To determine the equilibrium constant expression for the given reaction
[tex]\[ CO(g) + 2H_2(g) \longleftrightarrow CH_3OH(g), \][/tex]
we need to follow certain rules:
1. Identify the balanced chemical equation.
2. Write the expression for the equilibrium constant ([tex]\(K_{eq}\)[/tex]) as the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their coefficients in the balanced equation.
Here's the balanced chemical equation again for clarity:
[tex]\[ CO(g) + 2H_2(g) \longleftrightarrow CH_3OH(g). \][/tex]
From this reaction, we can see that:
- The product is [tex]\(CH_3OH\)[/tex] and its coefficient is 1.
- The reactants are [tex]\(CO\)[/tex] and [tex]\(H_2\)[/tex]; their coefficients are 1 and 2, respectively.
According to the law of mass action, the equilibrium constant expression ([tex]\(K_{eq}\)[/tex]) for this reaction is written as:
[tex]\[ K_{eq} = \frac{[\text{product}]^{\text{coefficient of product}}}{[\text{reactant}_1]^{\text{coefficient of reactant}_1} \times [\text{reactant}_2]^{\text{coefficient of reactant}_2}}. \][/tex]
Plugging in the values from our reaction:
- The concentration of the product [tex]\(CH_3OH\)[/tex] is [tex]\([CH_3OH]\)[/tex].
- The concentration of the reactant [tex]\(CO\)[/tex] is [tex]\([CO]\)[/tex].
- The concentration of the reactant [tex]\(H_2\)[/tex] is [tex]\([H_2]\)[/tex], and since its coefficient is 2, it will be raised to the power of 2.
Therefore, the equilibrium constant expression [tex]\(K_{eq}\)[/tex] for the reaction is:
[tex]\[ K_{eq} = \frac{[CH_3OH]}{[CO] \times [H_2]^2}. \][/tex]
So, the correct answer is:
[tex]\[ K_{eq} = \frac{[CH_3OH]}{[CO] \times [H_2]^2}. \][/tex]
