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

Question

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

XY(g) X(g) + Y(g)    Kc = 5.53 x 10-3

A 2.00 gram sample 0f XY(Mw = 165g/mol) is placed in a container with a movable piston at 25oC. 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

Explanation / Answer

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

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