Question

The equilibrium constant kp for the thermal decomposition of no2 is 6.5 × 10–6 at 450°c. if a reaction vessel at this temperature initially contains 0.500 atm no2, what will be the partial pressure of no2 in the vessel when equilibrium has been attained

Answer 1

We let x be the pressure of each product at equilibrium, giving this ICE table:
2NO2 (g) ↔ 2NO (g) + O2 (g)
Initial pressure (atm): 0.500 0 0
Change (atm): -2x +2x +x
Equilibrium (atm): 0.500-2x 2x x

We can calculate x from Kp:
Kp = [NO]^2 [O2] / [NO2]^26.5x10^-6 = (2x)^2 (x) / (0.500-2x)^2

Approximating that 2x is negligible compared to 0.500 simplifies the equation to
6.5x10^-6 = (2x)^2 (x)/(0.500)^2 = 4x^3 / (0.500)^2

Then we solve for x:
x3 = (6.5x10^-6)(0.500)^2 / 4
x = 0.00741

The pressure of NO2 at equilibrium is therefore
Pressure of NO2 = 0.500-2x = 0.500 - 2(0.00741) = 0.4852 atm