Question

A gaseous material l, XY(g) , dissociates to some extent to produce X(g) and Y(g).

XY(g) ↔ X(g) + Y(g)    K_c = 5.53 x 10^-3

A 2.00 gram sample 0f XY(M_w = 165g/mol) is placed in a container with a movable piston at 25^oC. The atmospheric pressure is 0.967 atm. Assume the piston is massless and frictionless. As XY(g) dissociates the piston moves keeping the pressure constant. Assuming ideal behavior, Calculate

a- The volume of the container at equilibrium

b- The percent dissociation of XY

c- The density of the gas mixture at equilibrium

Answer 1

grams of XY=2g
moles of XY =grams /MW = 2/165 = 0.0121 mol

XY(g) ↔ X(g) + Y(g)    Kc = 5.53 x 10-3
0.0121-x   x          x

Kc= x*x / (0.0121-x) = 5.53 x 10-3

so,
x^2 +5.53 x 10-3x - 5.53 x 10-3 *0.0121 = 0

x= 5.87 *10^-3

now,
Ideal gas equation
PV=nRT
as P kept constant and temperature is constant, R is gas constant

V/n = constant
V1= 0.0121 * 0.0821 *298/0.967 = 0.306L
n1= 0.0121

V1/n1 =V2/n2

n2 = 0.0121-x +x+x = 0.0121+x = 0.0121+0.00587=0.01797

so,
a) V2 = 0.306 *0.01797 / 0.0121 =0.454L

b) Percent dissociation =(1- [XY]final/[XY] ) *100
          = (0.0121-0.00587 )/ 0.0121    *100
   =48.51%

c)
Density of gas mixture = mass/volume = 2grams / 0.454L = 4.405 g/L