Respuesta :
Answers:
p(NO) = 0; p(O₂) = 0.300 atm; p(NO₂) = 0.400 atm
Step-by-step explanation:
Step 1. Moles of NO
pV = nRT
n = (pV)/(RT)
p = 0.500 atm
R = 0.082 06 L·atm·K⁻¹mol⁻¹
T = (22 + 273.15)
T = 295.15 K
n = (0.500 × 6.00)/(0.082 06 × 295.15)
n = 3.00/24.22
n = 0.1239 mol
===============
Step 2. Moles of O₂
p = 2.50 atm
V = 1.50 L
n = (2.50 × 1.50)/24.22
n = 3.75/24.22
n = 0.1548 mol
===============
Step 3. Identify the limiting reactant
2NO + O₂ ⟶ 2NO₂
n/mol: 0.1239 0.1548
Calculate the moles of NO₂ we can obtain from each reactant.
From NO:
The molar ratio of NO₂: NO is 2:2
Moles of NO₂ = 0.1239 × 2/2
Moles of NO₂ = 0.1239 mol NO₂
From O₂:
The molar ratio of NO₂: O₂ is 2:1.
Moles of NO₂ = 0.1548 × 2/1
Moles of NO₂ = 0.3097 mol NO₂
NO is the limiting reactant because it gives the smaller amount of NO₂.
===============
Step 4. Moles of each species after reaction
2NO + O₂ ⟶ 2NO₂
I/mol: 0.1239 0.1548 0
C:/mol: -0.1239 -0.06193 +0.1239
F/mol: 0 0.0929 0.1239
===============
Step 5. Partial pressure of NO
p(NO) = 0
===============
Step 6. Partial pressure of O₂
pV = nRT
p = (nRT)/V
V= 6.00 + 1.50
V = 7.50 L
p = (0.0929 × 24.22)/7.50
p = 2.25/7.50
p = 0.300 atm
===============
Step 7. Partial pressure of O₂
p = (0.1239 × 24.22)/7.50
p = 3.00/7.50
p = 0.400 atm
p(NO) = 0; p(O₂) = 0.300 atm; p(NO₂) = 0.400 atm
The partial pressure of NO is 0 atm, and [tex]\rm NO_2[/tex] is 0.400 atm at the end of the reaction.
The concentration of NO and [tex]\rm O_2[/tex] initially are;
By ideal gas equation:
PV =nRT
P = pressure
V = volume
n = moles
R = constant = 0.0816 L.atm/mol.K
T = temperature = 22 [tex]\rm ^\circ C[/tex] = 273 + 22 K = 295 K
Moles of NO = [tex]\rm \frac{PV}{RT}[/tex]
Moles of NO = [tex]\rm \frac{0.5\;\times\;6}{0.0816\;\times\;295}[/tex]
Moles of NO = 0.1239 moles
Moles of [tex]\rm O_2[/tex] = [tex]\rm \frac{2.50\;\times\;1.50}{0.0816\;\times\;295}[/tex]
Moles of [tex]\rm O_2[/tex] = 0.1548 moles
The reaction will be:
[tex]\rm 2\;NO\;+\;O_2\;\rightarrow\;2\;NO_2[/tex]
2 moles of NO gives 2 moles of [tex]\rm NO_2[/tex]
So, 0.1239 moles of NO gives 0.1239 moles of [tex]\rm NO_2[/tex]
1 mole of [tex]\rm O_2[/tex] gives 2 moles of [tex]\rm NO_2[/tex]
0.1548 moles of [tex]\rm O_2[/tex] gives 0.3097 moles of [tex]\rm NO_2[/tex]
Since No produces smaller amount of [tex]\rm NO_2[/tex], it is the limiting factor.
Partial pressure of NO = 0 since it is completely used.
Partial pressure of [tex]\rm NO_2[/tex] = [tex]\rm \frac{nRT}{V}[/tex]
n = 0.1239 moles
Total volume = 6 + 1.50
V = 7.5 L
Partial pressure of [tex]\rm NO_2[/tex] = [tex]\rm \frac{0.0929\;\times\;0.0816\;295}{7.5}[/tex]
Partial pressure of [tex]\rm NO_2[/tex] = 0.400 atm
For more information, refer the link:
https://brainly.com/question/8408970?referrer=searchResults
