Question

Nitrogen dioxide decomposes according to the reaction 2 no2(g) ⇌ 2 no(g) + o2(g) where kp = 4.48 × 10−13 at a certain temperature. if 0.70 atm of no2 is added to a container and allowed to come to equilibrium, what are the equilibrium partial pressures of no(g) and o2(g)

Answer 1

Answer:Partial pressure of the oxygen gas at equilibrium:

$p_{O_2}=3.8\times 10^{-5} atm$

Partial pressure of nitrogen monoxide gas at equilibrium:

$p_{NO}=7.6\times 10^{-5} atm$

Explanation;

Initial Partial pressure of nitrogen dioxide gas $NO_2$= p = 0.70 atm

The value of $K_p$ for the reaction $=4.48\times 10^{-13}$

         $2NO_2\rightleftharpoons 2NO+O_2$

Initially     p

At eq'm  p-2x          2x   x

$K_p=\frac{[2x]^2[x]}{[p-2x]^2}$

$4.48\times 10^{-13}=\frac{4x^3}{[0.70-2x]^2}$

On solving for the x

$x=3.8\times 10^{-5} atm$

Partial pressure of the oxygen gas at equilibrium:

$p_{O_2}=x=3.8\times 10^{-5} atm$

Partial pressure of nitrogen monoxide gas at equilibrium:

$p_{NO}=2x=2\times 3.8\times 10^{-5} atm=7.6\times 10^{-5} atm$