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

Here's what I get.

Step-by-step explanation:

`\rm C(s) + O_(2)(g) \rightleftharpoons CO_(2)(g)`

The general formula for an equilibrium constant expression is

`K_(eq) = \frac{[\text{Products}]}{[\text{Reactants}]}`

For this reaction,

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