To determine the equilibrium constant expression for the given reversible reaction:
[tex]\[
CO (g) + 2 H_2 (g) \longleftrightarrow CH_3OH (g)
\][/tex]
we need to follow the general form for writing equilibrium constant expressions.
For a general reaction:
[tex]\[
aA + bB \longleftrightarrow cC + dD
\][/tex]
the equilibrium constant expression [tex]\( K_{eq} \)[/tex] is given by:
[tex]\[
K_{eq} = \frac{[C]^c [D]^d}{[A]^a [B]^b}
\][/tex]
where:
- [tex]\([C]\)[/tex] and [tex]\([D]\)[/tex] are the concentrations of the products,
- [tex]\([A]\)[/tex] and [tex]\([B]\)[/tex] are the concentrations of the reactants,
- [tex]\(c\)[/tex] and [tex]\(d\)[/tex] are the stoichiometric coefficients of the products,
- [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are the stoichiometric coefficients of the reactants.
Applying this to the given reaction:
[tex]\[
CO (g) + 2 H_2 (g) \longleftrightarrow CH_3OH (g)
\][/tex]
The reactants are [tex]\(CO\)[/tex] and [tex]\(H_2\)[/tex] with their stoichiometric coefficients being 1 and 2, respectively. The product is [tex]\(CH_3OH\)[/tex] with a stoichiometric coefficient of 1.
The equilibrium constant expression [tex]\( K_{eq} \)[/tex] for this reaction would be:
[tex]\[
K_{eq} = \frac{[CH_3OH]}{[CO][H_2]^2}
\][/tex]
Thus, the correct equilibrium constant expression for the given system is:
[tex]\[
K_{eq} = \frac{\left[ CH_3OH \right]}{[ CO ][ H_2 ]^2}
\][/tex]
