In: Chemistry
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.
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