Question

A $10litre$ box contains $O_3$ and $O_2$ at equilibrium at 2000 K. $K_p=4\times {10}^{14}$ 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 patial pressure of $O_3$ will be:

• A. $8\times {10}^{-5}atm$
• B. $11.3\times {10}^{-7}atm$
• C. $9.71\times {10}^{-6}atm$
• D. $8\times {10}^{-2}atm$

Answer 1

The correct option is B $11.3\times {10}^{-7}atm$

$K_p$, the pressure constant at equilibrium, is the ratio of the partial pressure of products to that of the partial pressure of reactant.

$K_p=4\times {10}^4$

$2O_3\left(g\right)\leftrightharpoons 3O_2\left(g\right)$

$K_p=\frac{(P{O}_2{)}^3}{(P{O}_3{)}^2}$

Since, $P{O}_2>>P{O}_3$

$TotalPressure=P{O}_2+P{O}_3=8$

$P{O}_2=8-P{O}_3\approx 8$

$K_p=\frac{8\times 8\times 8}{(P_{O_3}{)}^2}$

$P_{O_3}=11.3\times {10}^{-7}atm$

Hence, option (B) is correct.