A feedstock of pure n-butane is cracked at 750 K and 1.2 bar to produce olefins. Only two reactions have favorable equilibrium conversions at these conditions:C4 H10  → C2H4  + C2H6 (I) C4H10  →  C3 H6  + CH4 (II) If these reactions reach equilibrium, what is the product composition? With data

Question

A feedstock of pure n-butane is cracked at 750 K and 1.2 bar to produce olefins. Only two reactions have favorable equilibrium conversions at these conditions:

[latex] \mathrm{C}_4 \mathrm{H}_{10} ightarrow \mathrm{C}_2 \mathrm{H}_4+\mathrm{C}_2 \mathrm{H}_6 [/latex] (I)

[latex] \mathrm{C}_4 \mathrm{H}_{10} ightarrow \mathrm{C}_3 \mathrm{H}_6+\mathrm{CH}_4 [/latex] (II)

If these reactions reach equilibrium, what is the product composition?

With data from App. C and procedures illustrated in Ex. 14.4, the equilibrium constants at 750 K are found to be:

[latex] K_1=3.856 \quad ext { and } \quad K_{11}=268.4 [/latex]

Accepted Answer

Equations relating the product composition to the reaction coordinates are developed
as in Ex. 14.3. With a basis of 1 mol of n-butane feed, they become:

[latex] y_{\mathrm{C}_4 \mathrm{H}_{10}}=\frac{1-\varepsilon_{\mathrm{I}}-\varepsilon_{\mathrm{II}}}{1+\varepsilon_{\mathrm{I}}+\varepsilon_{\mathrm{II}}} [/latex]

[latex] y_{\mathrm{C}_2 \mathrm{H}_4}=y_{\mathrm{C}_2 \mathrm{H}_6}=\frac{\varepsilon_{\mathrm{I}}}{1+\varepsilon_{\mathrm{I}}+\varepsilon_{\mathrm{II}}} \quad y_{\mathrm{C}_3 \mathrm{H}_6}=y_{\mathrm{CH}_4}=\frac{\varepsilon_{\mathrm{II}}}{1+\varepsilon_{\mathrm{I}}+\varepsilon_{\mathrm{II}}}[/latex]

The equilibrium relations, by Eq. (14.40), are:

[latex] \prod_i\left(y_i ight)^{ u_{i, j}}=\left(\frac{P}{P^0} ight)^{- u_j} K_j [/latex] (14.40)

[latex] \frac{y_{\mathrm{C}_2 \mathrm{H}_4} y_{\mathrm{C}_2 \mathrm{H}_6}}{y_{\mathrm{C}_4 \mathrm{H}_{10}}}=\left(\frac{P}{P^{\circ}} ight)^{-1} K_{\mathrm{I}} \quad \frac{y_{\mathrm{C}_3 \mathrm{H}_6} y_{\mathrm{CH}_4}}{y_{\mathrm{C}_4 \mathrm{H}_{10}}}=\left(\frac{P}{P^{\circ}} ight)^{-1} K_{\mathrm{II}} [/latex]

Combining these equilibrium equations with the mole-fraction equations yields:

[latex] \frac{\varepsilon_{\mathrm{I}}^2}{\left(1-\varepsilon_{\mathrm{I}}-\varepsilon_{\mathrm{II}} ight)\left(1+\varepsilon_{\mathrm{I}}+\varepsilon_{\mathrm{II}} ight)}=\left(\frac{P}{P^{\circ}} ight)^{-1} K_{\mathrm{I}} [/latex] (A)

[latex] \frac{\varepsilon_{\mathrm{II}}^2}{\left(1-\varepsilon_{\mathrm{I}}-\varepsilon_{\mathrm{II}} ight)\left(1+\varepsilon_{\mathrm{I}}+\varepsilon_{\mathrm{II}} ight)}=\left(\frac{P}{P^{\circ}} ight)^{-1} K_{\mathrm{II}} [/latex] (B)

Dividing Eq. (B) by Eq. (A) and solving for εII yields:

[latex] \varepsilon_{\mathrm{II}}=\kappa \varepsilon_{\mathrm{I}} [/latex]

where [latex] \kappa \equiv\left(\frac{K_{\mathrm{II}}}{K_{\mathrm{I}}} ight)^{1 / 2} [/latex]

Combining Eqs. (A) and (C) to eliminate εII, and then solving for εI gives:

[latex] \varepsilon_{\mathrm{I}}=\left[\frac{K_{\mathrm{I}}\left(P^{\circ} / P ight)}{1+K_{\mathrm{I}}\left(P^{\circ} / P ight)(\kappa+1)^2} ight]^{1 / 2} [/latex] (E)

Substitution of numerical values in Eqs. (D), (E), and (C) yields:

[latex] \kappa=\left(\frac{268.4}{3.856} ight)^{1 / 2}=8.343 [/latex]

[latex] \varepsilon_{\mathrm{I}}=\left[\frac{(3.856)(1 / 1.2)}{1+(3.856)(1 / 1.2)(9.343)^2} ight]^{1 / 2}=0.1068 [/latex]

[latex] \varepsilon_{ ext {II }}=(8.343)(0.1068)=0.8914 [/latex]

The product-gas composition for these reaction coordinates is:

[latex] y_{\mathrm{C}_4 \mathrm{H}_{10}}=0.0010 \quad y_{\mathrm{C}_2 \mathrm{H}_4}=y_{\mathrm{C}_2 \mathrm{H}_6}=0.0534 \quad y_{\mathrm{C}_3 \mathrm{H}_6}=y_{\mathrm{CH}_4}=0.4461 [/latex]

For this simple reaction scheme, analytical solution is possible. However, this is unusual. In most cases the solution of multireaction-equilibrium problems requires numerical techniques.

Question 14.12: A feedstock of pure n-butane is cracked at 750 K and 1.2 bar...

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