For the cracking reaction, c3h8(g) ® c2h4(g) + ch4(g) the equilibrium conversions negligible at 300k, but becomes appreciable at temperatures above 500 k. for a pressure of 1 bar, determine: a) the fractional conversion of propane at 625 k b) the temperature at which the fractional conversion is 85%

a. Fraction conversion is 0.778,  b. Temperature = 645.22 K

Explanation:

Write the reaction as follows

C₃H₈(g) > C₂H₄(g) + CH₄(g)

Take the basis as 1 mole of feed C₃H₈

Let ε be the reaction coordinate.

Calculate the final moles of each component as follows

             C₃H₈(g)       ⇒       C₂H₄(g) +    CH₄(g)

n₀ (mol)     1                  0           0

n (mol)1 - ε                         ε                 ε

Write the expression for fractional conversion as follows

Fractional conversion of C₃H₈ = (n₀ – n)/n₀ = [1 – (1 - ε)]/1 = ε

Calculate the mole fraction of each component as follows

y(C₃H₈) = n(C₃H₈)/n(T) = (1 – ε)/1 + ε

y(C₂H₄) = n(C₂H₄)/n(T) = ε/1 + ε

y(CH₄) = n(CH₄)/n(T) = ε/1 + ε

From the table, calculate the enthalpy of the reaction and Gibbs free energy as follows

ΔH(298) = ΔHf°(products) – ΔHf°(reactants)

= ΔHf°(C₂H₄) + ΔHf°(CH₄) – ΔHf°(C₃H₈)

= 52510 - 74520 – (-104680) = 82670 J/mol

ΔG(298) =   ΔGf°(products) – ΔGf°(reactants)

= ΔGf°(C₂H₄) + ΔGf°(CH₄) – ΔGf°(C₃H₈)

= 68460 – 50460 – (-24290) = 42290 J/mol

Expression for heat capacity is as follow

Cp(298)/R = A + BT + CT² + DT⁻²

Write the heat capacity coefficients for the components as follows

ComponentsA              B        C  D

C₃H₈  1.21328.785 x 10⁻³-8.824 x 10⁻⁶0

C₂H₄  1.42414.394 x 10⁻³-4.392 x 10⁻⁶0

CH₄  1.7029.081 x 10⁻³-2.164 x 10⁻⁶0

ΔA = A(CH₄) + A(C₂H₄) – A(C₃H₈)

= 1.702 + 1.424 – 1.213 = 1.913

ΔB = B(CH₄) + B(C₂H₄) – B(C₃H₈)

= (9.081 + 14.394 – 28.785) x 10⁻³ = -5.31 x 10⁻³

ΔC = C(CH₄) + C(C₂H₄) – C(C₃H₈)

= (-2.164 – 4.392 – (-8.824)) x 10⁻⁶ = 2.268 x 10⁻⁶

a.Calculate the Gibbs free energy at 625 K as follows

ΔG = ΔH – TΔS

= ΔH(298) + ΔH(298 K to 625 K) – T(ΔS(298) + ΔS(298 K to 625 K))

= ΔH(298) – TΔS(298) + ΔH(298 K to 625 K) – TΔS(298 K to 625 K)

= ΔH(298) – T[{ΔH(298) – ΔG(298)}/T₀] + ΔH(298K to 625K) – TΔS(298K to 625K)

Calculate the enthalpy change from 298 K to 625 K as follows

ΔH(298 K to 625 K) = ∫(298,625) (Cp(298))dT

= R∫(298,625) (ΔA + ΔBT + ΔCT² + ΔDT⁻²)dT

= R[1.913T – 5.31 x 10⁻³T²/2 + 2.268 x 10⁻⁶T³/3](298,625)

= 8.314 [1.913 (625 - 298) – 2.655 x 10⁻³(625² - 298²) + 0.756 x 10⁻⁶(625³ - 2983³)]

= 8.314 (625.55 – 801.33 + 164.56)

= -93.283 J/mol

Calculate the entropy change from 298 K to 625 K as follows

ΔS(298 K to 625 K) = ∫(298,625) (Cp(298))dT/T

= R∫(298,625) (ΔA + ΔBT + ΔCT² + ΔDT⁻²)dT/T

= R[1.913/T – 5.31 x 10⁻³ + 2.268 x 10⁻⁶T + (0)T⁻²](298,625)

= R[1.913lnT – 5.31 x 10⁻³T + 2.268 x 10⁻⁶T²/2](298,625)

= 8.314[1.913ln(625/298) – 5.31 x 10⁻³(625 - 298) + 1.134 x 10⁻⁶(625² - 298²)]

= 8.314 (1..417 – 1.736 + 0.342) = 0.191 J/molK

Calculate the Gibbs free energy as follows

= ΔH(298) – T[{ΔH(298) – ΔG(298)}/T₀] + ΔH(298K to 625K) – TΔS(298K to 625K)

= 82670 – (625/298)(82670/-42290) – 93.283 – 119.375

= -2232.26 J/mol

Calculate the equilibrium constant as follows

ΔG = -RTlnK

k = exp(-ΔG/RT)

k = exp(2232.26/8.314 x 625)

k = 1.537

write the expression for the equilibrium for the reaction as follows

K = y(C₂H₄) x y(CH₄)/y(C₃H₈)

Substitute the corresponding values in the above equation to calculate the value of ε

K = y(C₂H₄) x y(CH₄)/y(C₃H₈)

1.537 = {ε/(1 + ε)}{ε/(1 + ε)}/{(1 - ε)/(1 + ε)}

ε²/(1 – ε²) = 1.537

ε = 0.778

Hence the fraction conversion is 0.778

b.Fractional conversion ε is 0.85

Calculate the equilibrium constant as follows

K = y(C₂H₄) x y(CH₄)/y(C₃H₈)

= ε²/(1 – ε²)

= 0.852/(1 – (0.85)²)

= 2.603

Now calculate the Gibbs free energy as follows

ΔG = -RTlnK

= -8.314 x 625 x ln2.603

= -4971.1 J/mol

By trial and error method calculate the temperature at which the Gibbs free energy is -4971.1 J/mol

T₀          T          ΔH     ΔS     ΔG

298        643  -174.37  0.061461-4670.39

298        644  -179.066  0.054163-4805.95

298        645  -183.782  0.046846-4941.50

298        645.22-184.822  0.045234-4971.32

298        645.5  -186.147  0.043181-5009.27

Therefore, the required temperature is 645.22 K

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