Chemistry, 16.02.2021 03:10 maysahdabest

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.

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Answer from: nano792001

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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, $K_{c}$ .

For the first way to represent equilibrium of CO formed, the $K_{c}$ is calculated

$2CO_{(g)}$ ⇄ $2C_{(s)}+O_{2}_{(g)}$

$K=\frac{[O_{2}]}{[CO]^{2}}$

in which the symbol [ ] is concentration of the compound.

In equilibrium constant, solids are not included.

Equilibrium constants for the other reactions:

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

$K_{1}=\frac{[CO]^{2}}{[O_{2}]}$

Comparing K₁ and K, the first one is the inverse of K, so writing in terms of K

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

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

$K_{2}=\frac{[CO]}{[O_{2}]^{1/2}}$

In terms of K, K₂ is

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

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

$K_{3}=\frac{[O_{2}]^{1/2}}{[CO]}$

This constant in terms of K will be

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

In conclusion, K₁, K₂ and K₃ in terms of K is $\frac{1}{K}$ , $\frac{1}{K^{1/2}}$ and $K^{1/2}$ , respectively.

Answer from: Quest

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation: if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)

Answer from: Quest

fe 55.85 g/mol and h2o 18.02 g/mol.

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