Question

8.
2AB (g)⇌ A2(g) +2B2(g) 5 mols of AB2, were heated in a closed vessel to constant temperature,
Equilibrium was attained by the time 2 moles of it dissociated. If the equilibrium pressure was 12 atm
the Kp of the reaction is
A) 1.72
B) 0.9
C) 0.27
D) 0.6​

Answer 1

Answer:

given = reaction =2AB (g)⇌ A2(g) +2B2(g)

             mole = 5mol of AB2

             pressure = 12 atm

to find = Kp

solution =

      1st find the mole of reaction,

            2AB (g)⇌ A2(g) +2B2(g)

initial   5mole

            -2x               x           2x

            5 - 2x           x            2x              ....(according to given conditions)

        2x = 2 mole

at the equallibrium mole AB2 = 3

∴ mole at A1 = 1 , B2 = 2

XAB2 =  3 / 3+ 2 +1 = 1/2

consider the pressue

PAB2 = 1/2 x 12 =  6atm

PA2 = 1/6 X 12 = 2 atm    

for Kp = (PA2) (PB2)²/ (PAB2)²

          =  2 X (4)²/ (6)² = 0.88

         = 0.88≅0.9

So the option B is correct. Kp of the reaction is0.9