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 100L tank with 33.mol of ethylene gas and 16.mol of water vapor. When the mixture has come to equilibrium he determines that it contains 26.8mol of ethylene gas and 9.8mol of water vapor. The engineer then adds another 8.0mol 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.

Answer 1

The value is
`k = 7.111 \  moles`

Step-by-step explanation:

Generally the reaction between ethylene and water vapor is

`CH_2CH_2_((g)) \ \ \   +\ \ \ \ H_2O\ \ \  \rightarrow\ \ \ \  C_2H_5OH_((g))`

From the question the initial number of moles of ethylene is 33.mol

the initial number of moles of water vapor is 16.mol

So

At initial

`CH_2CH_2_((g)) \ \ \   +\ \ \ \ H_2O\ \ \  \rightarrow\ \ \ \  C_2H_5OH_((g))`

33.mol 16.mol 0 mols

Now at first equilibrium the number of moles of ethylene is 26.8mol

the number of moles of water vapor is 9.8mol

So the number of moles of ethanol is 33-26.8 = 6.2 mol

So at equilibrium

`CH_2CH_2_((g)) \ \ \   +\ \ \ \ H_2O\ \ \  \rightarrow\ \ \ \  C_2H_5OH_((g))`

26.8mol 9.8mol 6.2 mol

Generally the equilibrium constant is mathematically represented as

`K =  ([C_2H_5OH])/([CH_2CH_2] [H_2O])`

Here
`[C_2H_5OH]` is the concentration of ethanol which is mathematically represented as

`[C_2H_5OH]  =  (6.2 \ mol)/(100L)`

=>
`[C_2H_5OH] =   0.062 mol/L`

Also

`[CH_2CH_2]  =  (26.8 \ mol)/(100L)`

`[CH_2CH_2]  =  0.268 mol/L`

Also

`[H_2O]  =  ( 9.8 \ mol)/(100L)`

`[H_2O]  =  0.098 mol/L`

So

`K =  (0.062)/(0.268*  0.098 )`

`K = 2.3606`

From the question we are told that 8 moles was added to ethylene

So volume of ethylene becomes 26.8 + 8 = 34.8 moles

So after the addition

`CH_2CH_2_((g)) \ \ \   +\ \ \ \ H_2O\ \ \  \rightarrow\ \ \ \  C_2H_5OH_((g))`

34.8mol 9.8mol 6.2 mol

At the second equilibrium

`CH_2CH_2_((g)) \ \ \   +\ \ \ \ H_2O\ \ \  \rightarrow\ \ \ \  C_2H_5OH_((g))`

(34.8- z)mol (9.8-z)mol ( 6.2+z) mol

Generally the equilibrium constant is mathematically represented as

`K =  ([C_2H_5OH])/([CH_2CH_2] [H_2O])`

Here

`[C_2H_5OH]` is now equal to

`[C_2H_5OH] =  (6.2+z)/(100)`

`[CH_2CH_2]=  ( 34.8- z )/(100)`

`[CH_2CH_2]=  ( 9.8-z )/(100)`

So m

`2.3606 =  (  (6.2+z)/(100))/([ (34.8- z )/(100)] [( 9.8-z  )/(100)])`

`2.3606 =  (  (6.2+z)/(100))/(( z^2 -44.6 z + 341.04)/(10000) )`

=>
`2.3606 = ( 620 - 100z)/(z^2 -44.6 z + 341.04)`

=>
`2.3606z^2 - 105.283 z + 805.05 =620 - 100z`

=>
`2.3606z^2 - 205.283 z + 185.059 = 0`

Multiply through by minus

=>
`2.3606z^2 -108.498z + 3336.7 = 0`

Solving this using quadratic equation

So
`z = 0.911`

Hence the number of moles of ethanol present at the second equilibrium is

`k = 0.911 +6.2`

`k = 7.111 \  moles`