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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)

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)

Grade:

4 Answers

Prajwal kr
49 Points
8 years ago

Moles of B2=9

Moles of AB3=2

K={81(18)}/2

 

729{mol/lit}2

Akash Kumar Dutta
98 Points
8 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
8 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
8 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

 


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