Question

Consider the following reaction: A(g)⇌2B(g) Find the equilibrium partial pressures of A and B for Kp=2.0 . 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.

Answer 1

Kp for A.<---> 2B since there is no A in the begining of reaction and only A the reverse reaction is considered.

2= [PB^]2/ [PA]

where PA and PB are partial pressure of A and B repsectively

for the reverse reaction Kp =1/2 = [PA]/ [PB]²

PB= 1 atm PA=0

At equilibrium let x drop in pressure

At equilibrium     PB= 1-x   PA= x, x/(1-x)²= 1/2

2x= 1+x²-2x therefore x2-4x+1=0 solving the quadratic equation x= 0.2679

Partial pressures at equilibrium PA= 0.2679 atm and PB= 1-0.2679=0.7321 atm