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:

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

Answer 2

Partial pressure is the individual pressure of the gas if alone in a mixture. The partial pressure of oxygen gas is $\rm 3.8 \times 10 ^{-13} atm$ and of nitrogen monoxide is $\rm 7.6 \times 10 ^{-5} atm$.

What is partial pressure?

If the individual gas from the mixture occupies the container alone then the pressure exerted by it is called partial pressure.

The initial partial pressure (p) of the nitrogen dioxide gas is 0.70 atm and Kp is $4.8 \times 10 ^{-13}$.

The reaction can be shown as,

$\rm 2NO_{2} \leftrightarrow 2NO + O_{2}$

The initial concentration of the reactant is p and at equilibrium, it is $\rm (p - 2x)$ and that of the product is 2x and x.

Kp from the reaction can be shown as,

$\begin{aligned}\rm K_{p} &= \rm \dfrac{[2x]^{2}[x]}{[p - 2x]^{2}}\\\\4.48 \times 10^{-13} & = \rm \dfrac{4x^{2}}{[0.70 - 2x]^{2}}\\\\\rm x &= 3.8 \times 10^{-5}\;\rm atm\end{aligned}$

The partial pressure of the oxygen $(\rm pO_{2})$ at equilibrium is $\rm x = 3.8 \times 10^{-5}\;\rm atm$.

The partial pressure of nitrogen monoxide $(\rm pNO)$ will be:

$\begin{aligned} \rm 2 x &= 2 \times 3.8 \times 10^{-5}\;\rm atm\\\\& = 7.6 \times 10^{-5}\;\rm atm\end{aligned}$

Therefore partial pressure of oxygen is $\rm 3.8 \times 10 ^{-13} atm$ and of nitrogen monoxide is $\rm 7.6 \times 10 ^{-5} atm$.

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