Question

$13.8g$ of $N_2{O}_4$ was placed in a $1L$ reaction vessel at $400K$ and allowed attain equibrium $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
The total pressure at equilibrium was found to be $9.15$ bar. Calculate $K_c,K_p$ and partial pressure at equilibrium.

Answer 1

pressure of $N_2{O}_4$ at 400k

$P=\frac{nRT}{V}=\frac{0.15\times 0.0821\times 400}{1}=4.9bar\approx 5$

$N_2{O}_4⟶2N{O}_2$

$5-x$ $2x$ (at equilibrium)

Total pressure = $5+x=9.15bar$ or $x=4.15bar$

So $K_p=\frac{(P_{N{O}_2}{)}^2}{P_{N_2{O}_4}}=\frac{(2\times 4.15{)}^2}{(0.85{)}^2}=95.34$

Now, $K_c=K_{p}R{T}^{\Delta n_g}=95.34\times 0.0821\times {400}^1=3130.96$