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For the reaction 2A(g)⇌B(g)+2C(g), a reaction vessel initially contains only A at a pressure of PA=270...

For the reaction 2A(g)⇌B(g)+2C(g), a reaction vessel initially contains only A at a pressure of PA=270 mmHg . At equilibrium, PA=55 mmHg .

Calculate the value of Kp. (Assume no changes in volume or temperature.)

Include many steps please.

Solutions

Expert Solution


                    pA                 pB                 pC               


initial             270.0               0                   0                 

change              -2x                 +1x                 +2x               

equilibrium         270.0-2x            +1x                 +2x               


Given at equilibrium,
pA = 55.0
270.0-2x = 55.0
x = 107.5

Equilibrium constant expression is
Kp = pB*pC^2/pA^2
Kp = (x)*(2x)^2/(270.0-2x)^2
Kp = (107.5)*(2*107.5)^2/(270.0-2*107.5)^2
Kp = 1.64*10^3

Answer: 1.64*10^3

queen_honey_blossom answered 1 year ago
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