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

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:

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

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

\prod_i\left(y_i\right)^{\nu_{i, j}}=\left(\frac{P}{P^0}\right)^{-\nu_j} K_j       (14.40)

\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}}\right)^{-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}}\right)^{-1} K_{\mathrm{II}}

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

\frac{\varepsilon_{\mathrm{I}}^2}{\left(1-\varepsilon_{\mathrm{I}}-\varepsilon_{\mathrm{II}}\right)\left(1+\varepsilon_{\mathrm{I}}+\varepsilon_{\mathrm{II}}\right)}=\left(\frac{P}{P^{\circ}}\right)^{-1} K_{\mathrm{I}}         (A)

\frac{\varepsilon_{\mathrm{II}}^2}{\left(1-\varepsilon_{\mathrm{I}}-\varepsilon_{\mathrm{II}}\right)\left(1+\varepsilon_{\mathrm{I}}+\varepsilon_{\mathrm{II}}\right)}=\left(\frac{P}{P^{\circ}}\right)^{-1} K_{\mathrm{II}}       (B)

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

\varepsilon_{\mathrm{II}}=\kappa \varepsilon_{\mathrm{I}}

where        \kappa \equiv\left(\frac{K_{\mathrm{II}}}{K_{\mathrm{I}}}\right)^{1 / 2}

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

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

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

\kappa=\left(\frac{268.4}{3.856}\right)^{1 / 2}=8.343

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

\varepsilon_{\text {II }}=(8.343)(0.1068)=0.8914

The product-gas composition for these reaction coordinates is:

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.