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 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

Explanation / Answer

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 = x2/ (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

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