Be sure to answer all parts. Nitrogen dioxide decomposes according to the reaction 2 NO2 (g) ⇌ 2 NO (g) + O2 (g) where Kp = 4.48 * 10-13 at a certain temperature. If 0.65 atm of NO2 is added to a container and allowed to come to equilibrium, what are the equilibrium partial pressures of NO (g) and O2 (g) ?

Partial Pressure of NO = 9.12 x 10^ (-5) atm

Partial Pressure of O2 = 4.56 x 10^ (-5) atm

Explanation:

2 NO2 (g) ⇌ 2 NO (g) + O2 (g)

Now, Kp = 4.48 * 10^ (-13)

the initial pressure of NO2 that was added is 0.65 atm.

Now,

Partial pressure = mole fraction x total pressure

Let's say Total pressure is x;

So Partial pressure of (NO) = (1+1) x = 2x

Similarly Partial pressure of O2 = x

Now Kp = [ (2x) ² (x) ] / [0.65]²

Since Kp = 4.48 * 10^ (-13)

Thus, 4.48 * 10^ (-13) = [ (2x) ² (x) ] / [0.65]²

Multiply both sides by 0.65²

So, 4.48 x 0.65² x 10^ (-13) = 2x³

2x³ = 1.899 x 10^ (-13)

x³ = [ 1.889 x 10^ (-13) ] / 2

= 0.945 x 10^ (-13)

So x = ∛0.945 x 10^ (-13) = 4.56 x 10^ (-5) atm

So, Partial Pressure of NO = 2x = 2 (4.56 x 10^ (-5) atm) = 9.12 x 10^ (-5) atm

Also Partial Pressure of O2 = x = 4.56 x 10^ (-5) atm