Nitric oxide (NO) reacts with molecular oxygen as follows:
2NO(g) + O2(g) �¨ 2NO2(g)
Initially NO and O2 are separated as shown below. When the valve is opened, the reaction quickly goes to completion. Determine what gases remain at the end of the reaction and calculate their partial pressures. Assume that the temperatures remains constant at 25 Celsius.
Calculate the partial pressure of NO, O2 and NO2.
Volume of NO= 4.00L
Pressure of No= 0.500atm
Pressure of O2= 1.97 atm
Volume of O2 = 1.00 L
PNO2= 2.00 atm x L/(4.00 L + 1.97 L)= 0.335 atm
PO2= 0.97 atm x L/ (4.00 L + 1.97 L)= 0.162 atm
PNO=????
Show step by step how do I get PNO.
The answer is 0
To determine the partial pressures of NO, O2, and NO2 at the end of the reaction, we need to use the ideal gas law equation:
PV = nRT
where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, let's calculate the number of moles for each gas using the ideal gas law:
For NO:
P(NO) = 0.500 atm
V(NO) = 4.00 L
T = 25 + 273.15 = 298.15 K (convert Celsius to Kelvin)
Substituting these values into the ideal gas law equation:
n(NO) = (P(NO) * V(NO)) / (R * T)
n(NO) = (0.500 atm * 4.00 L) / (0.0821 L·atm/mol·K * 298.15 K)
n(NO) ≈ 0.0808 moles
For O2:
P(O2) = 1.97 atm
V(O2) = 1.00 L
T = 25 + 273.15 = 298.15 K
Substituting these values into the ideal gas law equation:
n(O2) = (P(O2) * V(O2)) / (R * T)
n(O2) = (1.97 atm * 1.00 L) / (0.0821 L·atm/mol·K * 298.15 K)
n(O2) ≈ 0.0812 moles
Since the reaction goes to completion, all of the NO will react to form NO2. Therefore, the number of moles of NO2 at the end of the reaction will be twice the number of moles of NO:
n(NO2) = 2 * n(NO)
n(NO2) = 2 * 0.0808 moles
n(NO2) ≈ 0.1616 moles
Now, we can calculate the partial pressures of the gases. Since the reaction goes to completion, all of the NO and O2 will be consumed, and the remaining gas will be NO2:
Partial pressure of NO2 = n(NO2) * R * T / V(total)
V(total) = V(NO) + V(O2) = 4.00 L + 1.00 L = 5.00 L
Substituting the values:
Partial pressure of NO2 = (0.1616 moles * 0.0821 L·atm/mol·K * 298.15 K) / 5.00 L
Partial pressure of NO2 ≈ 0.783 atm
Since all the NO and O2 has reacted, the partial pressures of NO and O2 will be zero.
Therefore, at the end of the reaction:
Partial pressure of NO = 0 atm
Partial pressure of O2 = 0 atm
Partial pressure of NO2 = 0.783 atm
2NO + O2 ==> 2NO2
mols NO =?? Use PV = nRT
moles O2 = ?? Use PV = nRT
From moles determine the identity of the limiting reagent, use that determine how much of the other reagent remains unreacted, then use PV = nRT to determine pressure of the "other" reagent and pressure o the product. Post your work if you get stuck.