Estimate the compositions of the liquid and vapor phases when ethylene reacts with water to form ethanol at 200°C and 34.5 bar, conditions that assure the presence of both liquid and vapor phases. The reaction vessel is maintained at 34.5 bar by connection to a source of ethylene at this pressure.

Question

Estimate the compositions of the liquid and vapor phases when ethylene reacts with water to form ethanol at 200°C and 34.5 bar, conditions that assure the presence of both liquid and vapor phases. The reaction vessel is maintained at 34.5 bar by connection to a source of ethylene at this pressure. Assume no other reactions.

Accepted Answer

According to the phase rule (Sec. 14.8), the system has two degrees of freedom. Specification of both T and P therefore fixes the intensive state of the system, independent of the initial amounts of reactants. Material-balance equations are not applicable because the total amount of material in the system is not fixed. Thus, we cannot use equations that relate compositions to the reaction coordinate. Instead, phase equilibrium relations must provide a sufficient number of equations to allow solution for the unknown compositions.

The most convenient approach to this problem is to regard the chemical reaction as occurring in the vapor phase. Thus,

[latex] \mathrm{C}_2 \mathrm{H}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) ightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(g) [/latex]

and the standard states are those of the pure ideal gases at 1 bar. For these standard states, the equilibrium expression is Eq. (14.25), which in this case becomes:

[latex] \prod_i\left(\frac{\hat{f}_i}{P^0} ight)^{ u_i}=K [/latex] (14.25)

[latex] K=\frac{\hat{f}_{\mathrm{E} O \mathrm{OH}} P^{\circ}}{\hat{f}_{\mathrm{C}_2 \mathrm{H}_4} \hat{f}_{\mathrm{H}_2 \mathrm{O}}} [/latex] (A)

where the standard-state pressure P° is 1 bar (expressed in appropriate units). A general expression for ln K as a function of T is provided by the results of Ex. 14.4. For 200°C (473.15 K), this equation yields:

ln K = −3.473 K = 0.0310

The task now is to incorporate the phase-equilibrium equations [latex] \hat{f}_i^v=\hat{f}_i^l [/latex], into Eq. (A) and to relate the fugacities to the compositions in such a way that the equations can be readily solved. Equation (A) may be written:

[latex] K=\frac{\hat{f}_{\mathrm{EtOH}}^v P^{\circ}}{\hat{f}_{\mathrm{C}_2 \mathrm{H}_4}^v \hat{f}_{\mathrm{H}_2 \mathrm{O}}^v}=\frac{\hat{f}_{\mathrm{EtOH}}^l P^{\circ}}{\hat{f}_{\mathrm{C}_2 \mathrm{H}_4}^v \hat{f}_{\mathrm{H}_2 \mathrm{O}}^l} [/latex] (B)

The liquid-phase fugacities are related to activity coefficients by Eq. (13.2), and the vapor-phase fugacities are related to fugacity coefficients by Eq. (10.52):

[latex] \gamma_i \equiv \frac{\hat{f_i}}{x_i f_i} [/latex] (13.2)

[latex] \hat{\phi}_i \equiv \frac{\hat{f}_i}{y_i P} [/latex] (10.52)

[latex] \hat{f}_i^l=x_i \gamma_i f_i^l [/latex] (C)

[latex] \hat{f}_i^v=y_i \hat{\phi}_i P [/latex] (D)

Elimination of the fugacities in Eq. (B) by Eqs. (C) and (D) gives:

[latex] K=\frac{x_{\mathrm{EtOH}} \gamma_{\mathrm{EtOH}} f_{\mathrm{EtOH}}^l P^{\circ}}{\left(y_{\mathrm{C}_2 \mathrm{H}_4} \hat{\phi}_{\mathrm{C}_2 \mathrm{H}_4} P ight)\left(x_{\mathrm{H}_2 \mathrm{O}} \gamma_{\mathrm{H}_2 \mathrm{O}} f_{\mathrm{H}_2 \mathrm{O}}^l ight)} [/latex] (E)

The fugacity [latex] f_i^l [/latex] is for pure liquid i at the temperature and pressure of the system. However, pressure has small effect on the fugacity of a liquid, and to a good approximation, [latex] f_i^I=f_i^{\mathrm{sat}} [/latex] is at ; whence by Eq. (10.40),

[latex] f_i^l=\phi_i^{ ext {sat }} P_i^{ ext {sat }} [/latex] (F)

