Starting with 1 mol of `O_(2) : 2` mol of `SO_(2)`, the equilibrium for the formation of `SO_(3)(g)` was established a certain temperature. If V is the volume of the vessel and 2x is the number of moles of `SO_(3)` present, equilibrium constant for the reaction `2SO_(2(g)) + O_(2(g)) hArr 2SO_(3(g))` would be :

Options

A. `(x^(2)V)/((1-x)^(3))`
B. `(4x^(2))/((2-x)(1-x))`
C. `((1-x)^(3))/(2V)`
D. `(x^(2))/((2-x)(1-x))`

Answer

A

Explanation

`{:(,2SO_(2(g)),+,O_(2(g)),hArr,2SO_(3(g))),(t = 0,2,,1,,0),(t_("eqm"),[(2-2x)/(V)],,[(1-x)/(V)],,[(2x)/(V)]):}`
`K_(c) = ([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)]) = ([2x//V^(2)])/([(2-2x)/(V)]^(2)[(1-x)/(V)]) = (x^(2)V)/((1-x)^(3))`