Respuesta :
Answer:
1.59 atm
Explanation:
The reaction is:
[tex]N_{2}O_{3}(g) - - -> NO_{2}(g)+NO(g)[/tex]
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:
[tex]P_{total}=P_{NO_{2}}+P_{NO}[/tex]
we don't include [tex]N_{2}O_{3}[/tex] 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 [tex]N_{2}O_{3}[/tex] is:
[tex]n_{N_{2}O_{3}}=\frac{P_{1}V}{RT_{1}}[/tex]
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 [tex]N_{2}O_{3}[/tex].
[tex]n_{NO_{2}}=\frac{P_{1}V}{RT_{1}}[/tex]
[tex]n_{NO}=\frac{P_{1}V}{RT_{1}}[/tex]
The partial pressure of each gas is:
[tex]P_{NO_{2}}=\frac{n_{NO_{2}*R*T_{2}}}{V}[/tex]
[tex]P_{NO}=\frac{n_{NO}*R*T_{2}}{V}[/tex]
so total pressure is:
[tex]P_{total}=(n_{NO_{2}}+n_{NO})*\frac{R*T_{2}}{V}[/tex]
replacing the moles we get:
[tex]P_{total}=(2*\frac{P_{1}V}{RT_{1}})*\frac{R*T_{2}}{V}[/tex]
We know that T2=3*T1
replacing this value in the equation we get:
[tex]P_{total}=2*\frac{P_{1}V}{RT_{1}}*\frac{R*3T_{1}}{V}[/tex]
[tex]P_{total}=6*P_{1} = 6*0.265 atm = 1.59 atm[/tex]
We have that for the Question "Calculate the final pressure of the gas mixture, assuming that the container volume does not change."
it can be said that
- The final pressure of the gas mixture, assuming that the container volume does not change = [tex]1.59 atm[/tex]
From the question we are told
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).
After the temperature is tripled,
[tex]N_2O_3 ---> NO_2 + NO[/tex]
using,
[tex]\frac{P_1V_1}{n_1T_1} = \frac{P_2V_2}{n_2T_2}\\\\\frac{0.265}{n*T} = \frac{P}{2n*3T}\\\\P = 6*0.265 \\\\= 1.59 atm[/tex]
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