Question

Consider the following reversible reaction.

What is the equilibrium constant expression for the given system?

Answer 1

Answer: The expression of equilibrium constant for the given equation is
`K_(eq)=([CO_2])/([O_2])`

Step-by-step explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as
`K_(eq)`

For a general chemical reaction:

`aA+bB\rightleftharpoons cC+dD`

The expression for
`K_(eq)` is written as:

`K_(eq)=([C]^c[D]^d)/([A]^a[B]^b)`

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

`C(s)+O_2(g)\rightleftharpoons CO_2(g0`

The expression of
`K_(eq)` for above equation follows:

`K_(eq)=([CO_2])/([O_2])`

Hence, the expression of equilibrium constant is written above.

Answer 2

K = [CO2] / [O2]

Step-by-step explanation:

The formula is K = [product] / [reactant].

C is excluded because it is a solid. And the rule says when there's a solid with other gases, only the gases will be considered for the equilibrium constant expression