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

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"Mars is an excellent place to investigate this question because it is the most similar planet to Earth in the Solar System. Evidence suggests that Mars was once full of water, warmer and had a thicker atmosphere, offering a potentially habitable environment."-

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