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

Answer is to avoid experimental error