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

##### 1 Answer

Suppose we had a simple dissociation reaction:

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

Then the expression for

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

#P_i# is the partial pressure of#i# at equilibrium.

#=> 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

#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...