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:

if you mean hydrogen and oxygen then water is the product

conduction is the transfer of thermal energy (heat in transfer) due to collisions between the molecules in the object. collisions between adjacent atoms and molecules transfer kinetic energy from the warmer to the cooler object. the objects must be in physical contact.