1.00 atm N2O5 is placed in an evacuated vessel and reacts according to the equation 2 N2O5(g) ⇄ 4 NO2(g) + O2(g) When equilibrium is reached, the partial pressure of NO2 is found to be 1.60 atm. Calculate Kp for this reaction?

Kp is the equilibrium constant based on the pressure, and it depends only on the gas substances. It is the multiplication of partial pressure of the products (elevated by their coefficients) divided by the multiplication of partial pressure of the reactants (elevated by their coefficients).

Making an equilibrium table:

2N₂O₅(g) ⇄ 4 NO₂(g) + O₂(g)

1.00 0 0 Initial

-2x +4x x Reacts (the stoichiometry is 2:4:1)

1-2x 4x x Equilibrium

So:

4x = 1.60

x = 0.40 atm

pN₂O₅ = 1 - 2*0.40 = 0.20 atm

pNO₂ = 1.60 atm

pO₂ = 0.40 atm

[tex]Kp = \frac{(pNO2)^4*(pO2)}{(pN2O5)^2}[/tex]

[tex]Kp = \frac{(1.60)^4*0.40}{(0.20)^2}[/tex]

[tex]Kp = \frac{10.24*0.40}{0.04}[/tex]

Kp = 10.24