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Which will not appear in the equilibrium constant expression for the reaction below?

[tex]\[ \text{H}_2\text{O} (g) + \text{C} (s) \longleftrightarrow \text{H}_2 (g) + \text{CO} (g) \][/tex]

A. [tex]\([ \text{H}_2 \text{O} ]\)[/tex]
B. [C]
C. [tex]\([ \text{H}_2 ]\)[/tex]
D. [CO]



Answer :

GinnyAnswer
In the context of equilibrium constants, certain phases of matter are not included in the expression. For the reaction

[tex]\[ \text{H}_2\text{O}(\text{g}) + \text{C}(\text{s}) \leftrightarrow \text{H}_2(\text{g}) + \text{CO}(\text{g}) \][/tex]

we can write the equilibrium constant expression [tex]\(K_c\)[/tex] as follows:

[tex]\[ K_c = \frac{[\text{H}_2][\text{CO}]}{[\text{H}_2\text{O}]} \][/tex]

Here, [tex]\( [\text{H}_2\text{O}] \)[/tex], [tex]\( [\text{H}_2] \)[/tex], and [tex]\( [\text{CO}] \)[/tex] represent the molar concentrations of water vapor, hydrogen gas, and carbon monoxide gas, respectively.

As a rule in equilibrium constant expressions, the concentrations of pure solids and pure liquids do not appear. This is because their activities are considered constant and equal to 1, so they do not affect the equilibrium constant [tex]\(K_c\)[/tex].

In our reaction, carbon ([tex]\( \text{C} \)[/tex]) is a solid. Therefore, it will not appear in the equilibrium constant expression.

So, the answer to the question "Which will not appear in the equilibrium constant expression for the reaction below?" is:
[tex]\[ [\text{C}] \][/tex]

Answer:

Explanation:                                                  

    In the equilibrium constant expression for a reaction, the substances that are in the gaseous or aqueous state are included, while solids and liquids are not. So, the substance that will not appear in the equilibrium constant expression for the given reaction is option C: 2H2(g). This is because gases like H2(g) are included in the equilibrium constant expression, but solids like Pt(s) are not.