Equilibrium products at high T — dissociation (H₂O ⇌ OH + H; CO₂ ⇌ CO + ½O₂): Question bank

0. The toolbox (every symbol defined before you meet it)

Before any trap, here is the full vocabulary the questions use. Nothing below relies on memory from another page.

Definition The players and their units

$T$ — absolute temperature, in kelvin (K). "How hard the molecules are shaking."
$P$ — total gas pressure (the sum of all the pushes from every molecule), in bar.
$R$ — the gas constant, $8.314 J mol^{- 1} K^{- 1}$ . Just the conversion factor between energy and (temperature × amount).
$Δ H^{\circ}$ — standard enthalpy change of the reaction, in $J/mol$ (or kJ/mol). The heat cost/gain of the reaction at the reference pressure $p^{\circ} = 1 bar$ . For bond breaking $Δ H^{\circ} > 0$ (you must pay heat to snap a bond).
$Δ S^{\circ}$ — standard entropy change, in $J mol^{- 1} K^{- 1}$ . A measure of "spread-out-ness". Breaking one molecule into two pieces makes more particles → more ways to arrange them → $Δ S^{\circ} > 0$ .
$Δ G^{\circ}$ — standard Gibbs free energy change, $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$ , in J/mol. The "net reluctance" of the reaction at $p^{\circ}$ . Negative ⇒ the standard-state reaction wants to go forward.
The little circle $^{\circ}$ always means "measured at the reference pressure $p^{\circ} = 1 bar$ ". Drop the circle ( $Δ G$ ) and you mean the actual mixture, not the reference one. $Δ G^{\circ}$ is fixed once $T$ is fixed; $Δ G$ depends on how much of each species is actually present.

How the $C O_{2}$ formula is built (follow the picture)

Figure — Equilibrium products at high T — dissociation (H₂O ⇌ OH + H; CO₂ ⇌ CO + ½O₂)

WHAT: Figure s01 is a bookkeeping cartoon. Column by column it tracks moles of $C O_{2} ⇌ C O + \frac{1}{2} O_{2}$ as a fraction $α$ breaks. WHY: every partial pressure needs mole fractions, and mole fractions need the total mole count — which grows. WHAT IT LOOKS LIKE: starting from 1 mol $C O_{2}$ , we lose $α$ of it ( $1 - α$ left), gain $α$ of $C O$ and $α /2$ of $O_{2}$ . Adding the column: $n_{t o t} = (1 - α) + α + \frac{α}{2} = 1 + \frac{α}{2}$ .

Now each mole fraction is (its moles) $/ n_{t o t}$ , and each partial pressure is (mole fraction) $\times P$ : $p_{C O_{2}} = \frac{1 - α}{1 + α /2} P, p_{C O} = \frac{α}{1 + α /2} P, p_{O_{2}} = \frac{α /2}{1 + α /2} P$ Feed these into $K_{p} = \frac{p _{C O} p _{O_{2}}^{1/2}}{p _{C O_{2}}}$ (exponents are the coefficients $+ 1, + \frac{1}{2}, - 1$ ). The $(1 + α /2)$ factors mostly cancel between the $p_{C O} / p_{C O_{2}}$ ratio, leaving one under the $O_{2}$ term: $K_{p} = \frac{α}{1 - α} (\frac{α /2}{1 + α /2} P)^{1/2}$ The lone $P^{1/2}$ appears because the net coefficient sum is $Δ n = (1 + \frac{1}{2}) - 1 = + \frac{1}{2}$ .

The three graphs behind the traps

Figure s02 shows the free-energy line $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$ : a straight line that starts positive (bonds win) and crosses zero at high $T$ (entropy wins). Where it dips below zero, dissociation turns spontaneous.

Figure s03 is the van 't Hoff line: plot $ln K_{p}$ against $1/ T$ . Because $ln K_{p} = - \frac{Δ H ^{\circ}}{R} \cdot \frac{1}{T} + \frac{Δ S ^{\circ}}{R}$ , it is a straight line of slope $- Δ H^{\circ} / R$ . For endothermic dissociation the slope is negative, so as $T$ rises ( $1/ T$ shrinks) $ln K_{p}$ climbs — that is the van 't Hoff equation, derived by differentiating: $d ln K_{p} / d T = Δ H^{\circ} / R T^{2} > 0$ .

Figure s04 shows $α$ versus $P$ at fixed $T$ (fixed $K_{p}$ ): a falling curve. Squeeze harder → less dissociation, because the product side has more moles. This is the picture for every pressure trap below.

True or false — justify

True or false: at 298 K a flame's products are essentially pure CO₂ and H₂O.

