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Consider the following reaction (assume an ideal gas mixture): 2NOBr(g) <--> 2NO(g) + Br2(g) A 1.0-liter...

Consider the following reaction (assume an ideal gas mixture): 2NOBr(g) <--> 2NO(g) + Br2(g) A 1.0-liter vessel was initially filled with pure NOBr, at a pressure of 0.893 atm, at 300 K. After equilibrium was established, the partial pressure of NOBr was 0.242 atm. What is Kp for the reaction? Report answer to 4 decimal points.

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Answer #1

        2NOBr(g) <--> 2NO(g) + Br2(g)

initial 0.893 atm         0        0

change    0.651           0.651   0.3255

equil     0.242         0.651   0.3255


    Kp = pNO^2*pBr2/pNOBr^2

       = 0.651^2*0.3255/0.242^2

       = 2.3555

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