How to do this equilibrium derivation? (I'll post an answer showing the problem)

2 Answers
Apr 16, 2015

N_2O_4(g) <=> 2NO_2(g)N2O4(g)2NO2(g)
Assuming that n_0n0 is the beginning number of moles of N_2O_4N2O4 and we start with 00 moles of NO_2NO2, show that xi_(eq)ξeq, the extent of reaction, can be expressed as:

xi_(eq)/n_0 = (K_p/(K_p + 4P))^(1/2)ξeqn0=(KpKp+4P)12

where K_pKp is the equilibrium constant for partial pressures, and PP is the total pressure.

What I tried was:

K_p = (P_(NO_2))^2/(P_(N_2O_4)) = [((2xi_(eq))P)/(1-xi_(eq))]^2/[((1-xi_(eq))P)/(1-xi_(eq))] = [(2xi_(eq))^2P]/[1-xi_(eq)]Kp=(PNO2)2PN2O4=[(2ξeq)P1ξeq]2(1ξeq)P1ξeq=(2ξeq)2P1ξeq
K_p - K_pxi_(eq) = 4Pxi_(eq)^2KpKpξeq=4Pξ2eq
4Pxi_(eq)^2 + K_pxi_(eq) - K_p = 04Pξ2eq+KpξeqKp=0

But when I try the quadratic formula on that, I get:

xi_(eq) = (-K_p pm sqrt(K_p^2-4(4P)(-K_p)))/(8P)ξeq=Kp±K2p4(4P)(Kp)8P
= -K_p pm sqrt(K_p^2+16PK_p)/(8P)Kp±K2p+16PKp8P

Apr 16, 2015

Here's how you'd go about solving this problem.

Start with the ICE table for the equilibrium reaction

" "N_2O_(4(g)) rightleftharpoons 2NO_(2(g)) N2O4(g)2NO2(g)
I..........n_0n0..................0
C.......-epsilonε..............+2epsilon+2ε
E...n_0-epsilonn0ε..............2epsilon2ε

Express the mole fraction of the two species like this

chi_(N_2O_4) = (n_0 - epsilon)/(n_0 - epsilon + 2epsilon) = (n_0 - epsilon)/(n_0 + epsilon)χN2O4=n0εn0ε+2ε=n0εn0+ε

chi_(NO_2) = (2epsilon)/(n_0 + epsilon)χNO2=2εn0+ε

Now just use the definition of K_pKp to solve for epsilonε

K_p = P_(NO_2)^(2)/P_(N_2O_4) = ((2epsilon)^(2)/(n_0 + epsilon)^(2) * P^(2))/((n_0 - epsilon)/(n_0 + epsilon) * P)Kp=P2NO2PN2O4=(2ε)2(n0+ε)2P2n0εn0+εP

K_p = ((2epsilon)^(2) * P^(cancel(2)))/((n_0 + epsilon)^(cancel(2))) * cancel(n_o + epsilon)/((n_0 - epsilon) * cancel(P))

K_p = (4epsilon^(2) * P)/((n_o - epsilon)(n_o + epsilon)) = (4epsilon^(2) * P)/(n_0^(2) - epsilon^(2))

K_p(n_0^(2) - epsilon^(2)) = 4epsilon^(2) P

4epsilon^(2)P + epsilon^(2)K_p - n_0^(2)K_p = 0

epsilon^(2)(4P + K_p) = n_0^(2) K_p | : n_0^(2)

epsilon^(2)/n_0^(2) = K_p/((4P + K_p)) |sqrt

color(green)(epsilon/n_0 = (K_p/(4P + K_p))^(1/2))