Q>8 mole of a gas AB3 are introduced into a 1l vessel.It dissociates as:

2AB3(g)=A2(g)+3B2(g)

At equilibrium 2 mol of A2 are found to be present.The equilibrium constant of the reaction is:

(= is reverse reaction above)

Moles of B2=9

Moles of AB3=2

K={81(18)}/2

729{mol/lit}²

at eq.
A2=2 moles
B2=6 moles
AB3=8-4=4 moles
find partial pr.
and then find Eq cons

                 2AB3(g)       ----------->     A2(g)        +        3B2(g)

t=0             8                                      0                        0

t=t             8 - x                                 x/2                      3x/2

Now       

mole of A2 are 2