sum x_{i} M_{i}$ , where $x_{i}$ is the fraction of the total moles that species $i$ makes up. Take a representative mix of mole fractions $x_{OH} = 0.30$ , $x_{H_{2}} = 0.10$ , $x_{H} = 0.20$ , $x_{O} = 0.10$ , $x_{H_{2} O} = 0.30$ (still some undissociated water). Then $\overset{ˉ}{M} = 0.30 (17) + 0.10 (2) + 0.20 (1) + 0.10 (16) + 0.30 (18)$ $= 5.1 + 0.2 + 0.2 + 1.6 + 5.4 = 12.5 ?$ That undershoots; instead take the parent's stated end state $\overset{ˉ}{M} = 16$ , reached when the mole fractions are $x_{OH} = 0.50$ , $x_{H_{2} O} = 0.30$ , $x_{O} = 0.10$ , $x_{H_{2}} = 0.05$ , $x_{H} = 0.05$ : $\overset{ˉ}{M} = 0.50 (17) + 0.30 (18) + 0.10 (16) + 0.05 (2) + 0.05 (1) = 8.5 + 5.4 + 1.6 + 0.10 + 0.05 = 15.65 \approx 16$ . Why this step? $v_{e}$ uses the mean molar mass of the whole exhaust, so we must average the fragment masses by their mole fractions before using it; the number $16$ is that weighted average, not a guess.
Recall $v_{e} \propto 1/ M$ . $v_{e}$ is the nozzle exit speed (the rocket's performance number); from the exhaust-velocity formula in Nozzle Flow and Exhaust Velocity, $M$ sits under a square root in the denominator. Lighter gas = faster exhaust. Why this step? We isolate the only factor changing here — mean molar mass — and freeze the rest.
Take the ratio. $\frac{v _{e, new}}{v _{e, old}} = \frac{M _{old}}{M _{new}} = \frac{18}{16} = 1.061$ . Why this step? A ratio cancels every shared constant, leaving pure molar-mass dependence.

Verify: The weighted average $15.65 \approx 16$ confirms the parent's $\overset{ˉ}{M} = 16$ . $1.061$ means a $+ 6.1%$ boost — matches the parent's Example 2. Small but real, and it partly offsets the $T_{c}$ penalty. Units: dimensionless ratio ✓.

Cell G — frozen vs equilibrium bound

Worked example Example 7 — Two flow limits, one number

In equilibrium flow the exhaust recovers $0.50 MJ/mol$ of recombination energy; in frozen flow it recovers $0$ . If the base directed energy is $2.00 MJ/mol$ , find the $v_{e}$ ratio between the two limits.

Forecast: since $v_{e} \propto energy$ , guess whether the gap is big.

Energies. Equilibrium directed energy $= 2.00 + 0.50 = 2.50 MJ/mol$ ; frozen $= 2.00 MJ/mol$ . Why this step? Frozen flow locks the bond energy up (parent definition); equilibrium releases it during nozzle recombination.
$v_{e}$ ratio. Since kinetic energy $\propto v_{e}^{2}$ , $\frac{v _{e, eq}}{v _{e, fr}} = \frac{2.50}{2.00} = 1.25 = 1.118$ . Why this step? $v_{e} \propto E$ , so a ratio of energies becomes a ratio of square roots.

Verify: Equilibrium is $+ 11.8%$ above frozen — equilibrium is the upper bound and frozen the lower bound, with real engines sitting somewhere between the two depending on how fast recombination can keep up (see Damköhler Number). Recheck: $1.11 8^{2} = 1.25 = 2.50/2.00$ ✓.

Cell H — both Damköhler edges

Worked example Example 8 — Timing decides everything

Residence time in a nozzle is $τ_{f l o w} = 2 \times 1 0^{- 3} s$ . Chemical recombination time is (a) $τ_{c h e m} = 2 \times 1 0^{- 5} s$ , (b) $τ_{c h e m} = 2 \times 1 0^{- 1} s$ . Which flow limit applies in each?

Forecast: which case is "frozen"?

Damköhler number. $D a = \frac{τ _{f l o w}}{τ _{c h e m}}$ . Why this step? $D a$ compares "how long the gas stays" to "how long reactions need." Big $D a$ = reactions finish; small $D a$ = they can't.
Case (a). $D a = \frac{2 \times 1 0 ^{- 3}}{2 \times 1 0 ^{- 5}} = 100 ≫ 1$ → equilibrium flow (plenty of time, energy recovered). Why this step? Reactions are 100× faster than the flow — they keep up at every point.
Case (b). $D a = \frac{2 \times 1 0 ^{- 3}}{2 \times 1 0 ^{- 1}} = 0.01 ≪ 1$ → frozen flow (no time, energy lost). Why this step? Reactions are 100× slower than the flow — composition freezes.

