Chemistry: 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

While ethanol (CH3CH2OH is produced naturally by

Ответ:

katepotter2019

21.01.2022

Chemistry

$n_{CH_3CH_2OH }=11.32mol$

Explanation:

Hello,

In this case, the reaction is:

$CH_2CH_2+H_2O\rightleftharpoons CH_3CH_2OH$

Thus, the law of mass action turns out:

$K=\frac{[CH_3CH_2OH]_{eq}}{[CH_2CH_2]_{eq}[H_2O]_{eq}}$

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 result:

x=24mol-15.4mol=8.6mol

In such a way, the equilibrium constant is then:

$K=\frac{\frac{8.6mol}{50.0L} }{\frac{15.4mol}{50.0L}\frac{15.4mol}{50.0L}} =1.81$

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:

$K=\frac{\frac{8.6mol+x_2}{50.0L} }{\frac{15.4mol-x_2}{50.0L}\frac{15.4mol+12mol-x_2}{50.0L}} =1.81$

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

x_2=2.72mol

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

$n_{CH_3CH_2OH }=8.6mol+2.72mol=11.32mol$

Best regards.

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