Equilibrium and using ICE tables

Question

For the reaction 2 NO₂(g) ⇌ 2 NO(g) + O₂(g) Kp = 1.11 × 10⁻⁵ at 200 °C. A 2.50 L vessel at 200 °C is filled with NO₂(g) at an initial pressure of 4.00 atm and allowed to come to equilibrium. What will be the pressure (in atm) of NO(g) at equilibrium?

Answer 1

2 NO₂(g) ⇌ 2 NO(g) + O₂(g)

4.00 atm.......0 atm.....0 atm.....Initial

-2x..............+2x.........+x...........Change

4-2x..............2x...........x...........Equilibrium

Kp = 1.11x10^-5 = (NO)²(O₂) / (NO₂)²

1.11x10^-5 = (2x)²(x) / (4-2x)² (since Kp is low, ignore -2x in denominator to avoid using quadratic eq.)

1.11x10^-5 = (2x)²(x) / (4)²

1.11x10^-5 = 4x³ / 16

4x³ = 1.78x10^-4

x³ = 4.44x10^-5

x = 3.54x10^-2 atm (note: this is less than 1% of the 4 atm so above assumption was valid)

At equilibrium (see ICE table) we have the following pressure for NO(g):

NO = 2x = (2)(3.54x10^-2 atm) = 7.08x10^-2 atm