Question

A 10 litre box contains $O_3$ and $O_2$ at equilibrium at 2000K. $K_P=4\times {10}^{14}$ atm for $2O_3\left(g\right)\rightleftharpoons 3O_2\left(g\right)$ Assume that $p_{O{}_2}>>p_{O{}_3}$ and if total pressure is 8 atm, then moles of $O_3$ at equilibrium will be:

Take $\sqrt{5}=2.2$, $\sqrt{3}=1.7$, $\sqrt{2}=1.4$

• A. $1.1\times {10}^{-5}$
• B. $7\times {10}^{-8}$
• C. $4\times {10}^{-9}$
• D. $9\times {10}^{-9}$

Answer 1

The correct option is B

$7\times {10}^{-8}$

$PV=nRT$

$8\times 10=n\times 0.082\times 2000$

$n\approx 0.5=\sum n$

$k_p=\frac{(n{O}_2{)}^3}{(n{O}_3{)}^2}\times \frac{P}{\sum n}$

$P\propto n$

$p{O}_2>>p{O}_3\to n{O}_2>>n{O}_3$

$\therefore \sum n=n{O}_2$

$4\times {10}^{14}=\frac{\sum n^2}{(n{O}_3{)}^2}\times P$

$(n{O}_3{)}^2=\frac{{0.5}^2\times 8}{4\times {10}^{14}}$

$n{O}_3=0.5\times \sqrt{2}\times {10}^{-7}=7\times {10}^{-8}$