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

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

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Answer 1 · 2015-04-16T22:21:49

$N_{2} O_{4} (g) \Leftrightarrow 2 N O_{2} (g)$ $N_2O_4(g) <=> 2NO_2(g)$
Assuming that $n_{0}$ $n_0$ is the beginning number of moles of $N_{2} O_{4}$ $N_2O_4$ and we start with $0$ $0$ moles of $N O_{2}$ $NO_2$ , show that $ξ_{e q}$ $xi_(eq)$ , the extent of reaction, can be expressed as:

$\frac{ξ_{e q}}{n_{0}} = {(\frac{K_{p}}{K_{p} + 4 P})}^{\frac{1}{2}}$ $xi_(eq)/n_0 = (K_p/(K_p + 4P))^(1/2)$

where $K_{p}$ $K_p$ is the equilibrium constant for partial pressures, and $P$ $P$ is the total pressure.

What I tried was:

$K_{p} = \frac{{(P_{N O_{2}})}_{N O_{2}}^{2}}{P_{N_{2} O_{4}}} = \frac{{[\frac{(2 ξ_{e q}) P}{1 - ξ_{e q}}]}^{2}}{\frac{(1 - ξ_{e q}) P}{1 - ξ_{e q}}} = \frac{{(2 ξ_{e q})}_{e q}^{2} P}{1 - ξ_{e q}}$ $K_p = (P_(NO_2))^2/(P_(N_2O_4)) = [((2xi_(eq))P)/(1-xi_(eq))]^2/[((1-xi_(eq))P)/(1-xi_(eq))] = [(2xi_(eq))^2P]/[1-xi_(eq)]$
$K_{p} - K_{p} ξ_{e q} = 4 P ξ_{e q}^{2}$ $K_p - K_pxi_(eq) = 4Pxi_(eq)^2$
$4 P ξ_{e q}^{2} + K_{p} ξ_{e q} - K_{p} = 0$ $4Pxi_(eq)^2 + K_pxi_(eq) - K_p = 0$

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

$ξ_{e q} = \frac{- K_{p} \pm \sqrt{K_{p}^{2} - 4 (4 P) (- K_{p})}}{8 P}$ $xi_(eq) = (-K_p pm sqrt(K_p^2-4(4P)(-K_p)))/(8P)$
= $- K_{p} \pm \frac{\sqrt{K_{p}^{2} + 16 P K_{p}}}{8 P}$ $-K_p pm sqrt(K_p^2+16PK_p)/(8P)$

Answer 2 · 2015-04-16T23:46:07

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

Start with the ICE table for the equilibrium reaction

$N_{2} O_{4 (g)} ⇌ 2 N O_{2 (g)}$ $" "N_2O_(4(g)) rightleftharpoons 2NO_(2(g))$
I.......... $n_{0}$ $n_0$ ..................0
C....... $- ε$ $-epsilon$ .............. $+ 2 ε$ $+2epsilon$
E... $n_{0} - ε$ $n_0-epsilon$ .............. $2 ε$ $2epsilon$

Express the mole fraction of the two species like this

$χ_{N_{2} O_{4}} = \frac{n_{0} - ε}{n_{0} - ε + 2 ε} = \frac{n_{0} - ε}{n_{0} + ε}$ $chi_(N_2O_4) = (n_0 - epsilon)/(n_0 - epsilon + 2epsilon) = (n_0 - epsilon)/(n_0 + epsilon)$

$χ_{N O_{2}} = \frac{2 ε}{n_{0} + ε}$ $chi_(NO_2) = (2epsilon)/(n_0 + epsilon)$

Now just use the definition of $K_{p}$ $K_p$ to solve for $ε$ $epsilon$

$K_{p} = \frac{P_{N O_{2}}^{2}}{P_{N_{2} O_{4}}} = \frac{\frac{{(2 ε)}^{2}}{{(n_{0} + ε)}_{0}^{2}} \cdot P^{2}}{\frac{n_{0} - ε}{n_{0} + ε} \cdot P}$ $K_p = P_(NO_2)^(2)/P_(N_2O_4) = ((2epsilon)^(2)/(n_0 + epsilon)^(2) * P^(2))/((n_0 - epsilon)/(n_0 + epsilon) * P)$

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

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

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

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

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

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

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

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

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