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

1 Answer
Aug 12, 2017

Suppose we had a simple dissociation reaction:

AB(g) -> A(g) + B(g)

Then the expression for K_P would be:

K_P = (P_A P_B)/(P_(AB))

where P_i is the partial pressure of i at equilibrium.

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

=> K_P = x^2/(P_(AB) - x)

The fraction of dissociation is defined by

alpha = x/P_(AB).

Thus...

K_P = (alphaP_(AB))^2/(P_(AB) - alphaP_(AB))

= (alpha^2P_(AB)^2)/((1 - alpha)P_(AB))

= (alpha^2)/(1 - alpha)P_(AB)

And so...

K_P - K_Palpha = P_(AB)alpha^2

P_(AB)alpha^2 + K_Palpha - K_P = 0

This becomes a quadratic equation in alpha:

color(blue)(alpha) = (-K_P pm sqrt(K_P^2 - 4(P_(AB))(-K_P)))/(2P_(AB))

= (-K_P pm sqrt(K_P^2 + 4K_PP_(AB)))/(2P_(AB))

= -K_P/(2P_(AB)) pm (sqrt(K_P^2 + 4K_PP_(AB)))/(2P_(AB))

= -K_P/(2P_(AB)) pm sqrt(K_P^2/(4P_(AB)^2) + K_P/P_(AB))

= -K_P/(2P_(AB)) pm K_P/(P_(AB))sqrt(1/4 + P_(AB)/K_P)

= color(blue)(K_P/(P_(AB))[-1/2 pm sqrt(1/4 + P_(AB)/K_P)])

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