14-7. Butadiene can be formed by dehydrogenation of n-butane: This reaction is c

Question

14-7. Butadiene can be formed by dehydrogenation of n-butane: This reaction is carried out in the gas phase. A. Calculate the equilibrium constant of this reac- tion at K298 15 B. A closed-system reactor initially contains 1 mole of n-butane. If the reactor is maintained at T-298.15 K and P 100 kPa, determine the contents of the reactor at equilibrium. C. A closed-system reactor initially contains 1 mole of butadiene and 1 mole of hydrogen. The reactor is maintained at P = 100 kPa, and a constant temperature at which the reaction has Kr 50. Determine the contents of the reactor at equilibrium.

Explanation / Answer

14-7

For the given reaction

C4H10 <==> C4H6 + 2H2

A. dGrxn = dG(products) - dG(reactants)

               = (185.4) - (-17) = 202.4 kJ/mol

dGrxn = -RTlnK

R = gas constant

T = 298.15 K

so,

202400 = -8.314 x 298.15 lnK

equilibrium constant K = 3.46 x 10^-36

B. when 1 mole of butane is initially present

ICE chart

               C4H10 <==> C4H6 + 2H2

I                  1                    -         -

C                -x                  +x      +2x

E               1-x                   x        2x

So,

K = [C4H6][H2]^2/[C4H10]

with x being a small value,

3.46 x 10^-36 = (x)(2x)^2

x = 9.53 x 10^-13 mol

at equilibrium,

C4H10 = 1 mole

C4H6 = 9.53 x 10^-13 mol

H2 = 2 x 9.53 x 10^-13 = 1.91 x 10^-12 mol

C. with Kt = 50

ICE chart

    butadiene + 2H2 ---> butane

I          1             1             -

C        -x           -2x          +x

E       1-x          1-2x          x

So,

Kt = [butane]/[butadiene][H2]^2

50 = x/(1-x)(1-2x)^2   

-600x^3 + 400x^2 - 199x + 50 = 0

x = 0.375 mol

at equilibrium,

butadiene = 1 - 0.375 = 0.625 mol

H2 = 1 - 2 x 0.375 = 0.25 mol

butane = 0.375 mol

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