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

Apr 16, 2015

${N}_{2} {O}_{4} \left(g\right) \iff 2 N {O}_{2} \left(g\right)$
Assuming that ${n}_{0}$ is the beginning number of moles of ${N}_{2} {O}_{4}$ and we start with $0$ moles of $N {O}_{2}$, show that ${\xi}_{e q}$, the extent of reaction, can be expressed as:

${\xi}_{e q} / {n}_{0} = {\left({K}_{p} / \left({K}_{p} + 4 P\right)\right)}^{\frac{1}{2}}$

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

What I tried was:

${K}_{p} = {\left({P}_{N {O}_{2}}\right)}^{2} / \left({P}_{{N}_{2} {O}_{4}}\right) = {\left[\frac{\left(2 {\xi}_{e q}\right) P}{1 - {\xi}_{e q}}\right]}^{2} / \left[\frac{\left(1 - {\xi}_{e q}\right) P}{1 - {\xi}_{e q}}\right] = \frac{{\left(2 {\xi}_{e q}\right)}^{2} P}{1 - {\xi}_{e q}}$
${K}_{p} - {K}_{p} {\xi}_{e q} = 4 P {\xi}_{e q}^{2}$
$4 P {\xi}_{e q}^{2} + {K}_{p} {\xi}_{e q} - {K}_{p} = 0$

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

${\xi}_{e q} = \frac{- {K}_{p} \pm \sqrt{{K}_{p}^{2} - 4 \left(4 P\right) \left(- {K}_{p}\right)}}{8 P}$
= $- {K}_{p} \pm \frac{\sqrt{{K}_{p}^{2} + 16 P {K}_{p}}}{8 P}$

Apr 16, 2015

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

$\text{ } {N}_{2} {O}_{4 \left(g\right)} r i g h t \le f t h a r p \infty n s 2 N {O}_{2 \left(g\right)}$
I..........${n}_{0}$..................0
C.......$- \epsilon$..............$+ 2 \epsilon$
E...${n}_{0} - \epsilon$..............$2 \epsilon$

Express the mole fraction of the two species like this

${\chi}_{{N}_{2} {O}_{4}} = \frac{{n}_{0} - \epsilon}{{n}_{0} - \epsilon + 2 \epsilon} = \frac{{n}_{0} - \epsilon}{{n}_{0} + \epsilon}$

${\chi}_{N {O}_{2}} = \frac{2 \epsilon}{{n}_{0} + \epsilon}$

Now just use the definition of ${K}_{p}$ to solve for $\epsilon$

${K}_{p} = {P}_{N {O}_{2}}^{2} / {P}_{{N}_{2} {O}_{4}} = \frac{{\left(2 \epsilon\right)}^{2} / {\left({n}_{0} + \epsilon\right)}^{2} \cdot {P}^{2}}{\frac{{n}_{0} - \epsilon}{{n}_{0} + \epsilon} \cdot P}$

${K}_{p} = \frac{{\left(2 \epsilon\right)}^{2} \cdot {P}^{\cancel{2}}}{{\left({n}_{0} + \epsilon\right)}^{\cancel{2}}} \cdot \frac{\cancel{{n}_{o} + \epsilon}}{\left({n}_{0} - \epsilon\right) \cdot \cancel{P}}$

${K}_{p} = \frac{4 {\epsilon}^{2} \cdot P}{\left({n}_{o} - \epsilon\right) \left({n}_{o} + \epsilon\right)} = \frac{4 {\epsilon}^{2} \cdot P}{{n}_{0}^{2} - {\epsilon}^{2}}$

${K}_{p} \left({n}_{0}^{2} - {\epsilon}^{2}\right) = 4 {\epsilon}^{2} P$

$4 {\epsilon}^{2} P + {\epsilon}^{2} {K}_{p} - {n}_{0}^{2} {K}_{p} = 0$

${\epsilon}^{2} \left(4 P + {K}_{p}\right) = {n}_{0}^{2} {K}_{p} | : {n}_{0}^{2}$

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

$\textcolor{g r e e n}{\frac{\epsilon}{n} _ 0 = {\left({K}_{p} / \left(4 P + {K}_{p}\right)\right)}^{\frac{1}{2}}}$