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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]\[ Keq = \frac{[CO][H_2]^2}{[CH_3OH]} \][/tex]

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

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

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

Sagot :

GinnyAnswer
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]
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