Question

Consider the reaction. A(g)⇌2B(g) Find the equilibrium partial pressures of A and B for each of the different values of Kp. Assume that the initial partial pressure of B in each case is 1.0 atm and that the initial partial pressure of A is 0.0 atm. Make any appropriate simplifying assumptions. Find (solve for B and A) (a) Kp= 1.9 (b) Kp= 2.1×10−4 (c) Kp= 1.2×105

Answer 1

a)

A (g)     <---------------->   2 B (g)

0.0                                    1.0

x                                       1.0 - 2x

Kp = P²B / PA

1.9 = (1.0 - 2x)^2 / x

x = 0.195

equilibrium partial pressures :

PB = 0.61 atm

PA = 0.195 atm

b)

Kp = P²B / PA

2.1 x 10^-4 = (1.0 - 2x)^2 / x

x = 0.5

equilibrium partial pressures :

PB = 0.0 atm

PA = 0.50 atm

c)

Kp = P²B / PA

1.2 x 10^5 = (1.0 - 2x)^2 / x

x = 8.33 x 10^-6

equilibrium partial pressures :

PB = 1.0 atm

PA = 8.3 x 10^-6 atm