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
