Question

A sample of air consisting of $N_2$ and $O_2$ was heated to $2500$ K until the equilibrium
$N_{2\left(g\right)}$+$O_{2\left(g\right)}$ $\rightleftharpoons$$2N{O}_{\left(g\right)}$
was established with an equilibrium constant $K_c$=2.1 $\times {10}^{-3}$. At equilibrium, the mole% of NO was $1.8$. Estimate the initial composition of air in mole fraction of $N_2$ and $O_2$

Answer 1

$N_{2\left(g\right)}+O_{2\left(g\right)}\leftrightharpoons 2N{O}_{\left(g\right)}$

At $t=0$- $a$ $b$ $0$

At $t=$ equilibrium- $(a-x)$ $(b-x)$ $2x$

$K_C=\frac{(2x{)}^2}{(a-x)(b-x)}=2.1\times {10}^{-3}$ equation $1$

But $\frac{2x}{a-x+b-x+2x}=0.018$

$\therefore \frac{2x}{a+b}=0.018$ equation $2$

By solving $1$ and $2$ we get,

$\frac{a}{a+b}=0.79=$ Mole fraction of $N_2$

and $\frac{b}{a+b}=0.21=$ Mole fraction of $O_2$

Therefore, mole fraction of $N_2$ and $O_2$ are $0.79$ and $0.21$ respectively.