Question

A $20L$ box contains $O_3$ and $O_2$ at equilibrium at $202K$. $K_p=2\times {10}^{14}$ for $2O_3\rightleftharpoons 3O_2$. Assume that $p_{O_3}>>p_{O_2}$ and total pressure is $8atm$, then the partial pressure of $O_3$ is:

• A. exactly $1.6\times {10}^{-6}atm$
• B. sufficiently less than $1.6\times {10}^{-6}atm$
• C. slightly more than $1.6\times {10}^{-6}atm$
• D. very slightly less than $1.6\times {10}^{-6}atm$

Answer 1

The correct option is D very slightly less than $1.6\times {10}^{-6}atm$

$K_p=\frac{P^3\left(o_2\right)}{P^2\left(o_3\right)}2o_3\rightleftharpoons 3o_2$

Total pressure $=8atm=P{o}_2+P{o}_3$

$P{o}_2=8-P{o}_3\simeq 8$

because $P{o}_3<<P{o}_2$ (neglected)

$K_p=\frac{(8{)}^3}{P_{o_3}^2}$

$2\times {10}^{14}=\frac{(8{)}^3}{P{o}_3^2}$

$P_{o_3}^2=\frac{(8{)}^3}{2\times {10}^{1}4}=2.56\times {10}^{-14}$

$=2.56\times {10}^{-12}$

$P_{o_3}^2=2.56\times {10}^{-12}$

$P{o}_3=1.6\times {10}^{-6}atm$

As $P{o}_3<<P{o}_2$

Thus very slightly less than $1.6\times {10}^{-6}$ atm