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 the following:

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

Solution :-

Lets first calculate the moles of the XY

Moles of XY = 2.00 g / 165 g /mol = 0.01212 mol XY

Now lets calculate the equilibrium moles of each species

XY ------- > X   +        Y

0.01212         0               0

-x                   +x             +x

0.01212-x       x               x

Kc= [X][Y]/[XY]

5.53*10^-3 = [x][x]/[0.01212-x]

5.53*10^-3 [0.01212-x] =x^2

Solving for x we get 0.005876 mol

Now lets calculate the equilibrium moles of the XY

[XY] eq = 0.01212 –x = 0.01212- 0.05876 = 0.006244 mol

Now total moles at the equilibrium are = 0.006244 mol + (2*0.005876 mol ) = 0.018 mol

a- The volume of the container at equilibrium

Solution :- T= 25 C +273 = 298 K

P = 0.967 atm

PV= nRT

V= nRT /P

   = 0.018 mol * 0.08206 L atm per mol K * 298 K / 0.967 atm

   = 0.455 L

So the volume of the container is 0.455 L

b- The percent dissociation of XY

% dissociation = (x/ initial moles of XY)*100%

                       = (0.005876 / 0.01212)*100 %

                      = 48.48 %

c- The density of t he gas mixture at equilibrium

Solution :- density = mass / volume

                                  = 2.0 g / 0.455 L

                                  = 4.395 g/L

So the density of the gas mixture is 4.395 g/L