Question 14.10: Estimate the compositions of the liquid and vapor phases whe...
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.
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,
\mathrm{C}_2 \mathrm{H}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(g)
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:
\prod_i\left(\frac{\hat{f}_i}{P^0}\right)^{\nu_i}=K (14.25)
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}}} (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 \hat{f}_i^v=\hat{f}_i^l , 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:
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} (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):
\gamma_i \equiv \frac{\hat{f_i}}{x_i f_i} (13.2)
\hat{\phi}_i \equiv \frac{\hat{f}_i}{y_i P} (10.52)
| \hat{f}_i^l=x_i \gamma_i f_i^l (C) | \hat{f}_i^v=y_i \hat{\phi}_i P (D) |
Elimination of the fugacities in Eq. (B) by Eqs. (C) and (D) gives:
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\right)\left(x_{\mathrm{H}_2 \mathrm{O}} \gamma_{\mathrm{H}_2 \mathrm{O}} f_{\mathrm{H}_2 \mathrm{O}}^l\right)} (E)
The fugacity f_i^l 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, f_i^I=f_i^{\mathrm{sat}} is at ; whence by Eq. (10.40),
f_i^l=\phi_i^{\text {sat }} P_i^{\text {sat }} (F)
In this equation \phi_i^{\text {sat }} is at is the fugacity coefficient of pure saturated liquid or vapor evaluated at T of the system and at p_i^{\text {sat }} is at, the vapor pressure of pure species i. The reasonable assumption that the vapor phase is an ideal solution allows substitution of \phi_{\mathrm{C}_2 \mathrm{H}_4} \text { for } \hat{\phi}_{\mathrm{C}_2 \mathrm{H}_4} \text {, where } \phi_{\mathrm{C}_2 \mathrm{H}_4} is the fugacity coefficient of pure ethylene at the system T and P. With this substitution and that of Eq. (F ), Eq. (E) becomes:
K=\frac{x_{\mathrm{EtOH}} \gamma_{\mathrm{EtOH}} \phi_{\mathrm{EtOH}}^{\text {sat }} P_{\mathrm{EtOH}}^{\text {sat }} P^{\circ}}{\left(y_{\mathrm{C}_2 \mathrm{H}_4} \phi_{\mathrm{C}_2 \mathrm{H}_4} P\right)\left(x_{\mathrm{H}_2 \mathrm{O}} \gamma_{\mathrm{H}_2 \mathrm{O}} \phi_{\mathrm{H}_2 \mathrm{O}}^{\text {sat }} P_{\mathrm{H}_2 \mathrm{O}}^{\text {sat }}\right)} (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 \sum_i y_i=1
y_{\mathrm{C}_2 \mathrm{H}_4}=1-y_{\mathrm{EtOH}}-y_{\mathrm{H}_2 \mathrm{O}} (H)
Eliminating y_{\mathrm{H}_2 \mathrm{O}} from Eq. (H) in favor of x_{\mathrm{EtOH}} \text { and } x_{\mathrm{H}_2 \mathrm{O}} using the vapor/liquid equilibrium relation, \hat{f}_i^v=\hat{f}_i^l , and combining the result with Eqs. (C ), (D), and (F ) then gives:
y_i=\frac{\gamma_i x_i \phi_i^{\text {sat }} P_i^{\text {sat }}}{\phi_i P} (I)
where \phi_i \text { replaces } \hat{\phi}_i because of the assumption that the vapor phase is an ideal solution. Equations (H) and (I) yield:
y_{\mathrm{C}_2 \mathrm{H}_4}=1-\frac{x_{\mathrm{EtOH}} \gamma_{\mathrm{EtOH}} \phi_{\mathrm{EtOH}}^{\text {sat }} P{ }_{\mathrm{EtOH}}^{\text {sat }}}{\phi_{\mathrm{EtOH}} P}-\frac{x_{\mathrm{H}_2 \mathrm{O}} \gamma_{\mathrm{H}_2 \mathrm{O}} \phi_{\mathrm{H}_2 \mathrm{O}}^{\text {sat }} P_{\mathrm{H}_2 \mathrm{O}}^{\text {sat }}}{\phi_i P}
Because ethylene is far more volatile than ethanol or water, we can assume to a good approximation that x_{C_2 H_4} = 0 . Then,
x_{\mathrm{H}_2 \mathrm{O}}=1-x_{\mathrm{EtOH}} (K)
Equations (G), (J ), and (K) are the basis for solution of the problem. The primary variables in these equations are: x_{\mathrm{H}_2 \mathrm{O}}, x_{\mathrm{EtOH}}, \text { and } y_{\mathrm{C}_2 \mathrm{H}_4} . Other quantities are either given or determined from correlations of data. The values of P_i^{\text {sat }} are:
P_{\mathrm{H}_2 \mathrm{O}}^{\text {sat }}=15.55 \quad P_{\mathrm{EtOH}}^{\text {sat }}=30.22 \mathrm{bar}
