Question

The equilibrium for the dissociation of XY₂ is given as,

\[\ce{2 XY2 (g) <=> 2 XY (g) + Y2 (g)}\]

if the degree of dissociation x is so small compared to one. Show that 2 K_p = PX³ where P is the total pressure and K_p is the dissociation equilibrium constant of XY₂.

Answer 1

\[\ce{2 XY2 (g) <=> 2 XY (g) + Y2 (g)}\]

	XY2	XY	Y2
Initial no. of. moles	1	-	-
No. of. moles dissociated	x	-	-
No. of. moles at equilibrium	(1 - x) 1	x	`"x"/2`

Total no. of moles = `1 - x + x + x/2 = 1 + x/2 ≅ 1`

[∵ Given that x << 1; 1 - x ≅ 1 and 1 + `x/2` ≅ 1]

`"K"_"p" = (["P"_"xy"]^2["P"_("Y"_2)])/["P"_("XY"_2)]^2`

`= ((x/1 xx "P")^2((x//2)/1 xx "P"))/(1/1 xx "P")^2`

`"K"_"P" = (x^2"P"^2 xx "P")/(2"P"^2)`; 2K_p = x³P