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

This page assumes nothing. We build every letter, arrow and ratio the parent note throws at you, in an order where each one only needs the ones before it. If you meet a symbol on the parent page and feel lost, it is defined here.

0. The plain-English cast of characters

Before any formula, here is what the parent topic is about, in words:

We have gas molecules that can split into smaller ones: this is called dissociation.
Splitting and re-joining reach a balance called equilibrium — both directions happen at equal rates, so amounts stop changing.
We measure "how far the split went" with a number, and we predict that number from energy and disorder.

Let us now earn each symbol, one at a time.

1. The reversible arrow $⇌$ — what "equilibrium" looks like

Picture it. Imagine a room with two doors between "left" and "right". People wander both ways. At first everyone starts on the left, so the net flow is left→right. As the right fills up, more people wander back. Eventually just as many cross each way per minute — the headcount in each room stops changing, even though people keep moving.

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

Why the topic needs this. Combustion products don't sit at "fully split" or "fully joined" — they settle at a balance somewhere in between. The single arrow $\to$ of a normal reaction ("goes to completion") would hide that. The double arrow is the whole reason we need a number to say where the balance lands.

2. Moles $n$ and the degree of dissociation $α$

Picture it. Start with $1$ scoop of $C O_{2}$ . A slice of width $α$ breaks off; the leftover intact part is width $1 - α$ . Every product amount is measured off that same $α$ slice.

Why the topic needs this. The whole "Worked example A" is just: express every amount using one unknown $α$ , then solve for it. The magic of $α$ is that one number captures the entire state of the split.

3. Stoichiometric coefficient $ν_{i}$ — the recipe numbers (with signs)

Picture it. In $C O_{2} ⇌ CO + \frac{1}{2} O_{2}$ :

species	role	$ν_{i}$
$C O_{2}$	reactant	$- 1$
$C O$	product	$+ 1$
$O_{2}$	product	$+ \frac{1}{2}$

Add them: $\sum_{i} ν_{i} = - 1 + 1 + \frac{1}{2} = + \frac{1}{2}$ . That single sum $+ \frac{1}{2}$ is the "net mole change" the parent page keeps invoking.

Why the topic needs this. Two big claims on the parent page ride entirely on $\sum ν_{i}$ :

The exponents in $K_{p}$ are the $ν_{i}$ (that's why $O_{2}$ gets a $\frac{1}{2}$ power).
Because $\sum ν_{i} = + \frac{1}{2} > 0$ , pressure suppresses dissociation.

4. Partial pressure $p_{i}$ , total pressure $P$ , mole fraction $x_{i}$

Picture it. Think of $P$ as one pie. The mole fraction $x_{i}$ is the size of species $i$ 's slice; its partial pressure is that slice's area. More molecules of a kind → bigger slice → bigger $p_{i}$ .

Why the topic needs this. The equilibrium number is built from partial pressures, but we know amounts (moles, via $α$ ). The bridge $p_{i} = x_{i} P$ is exactly how the parent page converts its mole table into pressures — and it is where the total pressure $P$ sneaks into $K_{p}$ .

5. The equilibrium constant $K_{p}$ — the "how far it split" number

Read it slowly. Products (with $+ ν_{i}$ ) sit upstairs (numerator); reactants (with $- ν_{i}$ ) sit downstairs (denominator, because a negative exponent flips a factor down). For $C O_{2} ⇌ CO + \frac{1}{2} O_{2}$ : $K_{p} = \frac{p _{C O}^{+ 1} p _{O_{2}}^{+ 1/2}}{p _{C O_{2}}^{+ 1}} .$

Why the topic needs this. $K_{p}$ is the target of every calculation. Big $K_{p}$ → mixture sits far to the product (split) side. Tiny $K_{p}$ → barely any splitting. It is the one number linking thermodynamics (energy) to composition (how much $O H$ , $C O$ you actually get).

6. Energy words: enthalpy $Δ H^{\circ}$ , entropy $Δ S^{\circ}$ , Gibbs $Δ G^{\circ}$

These three drive why $K_{p}$ is what it is. The little circle $^{\circ}$ means "measured at the standard reference conditions"; the $Δ$ ("delta") means "change: products minus reactants".

