Question

The synthesis of methanol from carbon monoxide and hydrogen gas is described by the following chemical equation: CO(g)+2H2(g)⇌CH3OH(g) The equilibrium constant for this reaction at 25 ∘C is Kc=2.3×104. In this tutorial, you will use the equilibrium-constant expression to find the concentration of methanol at equilibrium, given the concentration of the reactants. Determine the expression for the equilibrium constant, Kc, for the reaction by identifying which terms will be in the numerator and denominator.

Answer 1

Answer:

$Kc=\frac{[CH_{3}OH]}{[CO].[H_{2}]^{2} }$

Explanation:

Let's consider the following reaction.

CO(g) + 2 H₂(g) ⇌ CH₃OH(g)

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

$Kc=\frac{[CH_{3}OH]}{[CO].[H_{2}]^{2} }$

If you wanted to calculate [CH₃OH], suppose that the molar concentrations for CO and H₂ at equilibrium are [CO] = 0.04 M and [H₂] = 0.08 M. The concentration of methanol at equilibrium is:

$Kc= 2.3 \times 10^{4} = \frac{[CH_{3}OH]}{[CO].[H_{2}]^{2} } =\frac{[CH_{3}OH]}{(0.04).(0.08)^{2} }\\ \ [CH_{3}OH]=5.9M$