While ethanol (CH3CH2OH is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene CH2CH2) with water vapor at elevated temperatures. A chemical engineer studying this reaction fills a 50.0 L tank at 22. °C with 24. mol of ethylene gas and 24. mol of water vapor. He then raises the temperature considerably, and when the mixture has come to equilibrium determines that it contains 15.4 mol of ethylene gas and 15.4 mol of water vapor The engineer then adds another 12. mol of water, and allows the mixture to come to equilibrium again. Calculate the moles of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.

Thus, since at the beginning there are 24 moles of ethylene and once the equilibrium is reached, there are 15.4 moles of ethylene, the change [tex]x[/tex] result:

[tex]x=24mol-15.4mol=8.6mol[/tex]

In such a way, the equilibrium constant is then:

[tex]K=\frac{\frac{8.6mol}{50.0L} }{\frac{15.4mol}{50.0L}\frac{15.4mol}{50.0L}} =1.81[/tex]

Thereby, the initial moles for the second equilibrium are modified as shown on the denominator in the modified law of mass action by considering the added 12 moles of water:

[tex]K=\frac{\frac{8.6mol+x_2}{50.0L} }{\frac{15.4mol-x_2}{50.0L}\frac{15.4mol+12mol-x_2}{50.0L}} =1.81[/tex]

Thus, the second change, finally result (solving by solver or quadratic equation):

[tex]x_2=2.72mol[/tex]

Finally, such second change equals the moles of ethanol after equilibrium based on the stoichiometry:

[tex]n_{CH_3CH_2OH }=8.6mol+2.72mol=11.32mol[/tex]

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