True — at room temperature the dissociation

K_{p}

is astronomically tiny, so

α \approx 0

. The trap is applying this to a 3000 K flame, where

K_{p}

is no longer negligible.

True or false:

K_{p}

depends on both temperature and total pressure.

False —

K_{p}

is a function of

T

only, since it is built from

Δ G^{\circ} (T) = Δ H^{\circ} - T Δ S^{\circ}

. Pressure changes the degree of dissociation

α

, not

K_{p}

itself.

True or false: raising the chamber pressure lowers the degree of dissociation of CO₂.

True — the reaction

C O_{2} ⇌ C O + \frac{1}{2} O_{2}

has net mole change

\sum ν_{i} = + \frac{1}{2} > 0

, so squeezing the gas shifts equilibrium back toward the fewer-mole side (

C O_{2}

); this is the falling curve of figure s04.

True or false: because dissociation is endothermic, it can never happen spontaneously.

False — spontaneity is set by

Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}

, not

Δ H^{\circ}

alone. Since

Δ S^{\circ} > 0

(more particles), high enough

T

makes

Δ G^{\circ}

negative (figure s02) and the reaction proceeds.

True or false: dissociation raises the adiabatic flame temperature by releasing radicals.

False — bond breaking absorbs energy (

Δ H^{\circ} > 0

), so it lowers the achievable temperature. Radicals are a symptom, not an energy source.

True or false: if

Δ G^{\circ} > 0

then

K_{p} < 1

True — from

Δ G^{\circ} = - R T ln K_{p}

, a positive

Δ G^{\circ}

forces

ln K_{p} < 0

, hence

K_{p} < 1

. Dissociation reactions have large positive

Δ G^{\circ}

, giving very small

K_{p}

True or false: a small

K_{p}

(like

1 0^{- 3}

) means dissociation is completely negligible in a rocket.

False — even

K_{p} \sim 1 0^{- 3}

produces a few-percent radical pool, enough to steal noticeable energy and to matter for nozzle chemistry.

True or false: raising temperature increases

K_{p}

for every reaction.

False — only for endothermic reactions (

Δ H^{\circ} > 0

), by van 't Hoff

d ln K_{p} / d T = Δ H^{\circ} / R T^{2}

(positive slope in figure s03). Exothermic reactions have

K_{p}

that falls with

T

True or false:

K_{p}

and

K_{c}

are the same number for CO₂ dissociation.

False — they differ because

K_{p} = K_{c} (R T / p^{\circ})^{Δ n}

with

Δ n = + \frac{1}{2} \neq = 0

, so they are related by a factor of

R T / p^{\circ}

Spot the error

Spot the error: "

K_{p} = \frac{p _{C O} p _{O_{2}}}{p _{C O_{2}}}

for

C O_{2} ⇌ C O + \frac{1}{2} O_{2}

The

O_{2}

exponent must be its stoichiometric coefficient

\frac{1}{2}

, so it is

p_{O_{2}}^{1/2}

, not

p_{O_{2}}

. Dropping the half throws the computed

α

off badly.

Spot the error: "

Δ G = - R T ln K_{p}

, so if I heat the gas

K_{p}

stays fixed because

Δ G

is a constant."

Two mistakes. The relation uses the standard value:

Δ G^{\circ} = - R T ln K_{p}

(with the circle), not the actual-mixture

Δ G

(which is zero at equilibrium). And

Δ G^{\circ} (T) = Δ H^{\circ} - T Δ S^{\circ}

is not constant — it changes with

T

, which is exactly what makes

K_{p}

temperature-dependent.

Spot the error: "Since

K_{p}

only depends on

T

, changing

P

can't change any partial pressure."

The individual

p_{i} = x_{i} P

and the mole fractions both shift with

P

; only their particular combination

\prod (p_{i} / p^{\circ})^{ν_{i}}

stays pinned to

K_{p}

Spot the error: "Total moles stay at 1 because we start with 1 mol CO₂."

Dissociation creates particles:

n_{t o t} = 1 + α /2 > 1

(figure s01). Forgetting this growth corrupts every mole fraction.

Spot the error: "

K_{p}

has units of pressure, so I'll report it in bar."

In the definition each

p_{i}

is divided by the standard pressure

p^{\circ} = 1 bar

, so

K_{p}

is dimensionless. The reference pressure is what lets us take its logarithm.

Spot the error: "Because both dissociations absorb heat, doubling one automatically doubles the other's extent."

Each equilibrium has its own

Δ G^{\circ} (T)

and its own

K_{p}

; H₂O and CO₂ dissociate to different degrees at the same temperature.

