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

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