Certainly! Let's consider the reversible reaction:
[tex]\[ CO (g) + 2 H_2 (g) \rightleftharpoons CH_3OH (g) \][/tex]
To determine the equilibrium constant expression ([tex]\( K_{\text{eq}} \)[/tex]), we need to use the general formula for [tex]\( K_{\text{eq}} \)[/tex] of a reaction of the form:
[tex]\[ aA + bB \rightleftharpoons cC + dD \][/tex]
The equilibrium constant is given by:
[tex]\[ K_{\text{eq}} = \frac{[C]^c [D]^d}{[A]^a [B]^b} \][/tex]
For our specific reaction:
[tex]\[ CO (g) + 2 H_2 (g) \rightleftharpoons CH_3OH (g) \][/tex]
- [tex]\( A \)[/tex] = [tex]\( CO \)[/tex] with a coefficient of 1
- [tex]\( B \)[/tex] = [tex]\( H_2 \)[/tex] with a coefficient of 2
- [tex]\( C \)[/tex] = [tex]\( CH_3OH \)[/tex] with a coefficient of 1
- There are no other products.
According to the formula, the equilibrium constant expression will be:
[tex]\[ K_{\text{eq}} = \frac{[CH_3OH]^1}{[CO]^1[H_2]^2} \][/tex]
Simplified, it appears as:
[tex]\[ K_{\text{eq}} = \frac{[CH_3OH]}{[CO][H_2]^2} \][/tex]
Thus, the correct expression for the equilibrium constant for the given reaction is:
[tex]\[ K_{\text{eq}} = \frac{[CH_3OH]}{[CO][H_2]^2} \][/tex]