Spot the error: "Van 't Hoff gives

ln (K_{2} / K_{1}) = + \frac{Δ H ^{\circ}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})

The sign is wrong: it is

- \frac{Δ H ^{\circ}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})

, the integrated form of the negative-slope line in figure s03. With

T_{2} > T_{1}

and

Δ H^{\circ} > 0

this correctly gives a positive

ln (K_{2} / K_{1})

, i.e. more dissociation.

Why questions

Why does breaking a bond become favourable at high temperature even though it costs enthalpy?

Because the

- T Δ S^{\circ}

term grows with

T

, and dissociation increases entropy (more particles,

Δ S^{\circ} > 0

). Eventually this entropy payoff outweighs the enthalpy cost and

Δ G^{\circ}

dips below zero (figure s02).

Why does the

P^{1/2}

factor appear in the CO₂

K_{p}

expression but not

P^{1}

P^{0}

The exponent on

P

equals the net stoichiometric change

Δ n = (1 + \frac{1}{2}) - 1 = + \frac{1}{2}

. Each species' pressure carries a factor of

P

, and they combine to leave

P^{1/2}

(see the build in section 0).

Why do high-pressure rocket chambers dissociate less than low-pressure ones?

With

Δ n > 0

, higher

P

opposes the mole-increasing direction (Le Chatelier), pushing equilibrium back toward the intact molecule and lowering

α

— the falling curve of figure s04 — while

K_{p}

stays constant.

Why is atomic

H

O H

produced even when there is plenty of oxygen for "complete" combustion?

Completeness is a stoichiometric ideal; thermodynamics at flame temperature independently sets a non-zero

K_{p}

for dissociation, so a radical fraction always coexists with the "finished" products.

Why must a nozzle designer care whether the flow is frozen or in equilibrium?

If radicals re-combine as the gas cools and expands (Rocket Nozzle Frozen vs Equilibrium Flow), they release energy back into the flow; assuming the wrong regime mis-predicts thrust and exhaust composition.

Why can't we just use the room-temperature

Δ G^{\circ}

when analysing a 3000 K flame?

Δ G^{\circ}

(and hence

K_{p}

) is strongly temperature-dependent; the 298 K value gives a vanishing

K_{p}

and would predict zero dissociation, badly underestimating radical content.

Edge cases

Edge case: what is the degree of dissociation

α

in the limit

T \to 0

α \to 0

— as

T

falls,

Δ G^{\circ} \to Δ H^{\circ} > 0

, so

K_{p} \to 0

and the molecule stays fully bonded.

Edge case: what happens to

α

T \to

very large (say beyond flame temperatures)?

α \to 1

—

K_{p}

grows without bound as the

- T Δ S^{\circ}

term dominates, so dissociation approaches completeness.

Edge case: what happens to

α

P \to 0

(very low pressure), at fixed

T

α \to 1

— with

Δ n > 0

, dropping the pressure toward zero pulls equilibrium hard toward the more-numerous product side, so a rarefied gas dissociates almost completely (left end of figure s04).

Edge case: for the CO₂ formula

K_{p} = \frac{α}{1 - α} (\frac{α /2}{1 + α /2} P)^{1/2}

, what happens as

α \to 1

The

\frac{α}{1 - α}

factor diverges to infinity, correctly signalling that full dissociation requires an arbitrarily large

K_{p}

(i.e. extreme temperature) or vanishing

P

Edge case: if a reaction had

Δ n = 0

(equal moles both sides), how would

P

affect

α

It wouldn't — with no net mole change the

P

factors cancel in

K_{p}

, so pressure has no effect on the degree of dissociation, and

K_{p} = K_{c}

as well.

Edge case: at

α = 0

exactly, is the equilibrium expression still defined?

The forward extent is zero, giving

K_{p} = 0

; this is the degenerate no-dissociation limit reached only as

T \to 0

, not a physical flame state.

Edge case: could increasing pressure ever increase dissociation?

Only if

Δ n < 0

(fewer moles on the product side); for the dissociations here

Δ n > 0

, so pressure always suppresses

α

Recall One-line summary of the whole trap set

$K_{p}$ is $T$ -only (and $= K_{c} (R T / p^{\circ})^{Δ n}$ ); $α$ answers to both $T$ and $P$ ; heat breaks bonds (endothermic, entropy-driven); squeeze remakes them when the product side has more moles.

0. The toolbox (every symbol defined before you meet it)

How the CO2​ formula is built (follow the picture)

The three graphs behind the traps

True or false — justify

Spot the error

Why questions

Edge cases

How the $C O_{2}$ formula is built (follow the picture)