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Consider the following reversible reaction:

[tex]\[ C (s) + O_2 (g) \leftrightarrow CO_2 (g) \][/tex]

What is the equilibrium constant expression for the given system?

A. [tex]\( K_{eq} = \frac{[CO_2]}{[C][O_2]} \)[/tex]

B. [tex]\( K_{eq} = \frac{[CO_2]}{[O_2]} \)[/tex]

C. [tex]\( K_{eq} = \frac{[C][O_2]}{[CO_2]} \)[/tex]

D. [tex]\( K_{eq} = \frac{[O_2]}{[CO_2]} \)[/tex]


Sagot :

To determine the equilibrium constant expression for the given reversible reaction:

[tex]\[ C(s) + O_2(g) \leftrightarrow CO_2(g) \][/tex]

we need to consider the states of the reactants and products involved.

1. Pure Solids and Liquids in Equilibrium Expressions:
For equilibrium expressions, we exclude concentrations of pure solids and pure liquids. This is because their concentrations do not change during the reaction and are therefore considered constant.

2. Components in the Reaction:
- Solid carbon (C) is in the solid state (s).
- Oxygen ([tex]\(O_2\)[/tex]) is in the gaseous state (g).
- Carbon dioxide ([tex]\(CO_2\)[/tex]) is in the gaseous state (g).

3. Writing the Equilibrium Constant Expression:
Since solid carbon is a pure solid, it will not appear in the equilibrium constant expression. The equation involves the concentrations of [tex]\(O_2\)[/tex] and [tex]\(CO_2\)[/tex].

The correct equilibrium constant expression (K_eq) will only include the gaseous components:

[tex]\[ K_{eq} = \frac{[CO_2]}{[O_2]} \][/tex]

Therefore, the correct choice is:

[tex]\[ K_{eq} = \frac{\left[ CO_2 \right]}{\left[ O_2 \right]} \][/tex]

or equivalently,

[tex]\[ K_{eq} = \frac{[CO_2]}{[O_2]} \][/tex]

Hence, the correct answer is:

[tex]\[ \boxed{2} \][/tex]
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