Question

Consider the following reversible reaction.
C(s)+O2(g)<—>CO2(g)
What is the equilibrium constant expression for the given system?
[CO]
[CO]
[CO]

Answer 1

Answer:

Here's what I get.  

Explanation:

$\rm C(s) + O_{2}(g) \rightleftharpoons CO_{2}(g)$

The general formula for an equilibrium constant expression is

$K_{eq} = \dfrac{[\text{Products}]}{[\text{Reactants}]}$

For this reaction,  

$K_{eq} = \dfrac{[\text{CO}_{2}]} {[\text{O}_{2}]}$