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.
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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