Question

4) Let’s combine multiple equilibrium. C2H6(g) <--> C2H4(g) + H2(g) kp = 2.8x10-10 N2(g) + 3...

4) Let’s combine multiple equilibrium.

C2H6(g) <--> C2H4(g) + H2(g) kp = 2.8x10-10

N2(g) + 3 H2(g) <--> 2NH3(g) kp = 1.64x10-4

If C2H6 = 10atm, N2 = 10atm and H2 = 50atm

find the concentrations at equilibrium.

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Homework Answers

Answer #1

To solve the multiple equilibrium, add both the equations,

we get,

C2H6(g) + N2 (g) + 2H2 (g) -----> C2H4 (g) + 2NH3 (g)

Kp = Kp1*Kp2

= (2.8*10-10)(1.64*10-4)

= 4.60*10-14

ICE table is:

PC2H6 PN2 PH2 PC2H4 PNH3
Initial 10 10 50 0 0
Change -x -x -2x +x +2x
Equilibrium 10-x 10-x 50-2x x 2x

Kp = (PC2H4)(PNH3)2/[(PC2H6)(PN2)(PH2)2]

4.60*10-14 = x.(2x)2/[(10-x)(10-x)(50-2x)2]

Taking square root on both the sides.

2.14*10-7 = 2x2/[(10-x)(50-2x)]

2.14*10-7 = 2x2/[500-3x+2x2]

1.07*10-7 [500-3x+2x2] = x2

(5.35*10-5) - (3.21*10-7)x + (2.14*10-7)x2 = x2

x2 + (3.21*10-7)x​ - (5.35*10-5)​ = 0

x = 0.0073 atm

So, at equilibrium,

PC2H6 = 10-x = 10-0.0073 = 9.9927 atm

PN2 = 10-x = 10-0.0073 = 9.9927 atm

PH2 = 50-2x = 50-2(0.0073) = 49.9854 atm

PC2H4 = x = 0.0073 atm

PNH3 = 2x = 2(0.0073) = 0.0146 atm

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