The picture below shows two bulbs connected by a stopcock. the large bulb, with a volume of 6.00 l, contains nitric oxide at a pressure of 0.500 atm, and the small bulb, with a volume of 1.50 l, contains oxygen at a pressure of 2.50 atm. the temperature at the beginning and the end of the experiment is 22 °c. what are the partial gasses of no, and no2?

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 is 0.400 atm at the end of the reaction.

The concentration of NO and 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 = 273 + 22 K = 295 K

Moles of NO =

Moles of  NO =

Moles of NO = 0.1239 moles

Moles of =

Moles of = 0.1548 moles

The reaction will be:

2 moles of NO gives 2 moles of

So, 0.1239 moles of NO gives 0.1239 moles of

1 mole of gives 2 moles of

0.1548 moles of gives 0.3097 moles of

Since No produces smaller amount of , it is the limiting factor.

Partial pressure of NO = 0 since it is completely used.

Partial pressure of =

n = 0.1239 moles

Total volume = 6 + 1.50

V = 7.5 L

Partial pressure of =

Partial pressure of = 0.400 atm

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in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.