Question

When $36.6g{N}_2{O}_4\left(g\right)$ is introduced into a $1.0-litre$ flask at ${27}^{\circ }C$. The following equilibrium reaction occurs: $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right);K_p=0.1642atm$.
(a) Calculate $K_c$ of the equilibrium reaction?
(b) What are the number of moles of $N_2{O}_4$ and $N{O}_2$ at equilibrium?
(c) What is the total gas pressure in the flask at equilibrium?
(d) What is the percent dissociation of $N_2{O}_4$?

Answer 1

$36.6gof{N}_2{O}_{4}No.ofmoles=\frac{36.6}{92}\Rightarrow 0.40$

$Pv=nRT$

$P=\frac{n}{v}RT\Rightarrow \frac{36.6}{92\times 1}\times 0.084\times 300=9.80atm.$

$K_p=\frac{{\left(2x\right)}^2}{9.80-x}=0.1642$

$4x^2=1.61-0.1642x$

$4x^2+0.1642x-1.61=0$

$x=0.614$

$P_{N{O}_2}=1.228atm{P}_{N_2{O}_4}=9.186$

$\left(a\right)K_C=K_P{\left(RT\right)}^{-\Delta ng}\Delta ng=2-1=1$

$\Rightarrow 0.1642\times {\left(0.084\times 300\right)}^{-1}\Rightarrow 0.1642/24.63\Rightarrow 0.0067$

$\left(c\right)Totalpressure=9.186+1.228\Rightarrow 10.414atm$

$\left(b\right)n_{N_2{O}_4}=\frac{PV}{RT}\Rightarrow \frac{9.186\times 1}{0.084\times 300}\Rightarrow 0.373mol.$

$\Rightarrow N_{N{O}_2}=\frac{PV}{RT}=\frac{1.228}{0.084\times 300}\Rightarrow 0.050mol.$

$\left(d\right)\%dissociation\Rightarrow \frac{0.40-0.373}{0.40}\times 100\Rightarrow 0.0675\times 100\Rightarrow 6.75\%$