Question

Consider the following reversible reaction.

C(s) + O2(g) C02(9)
What is the equilibrium constant expression for the given system?
[CO]
o kor" (CO2)
[CO]
o KooTO]
[C]O2]
o Keg"
[CO2]
[02]
o Ker-1002]

Answer 1

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

Step-by-step explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric coefficients. It is represented by
`K_(eq)`

For a general chemical reaction:

`aA+bB\rightarrow cC+dD`

The
`K_(eq)` is written as:

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

The concentration of pure solid and pure liquid are taken as 1 in equilibrium expression.

The
`K_(eq)` for the given reaction is written as:

`C(s)+O_2(g)\rightarrow CO_2(g)`

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