# 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)

Prajwal kr
49 Points
9 years ago

Moles of B2=9

Moles of AB3=2

K={81(18)}/2

729{mol/lit}2

Akash Kumar Dutta
98 Points
9 years ago

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

Vikas TU
14149 Points
9 years ago

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

t=0             8                                      0                        0

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

Now

At equilibrium 2 mol of A2 are found

x/2    = 2

x = 4

t=t             4                                 2                      6

thus,

 equilibrium constant  K = (A2) (B2)3/(AB3)2 Neglecting the volume 1 litre.  K = 22 x 63 / 42 K = 4 x 216 / 16 K = 54

plz approve1

pushpendra bansal
19 Points
9 years ago

mole of A2 are 2

 mole of b2 are 6 mole of ab3 are 8-4=4 equlilibruim constant is equal to 2*6*6*6/4*4=27