Question 14.4: Calculate the equilibrium constant for the vapor-phase hydra...
Calculate the equilibrium constant for the vapor-phase hydration of ethylene at 145°C and at 320°C from data given in App. C.
Hydration of ethylene means reaction of ethylene with water vapor to produce ethanol vapor. Thus, we first determine values for ΔA, ΔB, ΔC, and ΔD for the reaction:
\mathrm{C}_2 \mathrm{H}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(g) \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(g)
The meaning of Δ is indicated by: \Delta=\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)-\left(\mathrm{C}_2 \mathrm{H}_4\right)-\left(\mathrm{H}_2 \mathrm{O}\right) the heat-capacity data of Table C.1:
ΔA = 3.518 − 1.424 − 3.470 = −1.376
\Delta B=(20.001-14.394-1.450) \times 10^{-3}=4.157 \times 10^{-3}
\Delta C=(-6.002+4.392-0.000) \times 10^{-6}=-1.610 \times 10^{-6}
\Delta D=(-0.000-0.000-0.121) \times 10^5=-0.121 \times 10^5
Values of \Delta H_{298}^{\circ} \text { and } \Delta G_{298}^{\circ} at 298.15 K for the hydration reaction are found from
the heat-of-formation and Gibbs-energy-of-formation data of Table C.4:
\Delta H_{298}^{\circ}=-235,100-52,510-(-241,818)=-45,792 \mathrm{~J} \cdot \mathrm{mol}^{-1}
\Delta G_{298}^{\circ}=-168,490-68,460-(-228,572)=-8378 \mathrm{~J} \cdot \mathrm{mol}^{-1}
For T = 145 + 273.15 = 418.15 K, values of the integrals in Eq. (14.18) are:
IDCPH(298.15, 418.15; −1.376, 4.157E-3, −1.610E-6, −0.121E+5) = −23.121
IDCPS(298.15, 418.15; −1.376, 4.157E-3, −1.610E-6, −0.121E+5) = −0.0692
Substitution of values into Eq. (14.18) for a reference temperature of 298.15 gives:
\frac{\Delta G^{\circ}}{R T}=\frac{\Delta G_0^{\circ}-\Delta H_0^{\circ}}{R T_0}+\frac{\Delta H_0^{\circ}}{R T}+\frac{1}{T} \int_{T_0}^T \frac{\Delta C_P^{\circ}}{R} d T-\int_{T_0}^T \frac{\Delta C_P^{\circ}}{R} \frac{d T}{T} (14.18)
\frac{\Delta G_{418}^{\circ}}{R T}=\frac{-8378+45,792}{(8.314)(298.15)}+\frac{-45,792}{(8.314)(418.15)}+\frac{-23.121}{418.15}+0.0692=1.9356
For T = 320 + 273.15 = 593.15 K,
IDCPH ( 298.15, 593.15; −1.376, 4.157E-3, −1.610E-6, −0.121E+5 ) = 22.632
IDCPS(298.15, 593.15; −1.376, 4.157E-3, −1.610E-6, −0.121E+5) = 0.0173
Thus, at 593.15 K:
\frac{\Delta G_{593}^{\circ}}{R T}=\frac{-8378+45,792}{(8.314)(298.15)}+\frac{-45,792}{(8.314)(593.15)}+\frac{22.632}{593.15}-0.0173=5.8286Finally,
@ 418.15 K : ln K = −1.9356 and K = 1.443 × 10^ −1
@ 593.15 K : ln K = −5.8286 and K = 2.942 × 10^ −3
Application of Eqs. (14.21), (14.22), and (14.24) provides an alternative solution to this example. By Eq. (14.21),
K_0=\exp \frac{8378}{(8.314)(298.15)}=29.366Moreover, \frac{\Delta H_0^{\circ}}{R T_0}=\frac{-45,792}{(8.314)(298.15)}=-18.473
With these values, the following results are readily obtained:
| K_0 | K_1 | K_2 | K | |
| 298.15 | 29.366 | 1 | 1 | 29.366 |
| 418.15 | 29.366 | 4.985 × 10−3 | 0.9860 | 1.443 × 10−1 |
| 593.15 | 29.366 | 1.023 × 10−4 | 0.9794 | 2.942 × 10−3 |
Clearly, the influence of K_1 is far greater than that of K_2. This is a typical result and is consistent with the observation that all of the lines on Fig. 14.2 are very nearly linear.
