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...