For the reaction 2AgBg2Cg a reaction vessel initially contai
For the reaction 2A(g)B(g)+2C(g), a reaction vessel initially contains only A at a pressure of PA=250mmHg.
At equilibrium, PA=71mmHg.
Calculate the value of Kp. (Assume no changes in volume or temperature.)
Solution
2A (g) ---------------- B(g) + 2 C(g)
250mm 0 0
-2x +x +2x
250 -2x +x +2x
At equilibrium
PA= 71mm
250 - 2x = 71
2x= 179
x= 89.5 mm
at equilibrium
partial pressure of B= PB= 89.5 atm
Partial pressure of C= 2x = 2x(89.5) = 179 atm
Kp = PB xP^2C/P^2A
Kp = (89.5) x(179)^2/(71)^2 =568.87
Kp = 568.87
