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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.

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okpalawalter8

Answer:

The value is           [tex]k = 7.111 \  moles  [/tex]

Explanation:

Generally the reaction between ethylene and water vapor is

[tex]CH_2CH_2_{(g)} \ \ \   +\ \ \ \ H_2O\ \ \  \rightarrow\ \ \ \  C_2H_5OH_{(g)}[/tex]

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

[tex]CH_2CH_2_{(g)} \ \ \   +\ \ \ \ H_2O\ \ \  \rightarrow\ \ \ \  C_2H_5OH_{(g)}[/tex]

  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

[tex]CH_2CH_2_{(g)} \ \ \   +\ \ \ \ H_2O\ \ \  \rightarrow\ \ \ \  C_2H_5OH_{(g)}[/tex]

26.8mol                     9.8mol                   6.2 mol

Generally the equilibrium constant is mathematically represented as

   [tex]K =  \frac{[C_2H_5OH]}{[CH_2CH_2] [H_2O]}[/tex]

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

  [tex][C_2H_5OH]  =  \frac{6.2 \ mol}{100L}[/tex]

=>  [tex][C_2H_5OH] =   0.062 mol/L [/tex]

Also

     [tex] [CH_2CH_2]  =  \frac{26.8 \ mol}{100L}[/tex]

      [tex] [CH_2CH_2]  =  0.268 mol/L[/tex]

Also

     [tex][H_2O]  =  \frac{ 9.8 \ mol}{100L}[/tex]

      [tex] [H_2O]  =  0.098 mol/L[/tex]

So

      [tex]K =  \frac{0.062}{0.268*  0.098 }[/tex]

        [tex]K = 2.3606 [/tex]

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

[tex]CH_2CH_2_{(g)} \ \ \   +\ \ \ \ H_2O\ \ \  \rightarrow\ \ \ \  C_2H_5OH_{(g)}[/tex]

    34.8mol                9.8mol                    6.2 mol  

At the second equilibrium

[tex]CH_2CH_2_{(g)} \ \ \ �� +\ \ \ \ H_2O\ \ \  \rightarrow\ \ \ \  C_2H_5OH_{(g)}[/tex]

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

Generally the equilibrium constant is mathematically represented as

       [tex]K =  \frac{[C_2H_5OH]}{[CH_2CH_2] [H_2O]}[/tex]

Here

   [tex][C_2H_5OH][/tex]  is now equal to  

       [tex][C_2H_5OH] =  \frac{6.2+z}{100}[/tex]

       [tex][CH_2CH_2]=  \frac{ 34.8- z }{100}[/tex]

         [tex][CH_2CH_2]=  \frac{ 9.8-z }{100}[/tex]

So m

    [tex]2.3606 =  \frac{  \frac{6.2+z}{100}}{[ \frac{34.8- z }{100}] [\frac{ 9.8-z  }{100}]}[/tex]

[tex]2.3606 =  \frac{  \frac{6.2+z}{100}}{\frac{ z^2 -44.6 z + 341.04}{10000} }[/tex]

=>    [tex]2.3606 = \frac{ 620 - 100z}{z^2 -44.6 z + 341.04}[/tex]

=>      [tex]2.3606z^2 - 105.283 z + 805.05 =620 - 100z[/tex]

=>      [tex]2.3606z^2 - 205.283 z + 185.059 = 0[/tex]

Multiply through by minus

=>      [tex]2.3606z^2 -108.498z + 3336.7 = 0[/tex]

Solving this using quadratic equation

So  [tex]z = 0.911[/tex]

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

        [tex]k = 0.911 +6.2 [/tex]

        [tex]k = 7.111 \  moles  [/tex]