Question

9. A mixture of carbon monoxide, hydrogen, and methanol is at equilibrium. The balanced chemical equation is: CO(g) + 2 H2(g) ⇌ CH3OH(g). At equilibrium, the mixture contains 0.0960 M CO, 0.191 M H2, and 0.150 M CH3OH. What is the value for Kc?

Answer 1

The value for the equilibrium constant Kc is 42.83

Step-by-step explanation:

Step 1: Data given

At equilibrium, the mixture contains:

0.0960 M CO,

0.191 M H2,

and 0.150 M CH3OH.

Step 2: The balance equation

CO(g) + 2 H2(g) ⇌ CH3OH(g).

Step 3: Calculate the value for the equilibrium constant Kc

Kc = [CH3OH] / [CO]*[H2]²

⇒with Kc = the equilibrium constant = TO BE DETERMINED

⇒with [CH3OH] = 0.150 M

⇒with [CO] = 0.0960 M

⇒with [H2] = 0.191 M

Kc = 0.150 M / (0.0960 M *( 0.191 M)²)

Kc = 42.83

The value for the equilibrium constant Kc is 42.83

Answer 2

42.8

Step-by-step explanation:

Let's consider the following reaction at equilibrium.

CO(g) + 2 H₂(g) ⇌ CH₃OH(g)

The concentration equilibrium constant (Kc) is equal to the product of the concentration of the products raised to their stoichiometric coefficients divided by the product of the concentration of the reactants raised to their stoichiometric coefficients.

The concentration equilibrium constant for this reaction is:

`Kc = ([CH_3OH])/([CO]* [H_2]^(2) ) = (0.150)/(0.0960* 0.191^(2) ) = 42.8`