Question

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.

Answer 1

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 = $\frac{[CH_3OH]_{eq}}{[CO]_{eq}([H_2]_{eq})^2}$ = 26.7