Question

Consider the reaction to produce methanolCO(g) + 2H2 (g) <-----> CH3OHAn equilibrium mixture in a 2.00-L vessel is found to contain 0.0406 mol CH3OH, 0.170 mol CO, and 0.302 mol H2 at 500 K.Calculate Kc at this temperature.Does equilibrium favor reactants or products?

Answer 1

Answer : The value of
`K_c` of the reaction is 10.5 and the reaction is product favored.

Explanation : Given,

Moles of
`CH_3OH` at equilibrium = 0.0406 mole

Moles of
`CO` at equilibrium = 0.170 mole

Moles of
`H_2` at equilibrium = 0.302 mole

Volume of solution = 2.00 L

First we have to calculate the concentration of
`CH_3OH,CO\text{ and }H_2` at equilibrium.

`\text{Concentration of }CH_3OH=\frac{\text{Moles of }CH_3OH}{\text{Volume of solution}}=(0.0406mole)/(2.00L)=0.0203M`

`\text{Concentration of }CO=\frac{\text{Moles of }CO}{\text{Volume of solution}}=(0.170mole)/(2.00L)=0.085M`

`\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=(0.302mole)/(2.00L)=0.151M`

Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is,

`CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)`

The expression of equilibrium constant
`K_c` for the reaction will be:

`K_c=([CH_3OH])/([CO][H_2]^2)`

Now put all the values in this expression, we get :

`K_c=((0.0203))/((0.085)* (0.151)^2)`

`K_c=10.5`

Therefore, the value of
`K_c` of the reaction is, 10.5

There are 3 conditions:

When
`K_(c)>1`; the reaction is product favored.

When
`K_(c)<1`; the reaction is reactant favored.

When
`K_(c)=1`; the reaction is in equilibrium.

As the value of
`K_(c)>1`. So, the reaction is product favored.