$13.8 \,g$ of $N_{2}O_{4}$ was placed in a $1 \,L$ reaction vessel at $400 \,K$ and allowed to attain equilibrium.
$N_{2}O_{4(g)} \rightleftharpoons 2NO_{2(g)}$
The total pressure at equilibrium was found to be $9.15 \,bar$. Calculate $K_{c}$,$K_{p}$,and the partial pressures at equilibrium.

(N/A) $1$. Calculate initial pressure $(p_i)$ of $N_{2}O_{4}$ using $pV = nRT$:
$n = \frac{13.8 \,g}{92 \,g \,mol^{-1}} = 0.15 \,mol$
$p_i = \frac{nRT}{V} = \frac{0.15 \,mol \times 0.083 \,bar \,L \,mol^{-1} \,K^{-1} \times 400 \,K}{1 \,L} = 4.98 \,bar$
$2$. Set up the equilibrium table:
Reaction: $N_{2}O_{4(g)} \rightleftharpoons 2NO_{2(g)}$
Initial: $4.98 \,bar$,$0$
Equilibrium: $(4.98-x) \,bar$,$2x \,bar$
$3$. Solve for $x$ using total pressure:
$p_{\text{total}} = (4.98-x) + 2x = 4.98 + x = 9.15 \,bar$
$x = 9.15 - 4.98 = 4.17 \,bar$
$4$. Calculate partial pressures at equilibrium:
$p_{N_{2}O_{4}} = 4.98 - 4.17 = 0.81 \,bar$
$p_{NO_{2}} = 2 \times 4.17 = 8.34 \,bar$
$5$. Calculate $K_{p}$:
$K_{p} = \frac{(p_{NO_{2}})^2}{p_{N_{2}O_{4}}} = \frac{(8.34)^2}{0.81} = 85.87$
$6$. Calculate $K_{c}$ using $K_{p} = K_{c}(RT)^{\Delta n}$:
$85.87 = K_{c}(0.083 \times 400)^{2-1}$
$K_{c} = \frac{85.87}{33.2} \approx 2.586 \approx 2.6$