Consider the following reaction and the value of its equilibrium constant: FeO(s)+ CO(g)⇌ Fe(s)+ CO₂(g)K_eq = 0.67 If the equilibrium concentration of CO is measured at 0.40 M, what is the equilibrium concentration of CO₂?

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the given reaction is:

K_eq = [Fe][COS]/ [FeO][CO]

We are given K_eq = 0.67 and [CO] = 0.40 M.

Let the equilibrium concentration of COS be x M. At equilibrium, the concentration of Fe and FeO will be equal since stoichiometrically they have the same coefficient. Therefore, both [Fe] and [FeO] can be represented as y M (where y is some constant).

Now, we can write:

K_eq = [Fe][COS]/[FeO][CO]
0.67 = y*x/(y*(0.40))
0.67 = x/0.40
x = 0.67*0.40
x = 0.268 M

Therefore, the equilibrium concentration of COS is 0.268 M, which corresponds to choice D.

Consider the Following Reaction and the Value of Its Equilibrium