Question

A container of N2O3(g) has a pressure of 0.265 atm. When the absolute temperature of the N2O3(g) is tripled, the gas completely decomposes, producing NO2(g) and NO(g). Calculate the final pressure of the gas mixture, assuming that the container volume does not change.

Answer 1

Answer:

1.59 atm

Explanation:

The reaction is:

$N_{2}O_{3}(g) - - -> NO_{2}(g)+NO(g)$

The dalton's law tell us that the total pressure of a mixture of gases is the sum of the partial pressure of every gas.

So after the reaction the total pressure is:

$P_{total}=P_{NO_{2}}+P_{NO}$

we don't include $N_{2}O_{3}$ because it decomposed completely.

Assuming  ideal gases

PV=nRT

P= pressure, V= volume of the container, n= mol of gas, R=constant of gases and T=temperature.

so moles of $N_{2}O_{3}$ is:

$n_{N_{2}O_{3}}=\frac{P_{1}V}{RT_{1}}$

from the  reaction stoichiometry (1:1) we have that after the reaction the number of moles of each product is the same number of moles of $N_{2}O_{3}$.

$n_{NO_{2}}=\frac{P_{1}V}{RT_{1}}$

$n_{NO}=\frac{P_{1}V}{RT_{1}}$

The partial pressure of each gas is:

$P_{NO_{2}}=\frac{n_{NO_{2}*R*T_{2}}}{V}$

$P_{NO}=\frac{n_{NO}*R*T_{2}}{V}$

so total pressure is:

$P_{total}=(n_{NO_{2}}+n_{NO})*\frac{R*T_{2}}{V}$

replacing the moles we get:

$P_{total}=(2*\frac{P_{1}V}{RT_{1}})*\frac{R*T_{2}}{V}$

We know that T2=3*T1

replacing this value in the equation we get:

$P_{total}=2*\frac{P_{1}V}{RT_{1}}*\frac{R*3T_{1}}{V}$

$P_{total}=6*P_{1} = 6*0.265 atm = 1.59 atm$