Question

atm=

While ethanol (CH,CH,0H) is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene (CH,CH2) with water vapor at elevated temperatures. A chemical engineer studying this reaction fills a 500. mL flask with 4.7 atm of ethylene gas and 4.8 atm of water vapor. When the mixture has come to equilibrium he determines that it contains 1.6 atm of ethylene gas and 1.7 atm of water vapor. alb The engineer then adds another 1.6 atm of water, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 sinnificant dinits

Answer 1

Solution :-

C2H4 + H2O   ---- > CH3CH2OH

Using the difference in the pressure of the C2H4 we can find the equilibrium pressure of CH3CH2OH

Difference in the pressure of CH3CH2OH = 4.7 atm – 1.6 atm = 3.1 atm

Now using the pressure of each reactant and product we can find the equilibrium constant kp

Kp=[CH3CH2OH]/[C2H4][H2O]

Kp = [3.1]/[1.6][1.7]

Kp= 1.14

Now using the Kp we can find the new equilibrium pressure of ethanol after addition of another 1.6 atm water

After adding the water new pressure of the water = 1.7 atm + 1.6 atm = 3.3 atm

Now lets set up ICE table

C2H4 + H2O   ---- > CH3CH2OH

1.6atm   3.3 atm            3.1 atm

-x                  -x                      +x

1.6-x           3.3-x             3.1+x

Kp=[CH3CH2OH]/[C2H4][H2O]

1.14 = [3.1+x]/[1.6-x][3.3-x]

1.14= (x+3.1) / (x^2-4.9x+5.28)

Multiply both sides by x^2-4.9x+5.28

1.14x^2 – 5.586x + 6.0192 = x+3.1

subtract both sides by x+3.1

1.14x^2 – 5.586x + 6.0192-(x+3.1) = x+3.1 – ( x+3.1)

1.14x^2 – 6.586x+2.9192=0

Using the quadratic equation we get value of x as

X=0.4838

Therefore the equilibrium pressure of the ethanol is calculated as follows

[CH3CH2OH] = 3.1 atm + x

                         = 3.1 atm + 0.4838 atm

                         = 3.6 atm

Hence the pressure of ethanol after second equilibrium is 3.6 atm