Answer:
P(O2) = 0.300 atm
P(NO2) = 0.400 atm
P(NO) = 0 atm
Explanation:
Step 1: Data given
Volume of large bulb = 6.00L
Large bulb contains nitric oxide at 0.500 atm
Volume of the small bulb = 1.50 L
Small bulb contains oxygen at 2/50 atm
The initial temperature = 22.0°C
Step 2: The balanced equation
2 NO + O2 → 2 NO2
Step 3: Calculate moles of NO
P*V=n*R*T
n = (P*V)/(R*T)
⇒ with P = the pressure of NO = 0.500 atm
⇒ with V = the volume of NO = 6.00L
⇒ with R = the gas constant = 0.08206 L*atm/K*mol
⇒ with T = the temperature = 22.0 °C = 295 K
⇒ with n = the number of moles of NO
n(NO) = (0.500 *6.00)/(0.08206*295)
n(NO) = 0.124 moles
Step 4: Calculate moles of O2
n = (P*V)/(R*T)
⇒ with P = the pressure of O2 = 2.50 atm
⇒ with V = the volume of 02 = 1.50 L
⇒ with R = the gas constant = 0.08206 L*atm/K*mol
⇒ with T = the temperature = 22.0 °C = 295 K
⇒ with n = the number of moles of O2
n(O2) = (2.50*1.50)/(0.08206*295)
n(O2) = 0.155 moles
Step 5: Calculate the limiting reactant
For 2 moles NO consumed, we need 1 mole O2 to produce 2 moles of NO2
No is the limiting reactant. It will completely be consumed. (0.124 moles)
O2 is in excess. There will react 0.124/2 = 0.062 moles O2
There will remain 0.155 - 0.062 = 0.093 moles O2
Step 6: Calculate moles of NO2
For 2 moles NO consumed, we need 1 mole O2 to produce 2 moles of NO2
For 0.124 moles NO, we'll have 0.124 moles NO2
Step 7: Calculate partial pressure of O2
P*V= n*R*T
P = (n*R*T) / V
⇒ n = the number of moles O2 = 0.093 moles O2 remain
⇒ R = the gas constant = 0.08206 L*atm/K*mol
⇒ T = the temperature = 295 Kelvin
⇒ V = the volume = 6.00 + 1.50 = 7.50 L
P(O2) = 0.300 atm
Step 8: Calculate partial pressure of NO2
P = (n*R*T) / V
⇒ n = the number of moles NO2 = 0.124 moles
⇒ R = the gas constant = 0.08206 L*atm/K*mol
⇒ T = the temperature = 295 Kelvin
⇒ V = the volume = 6.00 + 1.50 = 7.50 L
P(NO2) = 0.400 atm
P(NO) = 0 atm because NO is completely consumed.
