Estimate the maximum conversion of ethylene to ethanol by vapor-phase hydration at 250°C and 35 bar for an initial steam-to-ethylene molar ratio of 5.

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The calculation of K for this reaction was treated in Ex. 14.4. For a temperature of 250°C or 523.15 K the calculation yields:

[latex] K=10.02 imes 10^{-3} [/latex]

The appropriate equilibrium expression is Eq. (14.26). This equation requires evaluation of the fugacity coefficients of the species present in the equilibrium mixture.

[latex] \prod_i\left(y_i \hat{\phi}_i ight)^{ u_i}=\left(\frac{P}{P^{\circ}} ight)^{- u} K [/latex] (14.26)

This can be accomplished with Eq. (10.64).

[latex] \ln \hat{\phi}_k=\frac{P}{R T}\left[B_{k k}+\frac{1}{2} \sum_i \sum_j y_i y_j\left(2 \delta_{i k}-\delta_{i j} ight) ight] [/latex] (10.64)

However, the calculations involve iteration because the fugacity coefficients are functions of composition. For purposes of illustration, we assume here that the reaction mixture can be treated as an ideal solution, which eliminates the need for iteration. This calculation would also serve as the first iteration of a more rigorous calculation using mixture fugacity coefficients from Eq. (10.64). With the assumption of ideal solution behavior, Eq. (14.26) reduces to Eq. (14.27), which requires fugacity coefficients of the pure gases of the reacting mixture at the equilibrium T and P. Because ν = [latex] u=\sum_i u_i=-1 [/latex], this equation becomes:

[latex] \prod_i\left(y_i \phi_i ight)^{ u_i}=\left(\frac{P}{P^0} ight)^{- u} K [/latex] (14.27)

[latex] \frac{y_{\mathrm{EtOH}} \phi_{\mathrm{EtOH}}}{y_{\mathrm{C}_2 \mathrm{H}_4} \phi_{\mathrm{C}_2 \mathrm{H}_4} y_{\mathrm{H}_2 \mathrm{O}} \phi_{\mathrm{H}_2 \mathrm{O}}}=\left(\frac{P}{P^{\circ}} ight)\left(10.02 imes 10^{-3} ight) [/latex] (A)

Computations based on Eq. (10.68) in conjunction with Eqs. (3.61) and (3.62) provide values represented by:

[latex] \phi=\exp \left[\frac{P_r}{T_r}\left(B^0+\omega B^1 ight) ight] [/latex] (10.68)

[latex] B^0=0.083-\frac{0.422}{T_r^{1.6}} [/latex] (3.61)

[latex] B^1=0.139-\frac{0.17}{T_r^{4.2}} [/latex] (3.62)

[latex] PHIB(TR,PR,OMEGA) = ϕ_i[/latex]

The results of these calculations are summarized in the following table:

	[latex] T _{c i} ∕K[/latex]	[latex] P_{ci} ∕bar[/latex]	[latex] ω_i [/latex]	[latex]T_{ri}[/latex]	[latex] P_{ri}[/latex]	[latex]B^0 [/latex]	[latex] B^1 [/latex]	[latex] ϕ_i [/latex]
[latex] C_2H_4 [/latex]	282.3	50.40	0.087	1.853	0.694	–0.074	0.126	0.977
[latex]H_2O [/latex]	647.1	220.55	0.345	0.808	0.159	–0.511	–0.281	0.887
EtOH	513.9	61.48	0.645	1.018	0.569	–0.327	–0.021	0.887

The critical data and values of [latex]ω_i[/latex] are from App. B. The temperature and pressure in all cases are 523.15 K and 35 bar. Substitution of values for [latex]ϕ_i[/latex] and for (P∕P°) into Eq. (A) gives:

[latex] \frac{y_{\mathrm{EtOH}}}{y_{\mathrm{C}_2 \mathrm{H}_4} y_{\mathrm{H}_2 \mathrm{O}}}=\frac{(0.977)(0.887)}{(0.827)}(35)\left(10.02 imes 10^{-3} ight)=0.367 [/latex] (B)

By Eq. (14.5),

[latex] y_i=\frac{n_i}{n}=\frac{n_{i_0}+ u_i \varepsilon}{n_0+ u \varepsilon} [/latex] (14.5)

[latex] y_{\mathrm{C}_2 \mathrm{H}_4}=\frac{1-\varepsilon_e}{6-\varepsilon_e} \quad y_{\mathrm{H}_2 \mathrm{O}}=\frac{5-\varepsilon_e}{6-\varepsilon_e} \quad y_{\mathrm{EtOH}}=\frac{\varepsilon_e}{6-\varepsilon_e} [/latex]

Substituting these into Eq. (B) yields:

[latex] \frac{\varepsilon_e\left(6-\varepsilon_e ight)}{\left(5-\varepsilon_e ight)\left(1-\varepsilon_e ight)}=0.367 \quad ext { or } \quad \varepsilon_e^2-6.000 \varepsilon_e+1.342=0 [/latex]

The solution to this quadratic equation for the smaller root is [latex]ε_e[/latex] = 0.233. Because the larger root is greater than unity, and would therefore correspond to a negative mole fraction of ethylene, it does not represent a physically possible result. The maximum conversion of ethylene to ethanol under the stated conditions is therefore 23.3%. To carry out a more rigorous calculation without assuming that the gases form an ideal solution, one would next evaluate the mixture fugacity coefficients from Eq. (10.64), use the resulting values in Eq. (B), and calculate a new value of [latex]ε_e[/latex] and then iterate until the value of [latex]ε_e[/latex] stops changing. However, this is rarely necessary in practice.

In this reaction, increasing the temperature decreases K and hence the conversion. Increasing the pressure increases the conversion. Equilibrium considerations therefore suggest that the operating pressure be as high as possible (limited by condensation) and the temperature as low as possible. However, even with the best catalyst known, the minimum temperature for a reasonable reaction rate is about 150°C. This is an instance where both equilibrium and reaction rate influence the commercial viability of a reaction process.

The equilibrium conversion is a function of temperature, T, pressure, P, and the steam-to-ethylene ratio in the feed, a. The effects of all three variables are shown in Fig. 14.4. The curves in this figure come from calculations like those illustrated in this example, except that a less precise relation for K as a function of T was used. Comparing the families of curves for different pressures, and the curves for
different steam-to-ethylene ratios at a given pressure, illustrates how increasing P or a allows one to reach a given equilibrium ethylene conversion at higher T.

Question 14.6: Estimate the maximum conversion of ethylene to ethanol by va...

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