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 ofi 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...