Question

1) The equilibrium constant, K, for the following reaction is 2.02×10-2 at 513 K. PCl5(g) <------ ------> PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 18.3 L container at 513 K contains 0.216 M PCl5, 6.60×10-2 M PCl3 and 6.60×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 8.25 L?

[PCl5] = ______M

[PCl3] = _______M

[Cl2] = ________M

2)The equilibrium constant, K, for the following reaction is 2.83×10-2 at 522 K. PCl5(g) <-------- ------>PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 8.68 L container at 522 K contains 0.223 M PCl5, 7.94×10-2 M PCl3 and 7.94×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the volume of the container is increased to 16.5 L?

[PCl5] = _______M

[PCl3] = ______M

[Cl2] = ________M

Answer 1

mixture is compressed at constant temperature to a volume of 8.25 L?

At Equilibrium, [PCl5] =0.216M, [PCl3] =6.6*10^-2 M = [Cl2]

Moles of gases = Molarity* Volume in L

Moles : PCl5= 0.216*18.3=3.95, PCl3=Cl2= 6.6*10^-2*18.3= 1.23

When the volume is compressed to 8.25L, the new concentrations are [PCl5]=3.95/8.25= 0.478M, [PCl3] =[Cl2]= 1.23/8.25=0.149M

For the reaction, PCl5(g)<-->PCl3(g)+Cl2(g)

Q= reaction coefficient = [PCl3][Cl2]/[PCl5]= 0.149*0.149/0.478=0.046>K, more of PCl5 will have to form so that Q decreases to become K, the equilibrium constant.

Let x= drop in concentration of PCl3 to reach equilibrium

At equilibrium [PCl5]= 0.478+x, [PCl3] =[Cl2]= 0.149-x

K= [PCl3][Cl2]/[PCl5]= (0.149-x)²/ (0.478+x)= 0.0202

When solved using excel, x=0.046 and at Equilibrium [PCl5]= 0.478+0.046=0.524

[PCl3]=[Cl2] =0.149-0.046 =0.103M

2.

At Equilibrium, [PCl5] =0.223M, [PCl3] =7.94*10^-2 M = [Cl2]

Moles of gases = Molarity* Volume in L

Moles : PCl5= 0.223*8.68=1.94, PCl3=Cl2= 7.94*10^-2*8.68= 0.69

When the volume is compressed to 16.5L, the new concentrations are [PCl5]=1.94/16.5=0.1175M, [PCl3] =[Cl2]= 0.69/16.5=0.042

For the reaction, PCl5(g)<-->PCl3(g)+Cl2(g)

Q= reaction coefficient = [PCl3][Cl2]/[PCl5]= 0.042*0.042/0.1175=0.015<K, more of PCl3 and Cl2 will have to form so that Q increases to become K, the equilibrium constant.

Let x= drop in concentration of PCl5 to reach equilibrium

At equilibrium [PCl5]= 0.1175-x, [PCl3] =[Cl2]= 0.042+x

K= [PCl3][Cl2]/[PCl5]= (0.042+x)²/ (0.1175-x)= 2.83*10^-2

When solved using excel, x=0.0125 and at Equilibrium [PCl5]= 0.1175-0.0125=0.105M

[PCl3]=[Cl2] =0.042+0.0125=0.0545M