Question

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Consider the following reversible reaction.
C(s)+o2(g)<->Co2(g)

What is the equilibrium constant expression for the given system?
A. KEq=[CO2]
[C][O2]

B. KEq=[CO2]
[O2}

C. KEq=[C][O2]
[CO2]

D. KEq= [O2]
[CO2]

Answer 1

`\text{B. }K_{\text{eq}} = \frac{\text{[CO$_(2)$]}}{\text{[O$_(2)$]}}`

Step-by-step explanation:

C(s) + O₂(g) ⇌ CO₂(g)

The equilibrium constant expression is always of the form

`K_{\text{eq}} = \frac{\text{[products]}}{\text{[reactants]}}`

We do not include solids in the expression. That automatically eliminates options A and C.

The products are the substances on the right of the equilibrium arrows and the reactants are on the left.

That eliminates Option D, which has the reactant in the numerator.

The equilibrium constant expression is

`K_{\text{eq}} = \frac{\text{[CO$_(2)$]}}{\text{[O$_(2)$]}}`