The equilibrium constant Kp for the thermal decomposition of NO2 is 6.5x 10^-6 a

Question

The equilibrium constant Kp for the thermal decomposition of NO2 is 6.5x 10^-6 at 450 degrees C.  

2 NO2 (g) <--> 2NO (g) + O2 (g)

If the reaction vessel at this temperature initially contains 0.500 atm NO2, what will be the partial pressure

of NO2, NO, and O2 in the vessel when equilibrium has been attained?

P NO2 = ?

P NO = ?

P O2 = ?

The equilibrium constant Kp for the thermal decomposition of NO2 is 6.5x 10^-6 at 450 degrees C. 2 NO2 (g) 2NO (g) + O2 (g) If the reaction vessel at this temperature initially contains 0.500 atm NO2, what will be the partial pressure of NO2, NO, and O2 in the vessel when equilibrium has been attained? P NO2 = ? P NO = ? P O2 = ?

Explanation / Answer

(K_p = [P_{O_2}][P_{NO_2}]^2/[NO_2]^2)

(6.5 imes 10^{-6} = [x]^2[x/2]/[0.5-x]^2)

(x = 14.524 imes 10^{-3}) atm

P NO2 = 0.4855 atm

P NO = 0.01452 atm

P O2 = 0.00726 atm

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