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