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

According to ideal gas equation:

PV = nRT

V = nRT/P

n = 2/165

P = 0.967 atm

T= 25 + 273 = 298 K

R = 0.0821 lit.atm/mol

V = [(2/165)*(0.0821*298)]/(0.967)

= 0.306 lit

b.

Kc = 5.53*10^-3

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

0.306 Lit of container has a gas of 2 grams

1 lit of container will have a gas of = 2/0.306 = 6.53 gr/Lit

5.53*10^-3 = [X][Y] /[XY]

[X][Y] = (5.53*10^-3)*6.53

= (0.036)gr/Lit

0.036 gr is present in 1 lit container

quantity of gas present in 0.306 lit container = 0.036*0.306 = 0.011 grams

Percentage of decomposition = (0.011*100)/2 = 0.55%

=