A 100 ml flask containing O₂ at 1.2 atm and 300 K, is connected to a 250 ml flask containg NO(g) at 0.6 atm and 300 K, by means of a narrow tube of negligible volume, when they combine quantitatively to form NO_2. Finally NO₂(g) dimerized, partially into N₂O₄(g) and pressure inside the flask was found to be 0.41 atm at the same temperature. Find percentage dimerizationof NO₂ as :

2NO₂(g) (g)₄O₂ N

To solve the problem of determining the percentage dimerization of NO2 into N2O4, we need to follow a systematic approach. We will start by calculating the initial moles of O2 and NO, then find out how they react to form NO2, and finally determine how much of the NO2 dimerizes into N2O4. Let’s break this down step by step.

Step 1: Calculate Initial Moles of O2 and NO

Using the ideal gas law, we can find the number of moles (n) of each gas. The ideal gas law is given by:

PV = nRT

Where:

• P = pressure in atm
• V = volume in liters
• n = number of moles
• R = ideal gas constant (0.0821 L·atm/(K·mol))
• T = temperature in Kelvin

For O2 in the 100 ml flask:

• P = 1.2 atm
• V = 0.1 L (100 ml)
• T = 300 K

Calculating moles of O2:

n(O2) = (1.2 atm * 0.1 L) / (0.0821 L·atm/(K·mol) * 300 K) ≈ 0.0049 moles

For NO in the 250 ml flask:

• P = 0.6 atm
• V = 0.25 L (250 ml)
• T = 300 K

Calculating moles of NO:

n(NO) = (0.6 atm * 0.25 L) / (0.0821 L·atm/(K·mol) * 300 K) ≈ 0.0061 moles

Step 2: Reaction of O2 and NO to Form NO2

The balanced chemical equation for the reaction is:

2 NO + O2 → 2 NO2

From the stoichiometry of the reaction, 2 moles of NO react with 1 mole of O2 to produce 2 moles of NO2. We need to determine the limiting reactant:

Using the moles calculated:

• From 0.0049 moles of O2, we can produce 0.0049 * 2 = 0.0098 moles of NO2.
• From 0.0061 moles of NO, we can produce 0.0061 moles of NO2 (since it’s a 1:1 ratio).

Thus, NO is the limiting reactant, and we will produce 0.0061 moles of NO2.

Step 3: Total Moles After Reaction

After the reaction, we have:

• 0.0061 moles of NO2 produced.
• Remaining O2 = 0.0049 - (0.0061/2) = 0.0049 - 0.00305 = 0.00185 moles.
• NO is completely consumed.

Total moles after the reaction:

Total moles = moles of NO2 + moles of remaining O2 = 0.0061 + 0.00185 = 0.00795 moles.

Step 4: Dimerization of NO2

Next, we consider the dimerization of NO2 into N2O4:

2 NO2 ⇌ N2O4

Let x be the amount of NO2 that dimerizes. Therefore, the moles of NO2 remaining will be:

• NO2 remaining = 0.0061 - x
• N2O4 formed = x/2

The total moles after dimerization will be:

Total moles = (0.0061 - x) + (x/2) = 0.0061 - x/2

We know the total pressure after dimerization is 0.41 atm. Using the ideal gas law again:

P = (nRT)/V

0.41 atm = [(0.0061 - x/2) * 0.0821 * 300] / (0.35 L)

Solving for x:

0.41 atm * 0.35 L = (0.0061 - x/2) * 0.0821 * 300

0.1435 = (0.0061 - x/2) * 24.63

0.1435 / 24.63 = 0.0061 - x/2

0.00583 = 0.0061 - x/2

x/2 = 0.0061 - 0.00583 = 0.00027

x = 0.00054 moles of NO2 dimerized.

Step 5: Calculate Percentage Dimerization

To find the percentage dimerization of NO2:

Percentage dimerization = (moles dimerized / initial moles of NO2) * 100

Percentage dimerization = (0.00054 / 0.0061) * 100 ≈ 8.85%

Thus, the percentage of dimerization of NO2 into N2O4 is approximately 8.85%.