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Question: In a study of the system, Cl2(g) + Br2(g) 2BrCl(g), several different reaction mixtures were prepared, placed in a 15.00 × 10–3 m3 container, and allowed to reach equilibrium at 350 K. Despite having different starting compositions, three of the four mixtures had an identical composition at equilibrium. Which one of the systems has a different equilibrium composition than the others?

Select 1:

a. System 1: 0.100 moles of Cl2, 0.000 moles of Br2 and 0.600 moles of BrCl

b. System 2: 0.250 moles of Cl2, 0.150 moles of Br2 and 0.300 moles of BrCl

c. System 3: 0.400 moles of Cl2, 0.300 moles of Br2 and 0.000 moles of BrCl

d. System 4: 0.300 moles of Cl2, 0.250 moles of Br2 and 0.250 moles of BrCl

Answer 1

System 1 : For the reaction, 2BrCl<->Br2+Cl2

Let x= moles of Cl2 formed to reach equilibrium.

So at equilibrium, Cl2= 0.1+x, Br2= x and ClBr= 0.6-2x

Kc= [Br2][Cl2]/ [BrCl]2 = x*(0.1+x)/ (0.6-2x)² = 0.142, when solved x= 0.105

So moles at equilibrium, Cl2= 0.1+0.105= 0.205 Br2=0.105 and ClBr= 0.6-2*0.105= 0.39

System 2 : For the reaction, Br2+Cl2<-->2ClBr, Kc= 1/0.142= 7.05

Let x= moles of Cl2 decomped to reach equilibrium.

So at equilibrium, Cl2= 0.25-x Br2= 0.15- x and ClBr= 0.3+2x

Kc= [BrCl]2/ [Br2][Cl2] = (0.3+2x)²/ (0.25-x)*(0.15-x) =7.05, when solved,x =0.0449

So moles at equilibrium, Cl2= 0.25-0.0449=0.2051 Br2= 0.15-0.0449= 0.105, ClBr= 0.3+2*0.0449 = 0.39

System 3 :

For the reaction, Br2+Cl2<-->2ClBr, Kc= 1/0.142= 7.05

Let x= moles of Cl2 decomped to reach equilibrium.

So at equilibrium, Cl2= 0.4-x Br2= 0.3- x and ClBr= 2x

Kc= [BrCl]2/ [Br2][Cl2] = (2x)²/ (0.4-x)*(0.3-x) =7.05, when solved, x =0.195

So moles at equilibrium, Cl2= 0.4-0.195=0.205 Br2= 0.3-0.195= 0.105= ClBr= 2*0.195= 0.39

For the 4^th system

For the reaction, Br2+Cl2<--->2ClBr, Kc= 1/0.142= 7.05

Let x= moles of Cl2 decomped to reach equilibrium.

So at equilibrium, Cl2= 0.3-x Br2= 0.25- x and ClBr= 2x

Kc= [BrCl]2/ [Br2][Cl2] = (2x)²/ (0.3-x)*(0.25-x) =7.05, when solved, x=0.1553

So moles at equilibrium, Cl2= 0.3-0.1552=0.15, Br2= 0.25-0.155= 0.095= ClBr= 2*0.1553= 0.3106

the composition of 4th system is different from rest of the systems.