2NO + O2 -> 2NO2
200. mL of NO at STP is reacted with 500. mL O2of at STP. Calculate the partial pressure of NO2 after the reaction goes to completion. Assume a constant volume.
Constant volume? Exactly what is the volume...200ml, 500ml, or 700ml. It would be interesting to see how this reaction was accomplished. What was the temperature of the vessel when completed?
Of NO2? it's not given.
That's all they gave me..
To calculate the partial pressure of NO2 after the reaction goes to completion, we need to determine the number of moles of NO and O2, and then use the coefficients from the balanced chemical equation to determine the moles of NO2 produced.
First, let's calculate the number of moles of NO and O2 using the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Since the gases are at STP (Standard Temperature and Pressure), we can use the following values:
- Temperature (T) = 273.15 K
- Pressure (P) = 1 atm (STP condition)
For NO:
- Volume (V) = 200 mL = 0.2 L
Using PV = nRT, we can rearrange the equation to solve for n (number of moles):
n(NO) = (P(NO) * V(NO)) / (R * T)
Substituting the values:
n(NO) = (1 atm * 0.2 L) / (0.08206 L * atm / mol * K * 273.15 K)
n(NO) ≈ 0.00924 mol (rounded to 5 decimal places)
Similarly, for O2:
- Volume (V) = 500 mL = 0.5 L
Using PV = nRT, we can solve for n (number of moles):
n(O2) = (P(O2) * V(O2)) / (R * T)
Substituting the values:
n(O2) = (1 atm * 0.5 L) / (0.08206 L * atm / mol * K * 273.15 K)
n(O2) ≈ 0.0246 mol (rounded to 4 decimal places)
According to the balanced chemical equation, 2 moles of NO react to produce 2 moles of NO2:
2NO + O2 -> 2NO2
Therefore, since the reaction goes to completion, the number of moles of NO2 produced is also 0.00924 mol.
Finally, let's calculate the partial pressure of NO2. Since the volume remains constant, the partial pressure of any gas is given by:
Partial Pressure = (moles of the gas) * (R) * (Temperature) / (Volume)
Using:
- n(NO2) = 0.00924 mol
- R = 0.08206 L * atm / mol * K
- T = 273.15 K
- V = 0.2 L (volume of NO)
Partial Pressure of NO2 = (0.00924 mol) * (0.08206 L * atm / mol * K) * (273.15 K) / (0.2 L)
Partial Pressure of NO2 ≈ 3.16 atm (rounded to 2 decimal places)
Therefore, the partial pressure of NO2 after the reaction goes to completion is approximately 3.16 atm.