Picture the tug-of-war. $Δ H^{\circ}$ pulls toward staying bonded (splitting costs heat). $- T Δ S^{\circ}$ pulls toward splitting (more disorder is favoured), and this pull grows as $T$ rises. At low $T$ the enthalpy pull wins; at high $T$ the entropy pull takes over. That crossover is the story of high-temperature dissociation.

Why the topic needs this. The parent's headline slogan — "heat literally pays the entropy to break bonds" — is exactly the moment $- T Δ S^{\circ}$ overtakes $Δ H^{\circ}$ . And these feed the master link next.

7. The gas constant $R$ , temperature $T$ , and the master link

What is $ln$ and $exp$ ? $ln$ (natural logarithm) asks " $e$ to what power gives this number?"; $exp (x) = e^{x}$ undoes it. We need them because $K_{p}$ can span many powers of ten (from $1 0^{- 20}$ to huge), and energy relates to $K_{p}$ multiplicatively — logs turn "multiply" into "add", which is exactly how the additive energy $Δ G^{\circ}$ maps onto the multiplicative $K_{p}$ .

Why this is the keystone. It fuses everything: energy/disorder ( $Δ G^{\circ}$ ) on the left, composition ( $K_{p}$ ) on the right. Plug in a temperature, get a number for how much your flame dissociates. Worked example B is nothing more than one substitution into this line.

Prerequisite map

Each arrow means "you need the tail before you can understand the head". Notice everything funnels into $K_{p}$ , and $K_{p}$ plus the mole-change $\sum ν_{i}$ together answer the topic's central question.

Where these lead

The pieces built here are the entry tickets to the rest of the vault:

The energy referee sits at the heart of Gibbs Free Energy and Spontaneity.
The $K_{p}$ machinery and its unit-free ratios: Equilibrium Constant Kp and Kc.
"Heat breaks, squeeze makes" as a qualitative rule: Le Chatelier's Principle.
How $K_{p}$ climbs with $T$ : Van 't Hoff Equation.
Why the flame ends up cooler than the clean estimate: Adiabatic Flame Temperature.
The fragments themselves ( $O H$ , $H$ , $O$ ): Combustion Radicals OH H O.
Why all this even matters downstream: Rocket Nozzle Frozen vs Equilibrium Flow.

Return to the parent whenever you're ready: parent topic.

Equipment checklist

Test yourself — cover the right side and try to answer before revealing.

What does the double arrow

⇌

mean?

The reaction runs both directions at once; at equilibrium the two rates are equal so amounts stop changing.

What is the degree of dissociation

α

, and its range?

The fraction of the original molecules that have split apart; a unitless number from

0

(none split) to

1

(all split).

What is the sign convention for the stoichiometric coefficient

ν_{i}

Positive for products, negative for reactants.

Compute

\sum_{i} ν_{i}

for

C O_{2} ⇌ CO + \frac{1}{2} O_{2}

- 1 + 1 + \frac{1}{2} = + \frac{1}{2}

How do you turn a mole fraction into a partial pressure?

p_{i} = x_{i} P

, where

x_{i} = n_{i} / n_{t o t}

and

P

is total pressure.

What does the symbol

\prod

mean in the definition of

K_{p}

"Multiply all these factors together" (the multiplication analogue of

\sum

Why is dissociation endothermic (

Δ H^{\circ} > 0

Breaking chemical bonds costs energy, so heat must be absorbed.

Which term in

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

grows with temperature and favours splitting?

The

- T Δ S^{\circ}

term (since

Δ S^{\circ} > 0

for dissociation).

State the master link between

Δ G^{\circ}

and

K_{p}

Δ G^{\circ} = - R T ln K_{p}

, equivalently

K_{p} = exp (- Δ G^{\circ} / R T)

Why do we use

ln

exp

to connect energy and

K_{p}

K_{p}

spans many powers of ten and combines multiplicatively; logs convert that into the additive energy

Δ G^{\circ}