Question

Carbon monoxide is a colorless, odorless gas that binds irreversibly to hemoglobin in our blood, causing suffocation and death.

CO is formed during incomplete combustion of carbon. One way to represent this equilibrium is
CO(g) ⇌ C(s) + 1/2O2(g)
We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above.
1) 2C(S) + O2(g) ⇌ 2CO(g) K1=____
2) 2CO(g) ⇌ 2C(s) + O2(g) K2=____
3) C(S) +1/2O2(g) ⇌ CO(g) K3=___

Answer 1

- The equilibrium constant for the reaction 2C(s) + O2(g) ⇌ 2CO(g) is K1 = (CO)^2 / (C)^2 * (O2).

- The equilibrium constant for the reaction 2CO(g) ⇌ 2C(s) + O2(g) is K2 = (C)^2 * (O2) / (CO)^2.

- The equilibrium constant for the reaction C(s) + 1/2O2(g) ⇌ CO(g) is K3 = (CO) / (C) * (O2)^(1/2).

When writing the equilibrium constant for a reaction, the stoichiometric coefficients are used as exponents in the expression. The equilibrium constant (K) represents the ratio of the concentrations of products to reactants at equilibrium.

Let's write the equilibrium constant for each of the three alternative reactions in terms of K, the equilibrium constant for the given reaction:

1) For the reaction: 2C(s) + O2(g) ⇌ 2CO(g)

The stoichiometric coefficients for CO and C are both 2, and for O2, it is 1. Therefore, the equilibrium constant for this reaction (K1) is:

K1 = (CO)^2 / (C)^2 * (O2)

2) For the reaction: 2CO(g) ⇌ 2C(s) + O2(g)

In this reaction, the stoichiometric coefficients for CO and C are both 2, and for O2, it is 1. Hence, the equilibrium constant for this reaction (K2) is:

K2 = (C)^2 * (O2) / (CO)^2

3) For the reaction: C(s) + 1/2O2(g) ⇌ CO(g)

The stoichiometric coefficients for CO and C are both 1, and for O2, it is 1/2. Thus, the equilibrium constant for this reaction (K3) is:

K3 = (CO) / (C) * (O2)^(1/2)