consider the following reversible reaction. c(s)+o₂(g)⇌co₂(g) what is the equilibrium constant expression…

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

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

Answer

Explanation:

Step1: Recall equilibrium - constant expression rule

For a reaction $aA + bB\rightleftharpoons cC + dD$, $K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$, where pure solids and pure liquids are not included in the expression.

Step2: Analyze the given reaction

The reaction is $C(s)+O_2(g)\rightleftharpoons CO_2(g)$. Carbon ($C$) is a solid. So, it will not be included in the equilibrium - constant expression.

Step3: Write the equilibrium - constant expression

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

Answer:

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