Question

The Kc for the following reaction is 15 at 220 ∘C.

CO(g)+2H₂(g)⇌CH₃OH(g)

If an equilibrium mixture contains 0.44 MCO and 0.15 M H₂, what is the molar concentration of CH₃OH?

Answer 1

The molar concentration of CH₃OH at equilibrium can be calculated using the equilibrium constant expression Kc and the given concentrations. For the reaction CO(g) + 2H₂(g) ⇌ CH₃OH(g) with Kc = 15, and given [CO] = 0.44 M and [H₂] = 0.15 M, the molar concentration of CH₃OH is 0.1485 M.

Step-by-step explanation:

To find the molar concentration of CH₃OH in the equilibrium mixture when the Kc for the reaction CO(g) + 2H₂(g) ⇌ CH₃OH(g) is 15 at 220 ℃, and the equilibrium concentrations of CO and H₂ are given, we can use the expression for the equilibrium constant:

Kc = [CH₃OH] / ([CO] * [H₂]2)

Rearranging this equation to solve for [CH₃OH] gives:

[CH₃OH] = Kc * [CO] * [H₂]2

Substituting the given values:

[CH₃OH] = 15 * 0.44 * (0.15)2

Calculating this:

[CH₃OH] = 15 * 0.44 * 0.0225 = 0.1485 M

Therefore, the equilibrium molar concentration of CH₃OH is 0.1485 M.