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.

26.7

Explanation:

The reaction that takes place is:

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

The initial concentrations for each species are:

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

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:

Finally we calculate the equilibrium constant:

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

CO + 2H₂   ⇄   CH₃OH

CO ⇒

H₂ ⇒  

CH₃OH ⇒

CO ⇒

H₂ ⇒  

CH₃OH ⇒ 0

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:

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

Find more information about Equilibrium constant here:

it's c. elements are placed in rows based on how may valencies electrons have.