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

Question

Chemistry

Dorian Morse

26 April, 11:08

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

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Rodgers · 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