You should get into the habit of using an arrow; otherwise we must guess which are products and which are reactants.
..........N2O4 ==> 2NO2
initial....5.3atm....0
change.....-p.....2p
equil..... 5.3-p....2p
Kp = 0.25 = (NO2O^2/(N2O4)
Substitute from the ICE chart into Kp expression and solve for p.
Then 2p = PNO2 and 5.3-p = PN2O4.
For #2,
...........N2O4 ==> 2NO2
initial......0........7
change......+p........-p
equil........p.......7-p
Substitute into Ka and solve for p.
At a particular temperature, Kp = 0.25 for the following reaction.
N2O4(g) 2 NO2(g)
(a) A flask containing only N2O4 at an initial pressure of 5.3 atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases.
(b) A flask containing only NO2 at an initial pressure of 7.0 atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases.
2 answers
Solutions