**Answer: 2 $atm^{-1}$**

*Initiallly, P= 3 atm*

*Total no. of moles = 2 + 1 = 3*

*PV = nRT*

*3 $ \times $ V = 3RT*

*V = RT*

*$ 2SO_2 + O_2 \rightleftharpoons 2SO_3 $*

*__2_____1__________at t=0*

*(2-2x)__(1-x)___(2x)__at t=t'*

*Total no. of moles at equilibrium = 3-x*

*PV = nRT*

*$\Rightarrow$ 2.5 $ \times $ V = (3-x)RT*

*$\Rightarrow$ 2.5 $ \times $ RT = (3-x)RT*

*$\Rightarrow$ x = 0.5*

*No. of moles at equilibrium*

*$ SO_2 $ = 2(1 - x) = 2(1 - 0.5) = 1*

*$ O_2 $ = 1 - x = 1 - 0.5 = 0.5*

*$ SO_3 $ = 2x = 2 $ \times $ 0.5 = 1*

*$ \therefore $ Total no. of moles at equilibrium = 2.5*

*We know $ \rho_A $ = $ x_A $P*

*$ \rho_{SO_2} = \frac{1}{2.5} \times $ 2.5 = 1 atm*

*$ \rho_{O_2} = \frac{0.5}{2.5} \times $ 2.5 = 0.5 atm*

*$ \rho_{SO_3} = \frac {1}{2.5} \times $ 2.5 = 1 atm*

*$K_\rho = \frac{(\rho_{SO_3})^2}{(\rho_{O_2}) (\rho_{SO_3})^2} = \frac{(1)^2}{(0.5)(1)^2} = 2 atm^{-1}$*