# Derive an expression for the fraction of dissociation of a diatomic gas molecule into two gaseous atoms?

Aug 12, 2017

Suppose we had a simple dissociation reaction:

$A B \left(g\right) \to A \left(g\right) + B \left(g\right)$

Then the expression for ${K}_{P}$ would be:

${K}_{P} = \frac{{P}_{A} {P}_{B}}{{P}_{A B}}$

where ${P}_{i}$ is the partial pressure of $i$ at equilibrium.

$A B$ would dissociate so that its partial pressure decreases by $x$ and the partial pressures of $A$ and $B$ would be $x$ each...

$\implies {K}_{P} = {x}^{2} / \left({P}_{A B} - x\right)$

The fraction of dissociation is defined by

$\alpha = \frac{x}{P} _ \left(A B\right)$.

Thus...

${K}_{P} = {\left(\alpha {P}_{A B}\right)}^{2} / \left({P}_{A B} - \alpha {P}_{A B}\right)$

$= \frac{{\alpha}^{2} {P}_{A B}^{2}}{\left(1 - \alpha\right) {P}_{A B}}$

$= \frac{{\alpha}^{2}}{1 - \alpha} {P}_{A B}$

And so...

${K}_{P} - {K}_{P} \alpha = {P}_{A B} {\alpha}^{2}$

${P}_{A B} {\alpha}^{2} + {K}_{P} \alpha - {K}_{P} = 0$

This becomes a quadratic equation in $\alpha$:

$\textcolor{b l u e}{\alpha} = \frac{- {K}_{P} \pm \sqrt{{K}_{P}^{2} - 4 \left({P}_{A B}\right) \left(- {K}_{P}\right)}}{2 {P}_{A B}}$

$= \frac{- {K}_{P} \pm \sqrt{{K}_{P}^{2} + 4 {K}_{P} {P}_{A B}}}{2 {P}_{A B}}$

$= - {K}_{P} / \left(2 {P}_{A B}\right) \pm \frac{\sqrt{{K}_{P}^{2} + 4 {K}_{P} {P}_{A B}}}{2 {P}_{A B}}$

$= - {K}_{P} / \left(2 {P}_{A B}\right) \pm \sqrt{{K}_{P}^{2} / \left(4 {P}_{A B}^{2}\right) + {K}_{P} / {P}_{A B}}$

$= - {K}_{P} / \left(2 {P}_{A B}\right) \pm {K}_{P} / \left({P}_{A B}\right) \sqrt{\frac{1}{4} + {P}_{A B} / {K}_{P}}$

$= \textcolor{b l u e}{{K}_{P} / \left({P}_{A B}\right) \left[- \frac{1}{2} \pm \sqrt{\frac{1}{4} + {P}_{A B} / {K}_{P}}\right]}$

I think that's about as simple as I can get it...