Question

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 75.0 L tank at 18. °C with 29. mol of ethylene gas and 16. mol of water vapor. He then raises the temperature considerably, and when the mixture has come to equilibrium determines that it contains 16. mol of ethylene gas and 3.0 mol of water vapor. 囲 The engineer then adds another 15. mol of ethylene, 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.

Answer 1

Answer:

$n_{C2H_5OH}^{eq}=14.234mol$

Explanation:

Hello,

In this case, the reaction is:

$C_2H_4+H_2O\rightleftharpoons CH_3CH_2OH$

Thus, the law of mass action turns out:

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

Thus, since at the beginning there are 29 moles of ethylene and once the equilibrium is reached, there are 16 moles of ethylene, the change $x$ result:

$[CH_2CH_2]_{eq}=29mol-x=16mol\\x=29-16=13mol$

In such a way, the equilibrium constant is then:

$Kc=\frac{\frac{x}{V} }{\frac{16mol}{V}* \frac{3mol}{V}} =\frac{\frac{13mol}{75.0L} }{\frac{16mol}{75.0L}* \frac{3mol}{75.0L}} =20.31$

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 15 moles of ethylene:

$Kc=\frac{\frac{13+x_2}{V} }{\frac{16+15-x_2}{V}* \frac{3-x_2}{V}} =20.31$

Thus, the second change, $x_2$ finally result (solving by solver or quadratic equation):

$x_2=1.234mol$

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

$n_{C2H_5OH}^{eq}=x+x_2=13mol+1.234mol\\n_{C2H_5OH}^{eq}=14.234mol$

Best regards.