The quantities \phi_i^{\text {sat }} \text { and } \phi_i are found from the generalized correlation represented by Eq. (10.68) with B^0 and B^1 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:
\phi=\exp \left[\frac{P_r}{T_r}\left(B^0+\omega B^1\right)\right] (10.68)
B^0=0.083-\frac{0.422}{T_r^{1.6}} (3.61)
B^1=0.139-\frac{0.172}{T_r^{4.2}} (3.62)
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,
\mathrm{C}_2 \mathrm{H}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(g)
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:
\prod_i\left(\frac{\hat{f}_i}{P^0}\right)^{\nu_i}=K (14.25)
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}}} (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 \hat{f}_i^v=\hat{f}_i^l , 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:
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} (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):
\gamma_i \equiv \frac{\hat{f_i}}{x_i f_i} (13.2)
\hat{\phi}_i \equiv \frac{\hat{f}_i}{y_i P} (10.52)
| \hat{f}_i^l=x_i \gamma_i f_i^l (C) | \hat{f}_i^v=y_i \hat{\phi}_i P (D) |
Elimination of the fugacities in Eq. (B) by Eqs. (C) and (D) gives:
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\right)\left(x_{\mathrm{H}_2 \mathrm{O}} \gamma_{\mathrm{H}_2 \mathrm{O}} f_{\mathrm{H}_2 \mathrm{O}}^l\right)} (E)
The fugacity f_i^l 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, f_i^I=f_i^{\mathrm{sat}} is at ; whence by Eq. (10.40),
f_i^l=\phi_i^{\text {sat }} P_i^{\text {sat }} (F)
In this equation \phi_i^{\text {sat }} is at is the fugacity coefficient of pure saturated liquid or vapor evaluated at T of the system and at p_i^{\text {sat }} is at, the vapor pressure of pure species i. The reasonable assumption that the vapor phase is an ideal solution allows substitution of \phi_{\mathrm{C}_2 \mathrm{H}_4} \text { for } \hat{\phi}_{\mathrm{C}_2 \mathrm{H}_4} \text {, where } \phi_{\mathrm{C}_2 \mathrm{H}_4} is the fugacity coefficient of pure ethylene at the system T and P. With this substitution and that of Eq. (F ), Eq. (E) becomes:
K=\frac{x_{\mathrm{EtOH}} \gamma_{\mathrm{EtOH}} \phi_{\mathrm{EtOH}}^{\text {sat }} P_{\mathrm{EtOH}}^{\text {sat }} P^{\circ}}{\left(y_{\mathrm{C}_2 \mathrm{H}_4} \phi_{\mathrm{C}_2 \mathrm{H}_4} P\right)\left(x_{\mathrm{H}_2 \mathrm{O}} \gamma_{\mathrm{H}_2 \mathrm{O}} \phi_{\mathrm{H}_2 \mathrm{O}}^{\text {sat }} P_{\mathrm{H}_2 \mathrm{O}}^{\text {sat }}\right)} (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 \sum_i y_i=1
y_{\mathrm{C}_2 \mathrm{H}_4}=1-y_{\mathrm{EtOH}}-y_{\mathrm{H}_2 \mathrm{O}} (H)
Eliminating y_{\mathrm{H}_2 \mathrm{O}} from Eq. (H) in favor of x_{\mathrm{EtOH}} \text { and } x_{\mathrm{H}_2 \mathrm{O}} using the vapor/liquid equilibrium relation, \hat{f}_i^v=\hat{f}_i^l , and combining the result with Eqs. (C ), (D), and (F ) then gives:
y_i=\frac{\gamma_i x_i \phi_i^{\text {sat }} P_i^{\text {sat }}}{\phi_i P} (I)
where \phi_i \text { replaces } \hat{\phi}_i because of the assumption that the vapor phase is an ideal solution. Equations (H) and (I) yield:
y_{\mathrm{C}_2 \mathrm{H}_4}=1-\frac{x_{\mathrm{EtOH}} \gamma_{\mathrm{EtOH}} \phi_{\mathrm{EtOH}}^{\text {sat }} P{ }_{\mathrm{EtOH}}^{\text {sat }}}{\phi_{\mathrm{EtOH}} P}-\frac{x_{\mathrm{H}_2 \mathrm{O}} \gamma_{\mathrm{H}_2 \mathrm{O}} \phi_{\mathrm{H}_2 \mathrm{O}}^{\text {sat }} P_{\mathrm{H}_2 \mathrm{O}}^{\text {sat }}}{\phi_i P}
Because ethylene is far more volatile than ethanol or water, we can assume to a good approximation that x_{C_2 H_4} = 0 . Then,
x_{\mathrm{H}_2 \mathrm{O}}=1-x_{\mathrm{EtOH}} (K)
Equations (G), (J ), and (K) are the basis for solution of the problem. The primary variables in these equations are: x_{\mathrm{H}_2 \mathrm{O}}, x_{\mathrm{EtOH}}, \text { and } y_{\mathrm{C}_2 \mathrm{H}_4} . Other quantities are either given or determined from correlations of data. The values of P_i^{\text {sat }} are:
