Equilibrium of Methanol Vapor Decomposition

  • #1
i_love_science
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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.
 

Answers and Replies

  • #2
Ygggdrasil
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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.
 
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  • #3
epenguin
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You actually stated Graham's law of diffusion as relevant but you didn't use it.
 

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