Foundations — Combustion of hydrocarbons (RP-1 - kerosene, methane) and hydrogen

Before you can read the parent note, you must be able to read its symbols. This page assumes nothing — not even what a subscript means. We build each idea on a picture, then the next idea leans on the one before it.

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1. Atoms, molecules, and what a chemical formula says

Read $C{H}_4$ like this:

• The letter $C$ = one carbon atom. The letter $H$ = a hydrogen atom.
• The small number below-right (the subscript) tells you how many of the atom just before it. So $H_4$ means four hydrogen atoms. No subscript means "just one" (the $C$ here).

So $C{H}_4$ = 1 carbon + 4 hydrogen, all bonded into one molecule (methane).

Recall What does the subscript in

$C_{12}{H}_{26}$ tell you? Twelve carbon atoms and twenty-six hydrogen atoms in one molecule. This is dodecane, the stand-in for kerosene / RP-1.

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2. The subscript letters $x$ and $y$ (the general fuel $C_x{H}_y$)

Think of $x$ and $y$ as blank boxes on the recipe card:

Fuel	$x$ (carbons)	$y$ (hydrogens)	Formula
Methane	$1$	$4$	$C{H}_4$
Dodecane (RP-1)	$12$	$26$	$C_{12}{H}_{26}$
Hydrogen	$0$	$2$	$H_2$

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3. The oxidizer $O_2$ and the reaction arrow $⟶$

So $C{H}_4+2O_2⟶C{O}_2+2H_{2}O$ reads in English: "One methane plus two oxygen molecules turn into one carbon dioxide plus two waters."

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4. Coefficients — the big numbers in front

Recall How many oxygen atoms total in

$18.5O_2$? $18.5\times 2=37$ oxygen atoms. (This is exactly the count the RP-1 equation needs on the product side.)

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5. "Balanced" and why $\left(x+\frac{y}{4}\right)$ appears

Here is the counting logic behind the parent's coefficient $x+\frac{y}{4}$:

• Every carbon must end up in a $C{O}_2$ → you need $x$ of them. Each $C{O}_2$ carries 2 oxygen atoms → $2x$ oxygen atoms locked in carbon dioxide.
• Every 2 hydrogens make one $H_{2}O$ → you need $\frac{y}{2}$ waters. Each water carries 1 oxygen → $\frac{y}{2}$ oxygen atoms locked in water.
• Total oxygen atoms needed on the right: $2x+\frac{y}{2}$. Each $O_2$ supplies 2 atoms, so number of $O_2=\frac{2x+y/2}{2}=x+\frac{y}{4}$.

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6. Molar mass $M$ and the mole $n$

You get $M$ by adding the masses of the atoms in the formula:

Molecule	Atom masses	$M$ (g/mol)
$O_2$	$16+16$	$32$
$C{H}_4$	$12+4(1)$	$16$
$H_{2}O$	$2(1)+16$	$18$
$C{O}_2$	$12+2(16)$	$44$

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7. Enthalpy symbols: $\Delta H_f^{\circ }$, $\Delta H_c$, and the $\Sigma$ sign

• $\Delta H_f^{\circ }$ = enthalpy of formation: heat change to build one mole of a substance from its raw elements. For a pure element like $O_2$ this is zero (you built it from itself — no change).
• $\Delta H_c$ = enthalpy of combustion: heat released when one mole of fuel burns completely.
• ==$\Sigma$== (capital sigma) simply means "add up all of these." So $\sum \Delta H_f(\text{products})$ = "add the formation enthalpies of every product."

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8. Reading $v_e\propto \sqrt{T_c/M}$

Reading it in words: exhaust velocity gets bigger when the gas is hotter ($T_c$ up top) and when the gas is lighter ($M$ underneath). This is the payoff line connecting combustion chemistry to Specific Impulse $I_{sp}$.

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Prerequisite map

Each arrow means "you need the top box before the bottom one makes sense." Notice everything flows down from atoms and splits into two rivers — counting (balancing, O/F) and energy (enthalpy) — that rejoin at exhaust velocity.

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Equipment checklist

Cover the right side and test yourself. If any line stumps you, re-read its section above before starting the parent note.

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