consider the following reversible reaction.\nc(s)+o₂(g)⇌co₂(g)\nwhat is the equilibrium constant expression…

consider the following reversible reaction.\nc(s)+o₂(g)⇌co₂(g)\nwhat is the equilibrium constant expression for the given system?\n○ k_eq = \\frac{co₂}{co₂}\n○ k_eq = \\frac{co₂}{o₂}\n○ k_eq = \\frac{co₂}{co₂}\n○ k_eq = \\frac{o₂}{co₂}

consider the following reversible reaction.\nc(s)+o₂(g)⇌co₂(g)\nwhat is the equilibrium constant expression for the given system?\n○ k_eq = \\frac{co₂}{co₂}\n○ k_eq = \\frac{co₂}{o₂}\n○ k_eq = \\frac{co₂}{co₂}\n○ k_eq = \\frac{o₂}{co₂}

Answer

Explanation:

Step1: Recall equilibrium constant formula

For a reaction $aA + bB\rightleftharpoons cC + dD$, $K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$, where concentrations of reactants and products are used. Pure solids and pure liquids are not included in the equilibrium - constant expression because their concentrations are considered constant.

Step2: Analyze the given reaction

In the reaction $C(s)+O_2(g)\rightleftharpoons CO_2(g)$, carbon (C) is a solid. So, its concentration is not included in the $K_{eq}$ expression.

Step3: Write the equilibrium - constant expression

The equilibrium - constant expression for the reaction is $K_{eq}=\frac{[CO_2]}{[O_2]}$.

Answer:

$K_{eq}=\frac{[CO_2]}{[O_2]}$ (corresponding to the second option in the multiple - choice list)