Question

consider the following reversible reaction. c(s)+o₂(g)⇌co₂(g) what is the equilibrium constant expression for the given system? o k_eq = co₂/(co₂) o k_eq = co₂/o₂ o k_eq = co₂/co₂ o k_eq = o₂/co₂

Explanation:

Step1: Recall equilibrium - constant formula

For a general reaction $aA + bB
ightleftharpoons cC + dD$, the equilibrium - constant expression $K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$, where square - brackets denote molar concentrations and $a$, $b$, $c$, $d$ are stoichiometric coefficients.

Step2: Consider the states of reactants and products

In the reaction $C(s)+O_2(g)
ightleftharpoons CO_2(g)$, carbon ($C$) is a solid. The concentration of a solid is considered constant and is not included in the equilibrium - constant expression.

Step3: Write the equilibrium - constant expression

The equilibrium - constant expression for the given reaction is $K_{eq}=\frac{[CO_2]}{[O_2]}$ since the solid carbon is not part of the expression.

Answer:

$K_{eq}=\frac{[CO_2]}{[O_2]}$ (the second option)