Chemistry: 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) 1/2 O2(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.

D.Explanation:took it on edge...

Carbon monoxide is a colorless, odorless gas that

Ответ:

Arealbot

18.03.2021

The question is missing some parts. Here is the complete 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:

$2CO_{(g)}$ ⇄ $2C_{(s)}+O_{2}_{(g)}$

we could also write this reaction three other ways listed below. The equilibrium constant 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) $2C_{(s)}+O_{2}_{(g)}$ ⇄ $2CO_{(g)}$ K₁ =

2) $C_{(s)}+1/2O_{2}_{(g)}$ ⇄ $CO_{(g)}$ K₂ =

3) $CO_{(g)}$ ⇄ $C_{(s)}+1/2O_{2}_{(g)}$ K₃ =

1) $K_{1}=\frac{1}{K}$

2) $K_{2}=\frac{1}{K^{1/2}}$

3) $K_{3}=K^{1/2}$

Explanation: A chemical reaction can be reversible, i.e., can proceed in both directions: to the right of the arrow (forward) or towards the left of the arrow (backward).

When the rates of forward and backward reactions are the same, the reaction is in equilibrium. In that state, we can determine the equilibrium constant, $K_{c}$ .