The gas A₂ in the left flask allowed to react with gas B₂ present in right flask as 
${\mathrm{A}}_2\left(\mathrm{g}\right)+{\mathrm{B}}_2\left(\mathrm{g}\right) \rightleftharpoons 2\mathrm{AB}\left(\mathrm{g}\right); {\mathrm{K}}_{\mathrm{c}}=4 \mathrm{at} 27°\mathrm{C}.$
What is the concentration of AB when equilibrium is established?

Please find this answer

it is 1/3

This is the solution

Please find this answer

the correct answer is 0.66

The reaction starts only when the tap is open .
So, when the tap opens the volume of the actual mixture will be = 1+3 = 4 lt.

so, the initial concentration of a is = moles of a /vol. of solution (in lt)
                                                       = 2/4 = 0.5 mole
     the initial concentration of b is  = moles of b/vol.of solution (in lt)
                                                       = 4/4 = 1 mole .
                         
                                   a2 + b2 - 2ab
     at t=0                      0.5    1        0
      at t=equilibrium       0.5 -x     1-x       2x

                                    
                                              kc =  (2x)^2/(0.5-x)(1-x)
                                            given that kc = 4 
                                   solve this you get  x = 0.33 
       but as you see that at equilibrium the concentration of ab is 2x hence ,
                                                      the required answer is 2*0.33 = 0.66 moles                                                                        

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