P_{\mathrm{H}_2 \mathrm{O}}^{\text {sat }}=15.55 \quad P_{\mathrm{EtOH}}^{\text {sat }}=30.22 \mathrm{bar}
The quantities \phi_i^{\text {sat }} \text { and } \phi_i are found from the generalized correlation represented by Eq. (10.68) with B^0 and B^1 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:
\phi=\exp \left[\frac{P_r}{T_r}\left(B^0+\omega B^1\right)\right] (10.68)
B^0=0.083-\frac{0.422}{T_r^{1.6}} (3.61)
B^1=0.139-\frac{0.172}{T_r^{4.2}} (3.62)
| T_{ ci} ∕K | P_{ ci} ∕bar | ω_i | T -{r_i} | P_ {r_i} | P_{r_i}^{\text {sat }} | B^0 | B^1 | ϕ_i | ϕ_i^{sat} | |
| EtOH | 513.9 | 61.48 | 0.645 | 0.921 | 0.561 | 0.492 | –0.399 | –0.104 | 0.753 | 0.780 |
| H_2O | 647.1 | 220.55 | 0.345 | 0.731 | 0.156 | 0.071 | –0.613 | –0.502 | 0.846 | 0.926 |
| C_2H_4 | 282.3 | 50.40 | 0.087 | 1.676 | 0.685 | –0.102 | 0.119 | 0.963 |
Substitution of values so far determined into Eqs. (G), (J ), and (K) reduces these three equations to the following:
K=\frac{0.0493 x_{\mathrm{EtOH}} \gamma_{\mathrm{EtOH}}}{y_{\mathrm{C}_2 \mathrm{H}_4} x_{\mathrm{H}_2 \mathrm{O}} \gamma_{\mathrm{H}_2 \mathrm{O}}} (L)
y_{\mathrm{C}_2 \mathrm{H}_4}=1-0.907 x_{\mathrm{EtOH}} \gamma_{\mathrm{EtOH}}-0.493 x_{\mathrm{H}_2 \mathrm{O}} \gamma_{\mathrm{H}_2 \mathrm{O}}
x_{\mathrm{H}_2 \mathrm{O}}=1-x_{\mathrm{EtOH}}
The only remaining undetermined thermodynamic properties are \gamma_{\mathrm{H}_2 \mathrm{O}} \text { and } \gamma_{\mathrm{EtOH}} . Because of the highly nonideal behavior of a liquid solution of ethanol and water, these must be determined from experimental data. The required data, found from VLE measurements, are given by Otsuki and Williams ^9 From their results for the ethanol/water system, one can estimate values of \gamma_{\mathrm{H}_2 \mathrm{O}} \text { and } \gamma_{\mathrm{EtOH}} at 200°C. (Pressure has little effect on the activity coefficients of liquids.)
An intuitive ad hoc procedure for solution of the foregoing three equations is as follows:
1. Assume a value for for x_{EtOH} and calculate x_{\mathrm{H}_2 \mathrm{O}} by Eq. (K).
2. Determine \gamma_{\mathrm{H}_2 \mathrm{O}} \text { and } \gamma_{\mathrm{EtOH }} from data in the reference cited.
3. Calculate y_{\mathrm{C}_2 \mathrm{H}_4} by Eq. (M).
4. Calculate K by Eq. (L) and compare with the value of 0.0310 determined from standard- reaction data.
5. If the two values agree, the assumed value of x_{EtOH} is correct. If they do not agree, assume a new value of x_{EtOH} and repeat the procedure.
\text { If } x_{\mathrm{EtOH}}=0.06 \text {, then by Eq. }(K), x_{\mathrm{H}_2 \mathrm{O}}=0.94 \text {, } and from the reference cited,
\gamma_{\mathrm{EtOH}}=3.34 \quad \text { and } \quad \gamma_{\mathrm{H}_2 \mathrm{O}}=1.00
By Eq. (M),
y_{\mathrm{C}_2 \mathrm{H}_4}=1-(0.907)(3.34)(0.06)-(p .493)(1.00)(0.94)=0.355The value of K given by Eq. (L) is then:
K=\frac{(0.0493)(0.06)(3.34)}{(0.355)(0.94)(1.00)}=0.0296This result is in close enough agreement with the value, 0.0310, found from standard- reaction data to make further calculations unnecessary, and the liquid-phase composition is essentially as assumed \left(x_{\mathrm{EtOH}}=0.06, x_{\mathrm{H}_2 \mathrm{O}}=0.94\right) . The remaining vapor phase compositions ( y_{\mathrm{C}_2 \mathrm{H}_4} has already been determined as 0.356) are found by solution of Eq. (I) for y_{\mathrm{H}_2 \mathrm{O}} \text { or } y_{\mathrm{EtOH}} . All results are summarized in the following table.
| x_i | y_i | |
| EtOH | 0.060 | 0.180 |
| H_2O | 0.940 | 0.464 |
| C_2H_4 | 0.000 | 0.356 |
| \sum_i x_i=1.000 | \sum_i y_i=1.000 |
These results provide reasonable estimates of actual values, provided no other reactions take place.
^9H. Otsuki and F. C. Williams, Chem. Engr. Progr. Symp. Series No. 6, vol. 49, pp. 55–67, 1953.