Question

consider the following reversible reaction.
co(g) + 2h₂(g) ⇌ ch₃oh(g)
what is the equilibrium constant expression for the given system?
○ $k_{eq} = \frac{\text{co}\text{h}_2^2}{\text{ch}_3\text{oh}}$
○ $k_{eq} = \frac{\text{ch}_3\text{oh}}{\text{co}\text{h}_2^2}$
○ $k_{eq} = \frac{\text{co}\text{h}_2}{\text{ch}_3\text{oh}}$
○ $k_{eq} = \frac{\text{ch}_3\text{oh}}{\text{co}\text{h}_2}$

Explanation:

The equilibrium constant expression ($K_{eq}$) for a reversible reaction $aA + bB
ightleftharpoons cC + dD$ is given by $K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$, where the concentrations of products are in the numerator and reactants in the denominator, each raised to their stoichiometric coefficients. For the reaction $\text{CO}(g) + 2\text{H}_2(g)
ightleftharpoons \text{CH}_3\text{OH}(g)$, the product is $\text{CH}_3\text{OH}$ (stoichiometric coefficient 1) and reactants are $\text{CO}$ (stoichiometric coefficient 1) and $\text{H}_2$ (stoichiometric coefficient 2). So, $K_{eq}=\frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}$.

Answer:

B. $K_{eq} = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}$