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 K_p for reaction
A₂(g) + B₂(g) ⇌ 2AB(g)
(1) 12                  (2) 9                     (3) 4                   (4) 36

Dear Student,

The given reaction is :                                                                            
                              $A_2\left(g\right)\hspace{0.17em} + B_2\left(g\right) \stackrel{}{\rightleftharpoons } 2AB\left(g\right)$                                 
                           at t = 0          2             2                  0                      [concentration of A & B = $\frac{no. of moles }{volume }= \frac{2}{1}= 2$]       
                          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 : 
                                  $$\begin{gathered} K_c = \frac{[C{]}^2}{\left[A\right]\left[B\right]} \\ = \frac{3\times 3}{0.5 \times 0.5} \\ = 36 \end{gathered}$$
As, 
            $K_{p }= K_c (\hspace{0.17em}RT{)}^{{∆}_n} [{∆}_n = no. of moles of products - no. of moles of reac\tan t]$
And for the reaction Δ_n = 0 
Hence K_p = K_c  = 36

So, the correct option is (4) .

Regards.