# Equilibrium of Methanol Vapor Decomposition

• Chemistry
• i_love_science
In summary, the solution explains that the effused mixture will have 33.0 times as much H2 as CH3OH when the effusion rate ratio is multiplied by the equilibrium mole ratio of H2 to CH3OH. The person asking the question mistakenly set the equilibrium mole ratio as equal to 33.0 times, but this is incorrect because H2 effuses at a faster rate than CH3OH, resulting in a H2:CH3OH ratio greater than 33:1 in the effused gases. The relevant concept is Graham's law of diffusion.
i_love_science
Homework Statement
A 4.72-g sample of methanol (CH3OH) was placed in an otherwise empty 1.00-L flask and heated to 250°􏰀C to vaporize the methanol. Over time the methanol vapor decomposed by the following reaction:
CH3OH(g) <-> 34 CO(g) 􏰁+ 2H2(g)
After the system has reached equilibrium, a tiny hole is drilled in the side of the flask allowing gaseous compounds to effuse out of the flask. Measurements of the effusing gas show that it contains 33.0 times as much H2(g) as CH3OH(g). Calculate K for this reaction at 250°C.
Relevant Equations
equilibrium
graham's law of effusion
The solution says that when the effusion rate ratio is multiplied by the equilibrium mole ratio of H2 to CH3OH, the effused mixture will have 33.0 times as much H2 as CH3OH. I don't understand why.

I just set the equilibrium mole ratio of H2 to CH3OH as equal to 33.0 times, Why is this incorrect?

Thanks.

i_love_science said:
I just set the equilibrium mole ratio of H2 to CH3OH as equal to 33.0 times, Why is this incorrect?

If the mole ratio of H2 to CH3OH inside of the flask is 33:1, then the gasses effusing from the flask will have a H2:CH3OH ratio greater than 33:1 because H2 effuses from the flask at a faster rate than CH3OH.

i_love_science
You actually stated Graham's law of diffusion as relevant but you didn't use it.

## 1. What is the equilibrium of methanol vapor decomposition?

The equilibrium of methanol vapor decomposition refers to the point at which the rates of the forward and reverse reactions of methanol decomposition are equal. This means that the concentration of reactants and products remains constant over time.

## 2. What factors affect the equilibrium of methanol vapor decomposition?

The equilibrium of methanol vapor decomposition can be affected by factors such as temperature, pressure, and the presence of catalysts. Changes in these factors can shift the equilibrium in either the forward or reverse direction.

## 3. How is the equilibrium constant for methanol vapor decomposition determined?

The equilibrium constant for methanol vapor decomposition can be determined by measuring the concentrations of reactants and products at equilibrium and plugging them into the equilibrium expression. This constant represents the ratio of product concentrations to reactant concentrations at equilibrium.

## 4. What is the significance of the equilibrium of methanol vapor decomposition?

The equilibrium of methanol vapor decomposition is important because it allows us to predict the direction and extent of the reaction under different conditions. It also helps us understand the behavior of the system and optimize conditions for maximum product yield.

## 5. How can the equilibrium of methanol vapor decomposition be manipulated?

The equilibrium of methanol vapor decomposition can be manipulated by changing the temperature, pressure, or presence of catalysts. This can shift the equilibrium towards the desired direction and increase the yield of products. Additionally, the addition or removal of reactants or products can also affect the equilibrium position.

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