Question

XY2 dissociates as,XY2(g)⇌XY(g)+Y(g)Initial pressure of XY2 is 600 mm Hg.The total pressure at equilibrium is 800 mm Hg.Assuming volume of system to remain constant,the value of Kp is??

Answer 1

Kp =100, here the volume is constant so number of moles of reactant or product depends upon its partial pressure.

Answer 2

Final Answer : $K_p = 100 mm Hg$

Steps:
1) We have,
Since, Volume is constant .
$XY_2(g) \rightarrow XY(g) + Y(g) \\ \\ t=0 => 600. \rightarrow 0 \quad + 0 \\ \\ t=t_{eq. } => 600-x \rightarrow x.\quad + x$

2) Then at Equilibrium
$P_T = P_{XY_2}+ P_{XY} + P_Y \\ \\ => P_T = (600-x) + x + x \\ \\ => 800 = 600+ x \\ \\ => x = 200 mm \:Hg$

3) Now,
$K_p = \frac{x*x}{(600-x) } \\ \\ K_p = \frac{200*200}{400} = 100 mm\: Hg$

Hence, Value of $K_p = 100 \: mm Hg$