Suppose 100 mL of NO at STP is mixed with 400 mL of O2 at STP.
2 NO (g) + O2 (g)---> 2 NO2 (g)
After the reaction goes to completion, what is the partial pressure of NO2 in the resulting mixture of gases at STP?
I know the relation between pressure and volume is P1V1=P2V2, but I am not sure how to use it in this question.
I believe the way to approach this problem is through mole fraction. With all materials in the gaseous phase one can use volume as mols; first determine the limiting reagent. 100 mL NO will give you 100 x (2 mol NO2/2 mols NO) = 100 mL NO if we had all of the O2 needed.
For 400 mL O2 x (2 mols NO2/1 mol O2) = 800 mL NO2.
This means NO is the limiting reagent. Therefore, we will use 100 mL NO2 and there will be zero left, we will use 50 mL O2 (100 mL NO x (1 mol O2/2 mol NO) which will leave 350 mL O2 after reaction (400 mL-50 mL = 350 mL) and we will have 100 mL NO formed.
mols NO = 0
mol O2 = 0.350/22.4 = ?
mol NO2 = 0.1/22.4 = ?
ntotal = mols O2 + mols NO2.
mol fraction NO2 = XNO2 = nNO2/total mols
Then pNO2 = XNO2*Ptotal.
Ptotal is 760 mm. Solve for pNO2.
Note that the same answer is obtained by using volume as mols; i.e.,
100 mL + 350 mL = 450 mL.
XNO2 = 100/450 = 0.222 etc.
I used PV=nRT for both the gases but I am not getting the right answer. Please help
Is the total pressure 1 atm because the questions says STP?
To solve this question, you can use the ideal gas law equation, which is given by PV = nRT. In this case, we are assuming that the gases behave ideally. We can use the given equation, P1V1 = P2V2, to determine the partial pressure of NO2.
1. Convert the volumes of NO and O2 to the number of moles at STP:
- For NO: Since 1 mole occupies 22.4 L at STP, we can calculate the number of moles of NO in 100 mL (0.1 L) by dividing the volume by 22.4 L/mol: 0.1 L / 22.4 L/mol = 0.00446 mol.
- For O2: Similarly, we can calculate the number of moles of O2 in 400 mL (0.4 L): 0.4 L / 22.4 L/mol = 0.0179 mol.
2. Use stoichiometry to determine the number of moles of NO2 produced:
From the balanced equation, 2 moles of NO react with 1 mole of O2 to produce 2 moles of NO2. Since we have 0.00446 mol of NO and 0.0179 mol of O2, the limiting reactant is NO. Therefore, all the NO will be consumed and produce twice as many moles of NO2, which is 0.00892 mol of NO2.
3. Calculate the total volume of gases:
The total initial volume of gases is 100 mL + 400 mL = 500 mL, which can be converted to 0.5 L.
4. Apply the ideal gas law equation, PV = nRT, to find the partial pressure of NO2:
- Plug in the values: P1 = initial pressure of NO2 (unknown), V1 = initial volume of gases (0.5 L), n1 = initial number of moles of gases (0.00892 mol), T1 = initial temperature (STP, 273 K), R = ideal gas constant (0.0821 L·atm/K·mol).
- Rearrange the equation to solve for P1: P1 = (n1 * R * T1) / V1.
- Substitute the values: P1 = (0.00892 mol * 0.0821 L·atm/K·mol * 273 K) / 0.5 L.
- Calculate the result: P1 ≈ 12.36 atm.
Therefore, the partial pressure of NO2 in the resulting mixture of gases at STP is approximately 12.36 atm.