Question

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 0.30 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)?

Answer 1

Let; we have given,

Nitrogen dioxide decomposition reaction,

2NO_2(g) 2NO_(g) + O_2(g)

Where, Kp = 4.48x10^-13

Also given, P_NO2 =0.30 atm

Drawing an ICE table for the given equilibrium,

	2NO2	2NO	O2
I	0.30	0	0
C	-2x	+2x	+x
E	0.30-2x	+2x	+x

Now, Kp = (P_NO)² x (P_O2)/ (P_NO2)²

Substituting the equilibrium values of P_NO , P_O2 , P_NO2 , we get,

4.48x10^-13 = [2x]² x [ x] / [0.30-2x]² Since, x<<<<0.30 thus (0.30-2x) = 0.30

4.48x10^-13 = 4x³ / [0.30]²

4.48x10^-13 = 4x³ / 0.09

x³ = 1.008 x10^-14

x = 2.16x10^-5

So, Equillibrium partial pressure of O₂ (P_O2)= x = 2.2x10^-5 atm

Equillibrium partial pressure of NO(P_NO) = 2x = 4.3 x10^-5 atm