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

Volume of container can be calculated from PV= nRT

V= nRT/P= (2/165)*0.08206*(25+273.15)*0.967=0.31 L

Volume of container at equilibrium = 0.31L

Molar concenteation of A= (2/165)/0.31L= 0.039 molL

let X= Drop in concentration

Kc= x*y/xy

5.53*10^-3 = x^2/ (0.039-x)

-5.53*10-3x+0.000216 =x² or x² +5.53*10^-3x-0.000216=0 x =0.012189 moles/L

dissociated= 0.039-0.012189 =0.0269 mol/ L (since volume does not change)

% dissociation= 100*(0.0269/0.039)=68.82%

Moles of total gas at equilibrium = 0.0269 (moles of x remaining )+0.01289*2 ( moles of x and y) =0.051278

Density= 0.051278/0.31 L =0.165g/L