If K_p for the reaction ${\mathrm{N}}_2{\mathrm{O}}_4 \rightleftharpoons 2{\mathrm{NO}}_2$is 0.66 then what is the equilibrium pressure of N₂O₄. (Total pressure at equilibrium is 0.5 atm)
(1) 0.168                   (2) 0.322                   (3) 0.1                    (4) 0.5

Dear Student,

The given reaction is :
                                                                        $N_2{O}_4\stackrel{}{\rightleftharpoons }2N{O}_2$  
                                                      at t = 0       1               0                     initial mole
                                                    at eqm        1 - x            2x                  moles at eqm 
                           
 Mole fraction at                             eqm      $\frac{1-x}{1+x} \frac{2x}{1+x}$ 
As the total pressure at eqm = 0.5 atm.
So the partial pressure of $N_2{O}_4$ = ​$\left(\frac{1-x}{1+x}\right)\times 0.5 atm$
The partial pressure of $N{O}_2 = \left(\frac{2x}{1+x}\right)\times 0.5 atm$
Now, as the $$\begin{gathered} K_{p }= \frac{\left[p\right(N{O}_2){]}^2}{\left[p\right(N_2{O}_4\left)\right]} \\ 0.66 = \frac{\left({\displaystyle \frac{2x}{1+x}}\right)\left(\frac{2x}{1+x}\right) \times 0.5 \times 0.5 }{\left(\frac{1-x}{1+x}\right)\times 0.5 } \\ 0.66 = \frac{4x^2 \times 0.5 }{1-x^2} \\ 0.66 - 0.66x^2 = 2x^2 \\ 2.66x^2 = 0.66 \\ x = \sqrt{\frac{0.66}{2.66}} \\ = 0.5 \end{gathered}$$

Now put the value of x in partial pressure equation of ​$N_2{O}_4$ = 
                 $$\begin{gathered} = \frac{1-0.5 }{1+0.5}\times 0.5 \\ = \frac{0.5 \times 0.5}{1.5 } \\ = 0.166 atm \end{gathered}$$

So, the correct option is (1)

Regards.