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 0.50 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] = 0.50 M, and we need to find [COS]. Let x be the equilibrium concentration of COS. Therefore, we can write:
0.67 = [Fe][x]/[FeO][0.50]
We also know that the stoichiometry of the reaction is 1:1 between CO and COS. Therefore, at equilibrium, the concentration of COS will be equal to the concentration of CO that reacted with FeO. Since the total amount of FeO does not change, the amount of CO reacted must equal the amount of FeO reacted, and the equilibrium concentration of FeO must be (0.50 - x). Therefore, we can write:
[Fe][x] = [FeO][0.50 - x]      (using stoichiometry)
Simplifying and substituting the values we know, we get:
0.67 = x/(0.50 - x)
Solving for x, we get x = 0.34 M. Therefore, the equilibrium concentration of COS is 0.34 M, and the correct answer is (C).

Consider the Following Reaction and the Value of Its Equilibrium