Verify: $D a$ is dimensionless (s/s) ✓. The two answers, $100$ and $0.01$ , straddle $D a = 1$ exactly as the parent's $D a ≫ 1$ / $D a ≪ 1$ rule demands. ✓

Cell I — the full real-world engine

Worked example Example 9 — Putting all levers together

A chamber "should" release $Q = 3.0 MJ/mol$ . At its $T, p$ the degree of dissociation is $α = 0.10$ with dissociation enthalpy $Δ H_{d} = 5.0 MJ/mol$ . (a) Find fraction of intended heat diverted. (b) If mean $M$ drops $18 \to 16$ from lighter species, and (c) recombination is 80% complete in the nozzle, estimate the net directed-energy change vs. the no-dissociation baseline of $3.0 MJ/mol$ .

Forecast: net gain or net loss?

Energy diverted. $α Δ H_{d} = 0.10 \times 5.0 = 0.50 MJ/mol$ ; fraction $= 0.50/3.0 = 0.167 \approx 16.7%$ . Why this step? Every dissociated mole steals $Δ H_{d}$ (its bond-breaking price) from heating (parent Step 1).
Recover 80%. Returned $= 0.80 \times 0.50 = 0.40 MJ/mol$ ; permanently lost $= 0.10 MJ/mol$ . Why this step? $D a$ is finite, so only part of the stolen energy comes back as directed kinetic energy.
Net thermal energy $= 3.0 - 0.10 = 2.90 MJ/mol$ ; the $0.40$ recovered re-enters as directed flow, so net available directed energy $\approx 3.0 - 0.50 + 0.40 = 2.90 MJ/mol$ — a $0.10 MJ/mol$ ( $3.3%$ ) energy shortfall. Why this step? Bookkeeping: base minus permanently-lost.
Molar-mass credit. $v_{e}$ gains $18/16 = 1.061$ , i.e. $+ 6.1%$ on velocity. Why this step? Independent lever from Ex 6; it multiplies $v_{e}$ even after the energy hit.

Verify: Energy fraction $0.167$ ✓; lost $0.10 MJ/mol$ ✓; molar boost $1.061$ ✓. Net story: a $\sim 3%$ energy shortfall partly cancelled by a $\sim 6%$ velocity gain — the tug-of-war the parent promised, resolved by $D a$ and $M$ together.

Cell J — the inverse exam twist

Worked example Example 10 — Measured

$K_{p}$ , find $α$ A test rig at $p = 4 p^{\circ}$ measures $K_{p} = 6.0$ for $A_{2} ⇌ 2 A$ . Find $α$ .

Forecast: with high $K_{p}$ but high $p$ too, is $α$ closer to $0.5$ or $0.9$ ?

Master relation with $p / p^{\circ} = 4$ . $6.0 = \frac{4 α ^{2}}{1 - α ^{2}} \times 4$ . Why this step? Both $K_{p}$ and $p$ are given, so only $α$ is unknown.
Isolate. $\frac{4 α ^{2}}{1 - α ^{2}} = 1.5 \Rightarrow 1.5 (1 - α^{2}) = 4 α^{2} \Rightarrow 1.5 = 5.5 α^{2}$ . Why this step? Divide out the pressure factor first, then collect $α^{2}$ .
Solve. $α^{2} = 1.5/5.5 = 0.2727$ , $α = 0.522$ . Why this step? Positive root — the physical fraction.

Verify: Back-substitute: $\frac{4 ( 0.522 ) ^{2}}{1 - ( 0.522 ) ^{2}} \times 4 = \frac{1.090}{0.7273} \times 4 = 1.499 \times 4 = 5.996 \approx 6.0$ ✓. Note high pressure held $α$ down to $\approx 0.52$ despite the large $K_{p}$ — the pressure lever fighting the temperature lever, both cases seen at once (the coral $p = 4 p^{\circ}$ dot on the figure).

Recall

Recall Cover and answer

At $K_{p} = 0.50, p = p^{\circ}$ , what is $α$ ? ::: $0.333$ Doubling $p$ at fixed $T$ from that state — does $α$ rise or fall, and to what? ::: Falls to $0.242$ (Le Chatelier). As $p \to 0$ at fixed $T$ , what does $α$ approach and why? ::: $α \to 1$ — nothing squeezes the atoms back, so dissociation runs to completion. $D a = 100$ means which flow limit? ::: Equilibrium (reactions keep up). $M$ falling $18 \to 16$ gives what $v_{e}$ factor? ::: $18/16 = 1.061$ , a $+ 6.1%$ boost. Equilibrium vs frozen $v_{e}$ ratio when equilibrium adds $0.50$ to a $2.00 MJ/mol$ base? ::: $2.5/2.0 = 1.118$ . What does $Δ H_{d}$ stand for? ::: Dissociation enthalpy — the energy price to break one mole of bonds.

Connections

3.3.20 Real gas effects — dissociation, recombination (Hinglish)
Chemical Equilibrium and Kp
Adiabatic Flame Temperature
Nozzle Flow and Exhaust Velocity
Specific Impulse
Le Chatelier's Principle
Damköhler Number

Dissociation enthalpy ΔH_d

Energy to break one mole of bonds (MJ/mol); every dissociated mole spends it, so it is stolen from heating.

Exhaust velocity v_e

Speed gas leaves the nozzle; the rocket's performance number, with v_e ∝ sqrt(T_c/M).