In this equation [latex] \phi_i^{ ext {sat }} [/latex] is at is the fugacity coefficient of pure saturated liquid or vapor evaluated at T of the system and at [latex] p_i^{ ext {sat }} [/latex]is at, the vapor pressure of pure species i. The reasonable assumption that the vapor phase is an ideal solution allows substitution of [latex] \phi_{\mathrm{C}_2 \mathrm{H}_4} ext { for } \hat{\phi}_{\mathrm{C}_2 \mathrm{H}_4} ext {, where } \phi_{\mathrm{C}_2 \mathrm{H}_4} [/latex] is the fugacity coefficient of pure ethylene at the system T and P. With this substitution and that of Eq. (F ), Eq. (E) becomes:

[latex] K=\frac{x_{\mathrm{EtOH}} \gamma_{\mathrm{EtOH}} \phi_{\mathrm{EtOH}}^{ ext {sat }} P_{\mathrm{EtOH}}^{ ext {sat }} P^{\circ}}{\left(y_{\mathrm{C}_2 \mathrm{H}_4} \phi_{\mathrm{C}_2 \mathrm{H}_4} P ight)\left(x_{\mathrm{H}_2 \mathrm{O}} \gamma_{\mathrm{H}_2 \mathrm{O}} \phi_{\mathrm{H}_2 \mathrm{O}}^{ ext {sat }} P_{\mathrm{H}_2 \mathrm{O}}^{ ext {sat }} ight)} [/latex] (G)

where the standard-state pressure P° is 1 bar, expressed in the units of P. In addition to Eq. (G) the following expressions apply. Because [latex] \sum_i y_i=1 [/latex]
[latex] y_{\mathrm{C}_2 \mathrm{H}_4}=1-y_{\mathrm{EtOH}}-y_{\mathrm{H}_2 \mathrm{O}} [/latex] (H)

Eliminating [latex] y_{\mathrm{H}_2 \mathrm{O}} [/latex] from Eq. (H) in favor of [latex] x_{\mathrm{EtOH}} ext { and } x_{\mathrm{H}_2 \mathrm{O}} [/latex] using the vapor/liquid equilibrium relation, [latex] \hat{f}_i^v=\hat{f}_i^l [/latex] , and combining the result with Eqs. (C ), (D), and (F ) then gives:

[latex] y_i=\frac{\gamma_i x_i \phi_i^{ ext {sat }} P_i^{ ext {sat }}}{\phi_i P} [/latex] (I)

where [latex] \phi_i ext { replaces } \hat{\phi}_i [/latex] because of the assumption that the vapor phase is an ideal solution. Equations (H) and (I) yield:

[latex] y_{\mathrm{C}_2 \mathrm{H}_4}=1-\frac{x_{\mathrm{EtOH}} \gamma_{\mathrm{EtOH}} \phi_{\mathrm{EtOH}}^{ ext {sat }} P{ }_{\mathrm{EtOH}}^{ ext {sat }}}{\phi_{\mathrm{EtOH}} P}-\frac{x_{\mathrm{H}_2 \mathrm{O}} \gamma_{\mathrm{H}_2 \mathrm{O}} \phi_{\mathrm{H}_2 \mathrm{O}}^{ ext {sat }} P_{\mathrm{H}_2 \mathrm{O}}^{ ext {sat }}}{\phi_i P} [/latex]

Because ethylene is far more volatile than ethanol or water, we can assume to a good approximation that [latex]x_{C_2 H_4} = 0 [/latex]. Then,

[latex] x_{\mathrm{H}_2 \mathrm{O}}=1-x_{\mathrm{EtOH}} [/latex] (K)

Equations (G), (J ), and (K) are the basis for solution of the problem. The primary variables in these equations are: [latex] x_{\mathrm{H}_2 \mathrm{O}}, x_{\mathrm{EtOH}}, ext { and } y_{\mathrm{C}_2 \mathrm{H}_4} [/latex] . Other quantities are either given or determined from correlations of data. The values of [latex] P_i^{ ext {sat }} [/latex] are:

[latex] P_{\mathrm{H}_2 \mathrm{O}}^{ ext {sat }}=15.55 \quad P_{\mathrm{EtOH}}^{ ext {sat }}=30.22 \mathrm{bar} [/latex]

The quantities [latex]\phi_i^{ ext {sat }} ext { and } \phi_i [/latex] are found from the generalized correlation represented by Eq. (10.68) with [latex] B^0 and B^1[/latex] given by Eqs. (3.61) and (3.62). Computed results are represented by PHIB(TR,PR,OMEGA). With T = 473.15 K, P = 34.5 bar, and critical data and acentric factors from App. B, computations provide:

[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{0.172}{T_r^{4.2}} [/latex] (3.62)