Answered

Consider the following reversible reaction:
[tex]\[
CO (g) + 2H_2 (g) \longleftrightarrow CH_3OH (g)
\][/tex]

What is the equilibrium constant expression for the given system?

A. [tex]\[ K_{eq} = \frac{[CO][H_2]^2}{[CH_3OH]} \][/tex]

B. [tex]\[ K_{eq} = \frac{[CH_3OH]}{[CO][H_2]^2} \][/tex]

C. [tex]\[ K_{eq} = \frac{[CO][H_2]}{[CH_3OH]} \][/tex]

D. [tex]\[ K_{eq} = \frac{[CH_3OH]}{[CO][H_2]} \][/tex]



Answer :

GinnyAnswer
To find the equilibrium constant expression for the given chemical reaction, we need to follow these steps:

Step 1: Write down the balanced chemical equation for the reaction:
[tex]\[ CO(g) + 2H_2(g) \longleftrightarrow CH_3OH(g) \][/tex]

Step 2: Identify the reactants and the products. In this reaction:
- Reactants: CO and [tex]\(H_2\)[/tex]
- Product: [tex]\(CH_3OH\)[/tex]

Step 3: Recall the general form of the equilibrium constant expression for a reaction [tex]\(aA + bB \longleftrightarrow cC + dD\)[/tex], which is:
[tex]\[ Keq = \frac{[C]^c [D]^d}{[A]^a [B]^b} \][/tex]

Step 4: Apply the balanced equation to this general form. In our case:
- [tex]\(a = 1\)[/tex] for CO
- [tex]\(b = 2\)[/tex] for [tex]\(H_2\)[/tex]
- [tex]\(c = 1\)[/tex] for [tex]\(CH_3OH\)[/tex]
- Since there's no other product, there's no additional term in the numerator.

Step 5: Plug these values into the equilibrium constant expression:
[tex]\[ Keq = \frac{[CH_3OH]^1}{[CO]^1 [H_2]^2} \][/tex]

Simplifying the expression, we get:
[tex]\[ Keq = \frac{[CH_3OH]}{[CO][H_2]^2} \][/tex]

Step 6: Match this expression with the options provided:
- [tex]\(Keq = \frac{[CO][H_2]^2}{[CH_3OH]}\)[/tex]
- [tex]\(Keq = \frac{[CH_3OH]}{[CO][H_2]^2}\)[/tex]
- [tex]\(Keq = \frac{[CO][H_2]}{[CH_3OH]}\)[/tex]
- [tex]\(Keq = \frac{[CH_3OH]}{[CO][H_2]}\)[/tex]

The correct equilibrium constant expression matches:
[tex]\[ Keq = \frac{[CH_3OH]}{[CO][H_2]^2} \][/tex]

Thus, the correct choice is:
[tex]\[ \boxed{2} \][/tex]

Other Questions