Two moles of gas A2 are mixed with two moles of gas B2 in a flask of volume 1 lit. If at equilibrium 0.5 moles of A2 are obtained. Then find out Kp for reaction
A2(g) + B2(g) ⇌ 2AB(g)
(1) 12                  (2) 9                     (3) 4                   (4) 36

Dear Student,

The given reaction is :                                                                            
                                               A2(g)  + B2(g)  2AB(g)                                 
                           at t = 0          2             2                  0                      [concentration of A & B = no. of moles volume = 21= 2]       
                          at eqm          2-x          2-x                2x 


So, at eqm as concentration of A is given = 0.5 mol
            Hence, 2-x = 0.5 mol 
                           x = 1.5 mol
So, the remaining concentration of B = 0.5 mol 
And the formed concentration of C = 2x = 2 x 1.5 = 3 mol 

Now according to Law of Mass Action : 
                                  Kc = [C]2[A][B]     = 3×30.5 ×0.5     = 36
As, 
            Kp = Kc (RT)n                 [n = no. of moles of products - no. of moles of reactant] 
And for the reaction Δn = 0 
Hence K= K = 36

So, the correct option is (4) .

Regards.

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