# ?Ethylene oxide as a vapor and water as liquid, both at $$25^{\circ} \mathrm{C}$$ and

Chapter 14, Problem 14.39

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Ethylene oxide as a vapor and water as liquid, both at $$25^{\circ} \mathrm{C}$$ and 101.33 kPa, react to form a liquid solution containing ethylene glycol (1,2-ethanediol) at the same conditions:

$$\left\langle\left(\mathrm{CH}_{2}\right)_{2}\right\rangle \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CH}_{2} \mathrm{OH} \cdot \mathrm{CH}_{2} \mathrm{OH}$$

If the initial molar ratio of ethylene oxide to water is 3.0, estimate the equilibrium conversion of ethylene oxide to ethylene glycol.

At equilibrium the system consists of liquid and vapor in equilibrium, and the intensive state of the system is fixed by the specification of T and P. Therefore, one must first determine the phase compositions, independent of the ratio of reactants. These results may then be applied in the material-balance equations to find the equilibrium conversion.

Choose as standard states for water and ethylene glycol the pure liquids at 1 bar and for ethylene oxide the pure ideal gas at 1 bar. Assume any water present in the liquid phase has an activity coefficient of unity and that the vapor phase is an ideal gas. The partial pressure of ethylene oxide over the liquid phase is given by:

$$p_{i} / \mathrm{kPa}=415 x_{i}$$

The vapor pressure of ethylene glycol at  $$25^{\circ} \mathrm{C}$$ is so low that its concentration in the vapor phase is negligible.

Text Transcription:

25^circ C

langle(CH_2)_2 rangle O+H_2O rightarrow CH_2OH cdot CH_2OH

p_i/kPa=415x_i

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