A gaseous material l, XY(g) , dissociates to some extent to produce X(g) and Y(g
ID: 913451 • Letter: A
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
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
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