1) The equilibrium constant, K, for the following reaction is 2.02×10-2 at 513 K
ID: 691971 • Letter: 1
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
Explanation / Answer
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
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