Consider the following reversible reaction:

$CO(g)+2 H_2(g) \longleftrightarrow CH_3 OH(g)$

What is the equilibrium constant expression for the given system?

A$K_{e q}=\frac{\left[CH_3 OH\right]}{[CO]\left[H_2\right]^2}$
B$K_{e q}=\frac{[CO]\left[H_2\right]}{\left[CH_3 OH\right]}$
C$K_{e q}=\frac{\left[CH_3 OH\right]}{[CO]\left[H_2\right]}$
D$K_{e q}=\frac{[CO]\left[H_2\right]^2}{\left[CH_3 OH\right]}$

Question

delaney138 delaney138

01-11-2026
Chemistry

Answered

Consider the following reversible reaction:

$CO(g)+2 H_2(g) \longleftrightarrow CH_3 OH(g)$

What is the equilibrium constant expression for the given system?

A$K_{e q}=\frac{\left[CH_3 OH\right]}{[CO]\left[H_2\right]^2}$
B$K_{e q}=\frac{[CO]\left[H_2\right]}{\left[CH_3 OH\right]}$
C$K_{e q}=\frac{\left[CH_3 OH\right]}{[CO]\left[H_2\right]}$
D$K_{e q}=\frac{[CO]\left[H_2\right]^2}{\left[CH_3 OH\right]}$

Verified Answer :

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GinnyAnswer GinnyAnswer · Answer 1 · 2026-11-01T16:31:47-05:00

Identify the reactants ($CO$, $H_2$) and products ($CH_3OH$) in the reaction.
Determine the stoichiometric coefficients: $1$ for $CO$, $2$ for $H_2$, and $1$ for $CH_3OH$.
Apply the Law of Mass Action, which puts products in the numerator and reactants in the denominator, with concentrations raised to their stoichiometric coefficients.
Select the expression that matches the derived formula: $\boxed{K_{eq} = \frac{[CH_3OH]}{[CO][H_2]^2}}$.

Explanation

Understanding the Law of Mass Action

The problem asks for the equilibrium constant expression, $K_{eq}$, for the reversible chemical reaction given by:
\[CO(g) + 2H_2(g) \longleftrightarrow CH_3OH(g)\]
According to the Law of Mass Action, for any reversible reaction $aA + bB \longleftrightarrow cC + dD$, the equilibrium constant $K_{eq}$ is defined as the ratio of the product of the concentrations of the products raised to their respective stoichiometric coefficients to the product of the concentrations of the reactants raised to their respective stoichiometric coefficients. The formula is:
\[K_{eq} = \frac{[C]^c [D]^d}{[A]^a [B]^b}\]
where the brackets $[\cdot]$ denote the molar concentrations of the species at equilibrium.

Constructing the Expression

From the balanced chemical equation, we can identify the following components:

Reactants: $CO$ and $H_2$.
Products: $CH_3OH$.
Stoichiometric coefficients:
For $CO$: $1$
For $H_2$: $2$
For $CH_3OH$: $1$

Applying the Law of Mass Action formula:

Place the concentration of the product $CH_3OH$ in the numerator, raised to the power of its coefficient, which is $1$: $[CH_3OH]^1$.
Place the concentrations of the reactants $CO$ and $H_2$ in the denominator, raised to their respective coefficients, $1$ and $2$: $[CO]^1 \cdot [H_2]^2$.

Therefore, the equilibrium constant expression is:
\[K_{eq} = \frac{[CH_3OH]}{[CO][H_2]^2}\]

Final Conclusion and Selection

Comparing the derived expression with the provided options:

A$K_{eq} = \frac{[CH_3OH]}{[CO][H_2]^2}$

B$K_{eq} = \frac{[CO][H_2]}{[CH_3OH]}$

C$K_{eq} = \frac{[CH_3OH]}{[CO][H_2]}$

D$K_{eq} = \frac{[CO][H_2]^2}{[CH_3OH]}$

The expression matches option a. Thus, the correct equilibrium constant expression is $\boxed{K_{eq} = \frac{[CH_3OH]}{[CO][H_2]^2}}$.

Examples

The concept of equilibrium constants is fundamental in chemical engineering and industrial chemistry. For example, in the production of methanol ($CH_3OH$) from synthesis gas ($CO$ and $H_2$), knowing the equilibrium constant helps engineers determine the temperature and pressure conditions that will maximize the yield of methanol while minimizing energy waste. By setting up the correct $K_{eq}$ expression, they can calculate how much product is formed at a given state, ensuring that the process runs efficiently in a large-scale chemical plant.