At 200 °C, the equilibrium constant (Kp) for the reactionbelow is 2.40 × 103. 2N

Question

At 200 °C, the equilibrium constant (Kp) for the reactionbelow is 2.40 × 103. 2NO (g) N2 (g) + O2 (g) A closed vessel is charged with 36.1 atm of NO. Atequilibrium, the partial pressure of O2 is __________ atm.
I end up using the quadratic formula but my answer is a crazyorder of magnitude. Thanks for your help. At 200 °C, the equilibrium constant (Kp) for the reactionbelow is 2.40 × 103. 2NO (g) N2 (g) + O2 (g) A closed vessel is charged with 36.1 atm of NO. Atequilibrium, the partial pressure of O2 is __________ atm.
I end up using the quadratic formula but my answer is a crazyorder of magnitude. Thanks for your help.

Explanation / Answer

                                2NO (g) ------> N2 (g) + O2 (g) initialpressure             36.1                  0            0 change                         -2x                  +x           +x equb.pressure           36.1-2x               x             x Equilibrium constant Kp = p N2 * pO2 / p 2NO                         2.4*10^3 = x * x / ( 36.1 -2x)^2                 2400 = x^2 / ( 36.1-2x)^2 ===> x / (36.1-2x) = 2400 = 48.98              36.1 - 2x = 0.0204 x            x = 17.86 atm So the partyial pressure of O2 ,  pO2 = x = 17.86 atm                         2.4*10^3 = x * x / ( 36.1 -2x)^2                 2400 = x^2 / ( 36.1-2x)^2 ===> x / (36.1-2x) = 2400 = 48.98              36.1 - 2x = 0.0204 x            x = 17.86 atm So the partyial pressure of O2 ,  pO2 = x = 17.86 atm

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