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A reaction mixture in a 5.19 L flask at a certain temperature contains 26.9 g CO and 2.34 g H2. At equilibrium, the flask contains 8.65 g CH3OH. Calculate the equilibrium constant (Kc) for the reaction at this temperature.

Sagot :

Maacastrobridn

Answer:

26.7

Explanation:

The reaction that takes place is:

  • CO + 2H₂ ↔ CH₃OH

We convert the given masses to moles, using their respective molar masses:

  • CO ⇒ 26.9 g ÷ 28 g/mol = 0.961 mol
  • H₂ ⇒ 2.34 g ÷ 2 g/mol = 1.17 mol
  • CH₃OH ⇒ 8.65 g ÷ 32 g/mol = 0.270 mol

The initial concentrations for each species are:

  • CO ⇒ 0.961 mol / 5.19 L = 0.185 M
  • H₂ ⇒  1.17 mol / 5.19 L = 0.225 M
  • CH₃OH ⇒ 0

While the equilibrium concentration for CH₃OH, [CH₃OH]eq is:

  • 0.270 mol / 5.19 L = 0.052 M

We put the data in a table:

            CO      + 2H₂    ↔    CH₃OH

initial    0.185        0.225      ↔      0

eq          (0.185 - x) (0.225-2x)  ↔  x

We know that x = 0.052 M (That's the equilibrium concentration of CH₃OH).

We proceed to calculate [CO]eq and [H₂]eq:

  • [CO]eq = 0.185 - 0.052 = 0.133 M
  • [H₂]eq = 0.225 - 2*0.052 = 0.121 M

Finally we calculate the equilibrium constant:

  • Kc = [tex]\frac{[CH_3OH]_{eq}}{[CO]_{eq}([H_2]_{eq})^2}[/tex] = 26.7

The equilibrium constant (Kc) for the reaction at this temperature is 26.7

Chemical reaction:

CO + 2H₂   ⇄   CH₃OH

  • Conversion of moles:

CO ⇒ [tex]\frac{26.9 g}{28 g/mol} = 0.961 mol[/tex]

H₂ ⇒  [tex]\frac{2.34 g }{ 2 g/mol} =1.17 mol[/tex]

CH₃OH ⇒ [tex]\frac{8.65 g}{32 g/mol} = 0.270 mol[/tex]

  • The initial concentrations for each species are:

CO ⇒ [tex]\frac{0.961 mol}{5.19 L} = 0.185 M[/tex]

H₂ ⇒  [tex]\frac{1.17 mol}{5.19 L } = 0.225 M[/tex]

CH₃OH ⇒ 0

  • While the equilibrium concentration for CH₃OH, [CH₃OH]eq is:

[tex]\frac{0.270 mol}{ 5.19 L} = 0.052 M[/tex]

We put the data in a table:

                                  CO   +   2H₂   ⇄    CH₃OH

Initial                         0.185        0.225           0

Equilibrium              (0.185 - x) (0.225-2x)    x

x = 0.052 M (That's the equilibrium concentration of CH₃OH).

We proceed to calculate [CO]eq and [H₂]eq:

[tex][CO]_{eq}= 0.185 - 0.052 = 0.133 M\\\\\\[H_2]_{eq} = 0.225 - 2*0.052 = 0.121 M[/tex]

  • Calculation of  equilibrium constant:

[tex]K_c =\frac{[CH_3OH]_{eq}}{[CO]_{eq}[H_2_{eq}]^2} = 26.7[/tex]

Thus, the equilibrium constant (Kc) for the reaction at this temperature is 26.7

Find more information about Equilibrium constant here:

brainly.com/question/12270624

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