[Sophia7X]

A 4.72 g sample methanol is placed in a 1.00 L flask and heated to 250° C to vaporize.
After the system has reached equilibrium, a tiny hole is drilled in the side of the flask. Measurements of the effusing gas show that it contains 33 times as much H2 as CH3OH. Calculate  K.

(The methanol vapor over time decomposed by the following reaction:)
CH3OH(g)  CO(g) + 2 H2(g)
0.147 M                   0             0
-x                          +x         +2x
0.147-x                  x             2x

I did 33 = 2x/0.147-x, calculated x = 0.139 and got K = 1.27. But I don't like the sound of methanol spontaneously decomposing into carbon monoxide.

I have a feeling I should use the effusion rate (by Graham's law, H2 effuses 3.987x faster than CH3OH) but I have no idea how to use this number. Help?

[Borek]

I don't see anything wrong about your solution. Methanol is produced from the CO and H₂ on the industrial scale, nothing strange in the reverse reaction.

[curiouscat]

I have a feeling I should use the effusion rate (by Graham's law, H2 effuses 3.987x faster than CH3OH) but I have no idea how to use this number. Help?

You could correct for Grahams Law  but I'm not sure if or not the question intended this.

Rate_H2 / Rate_MeOH = sqrt( MW_MeOH / MW_H2 ) = sqrt (32 / 2)=4

So 4 times as much H2 would effuse out as MeOH from an equimolar solution. Hence (33/4) may be the molar ratio of (H2:MeOH) to use in your formula.

8.25 =2x/0.147-x

giving x=0.1183
giving K=0.23

Caveat emptor!   

[Sophia7X]

Curious, your way does give the right answer. The answer key says K = 0.2

However, I don't understand the meaning/purpose of dividing 33/4?

[Borek]

Then it is a matter of poorly worded problem (or my English fails me). Statement

Measurements of the effusing gas show that it contains 33 times as much H2 as CH3OH.

suggests molar ratio 33:1, not 33:4. Doesn't matter what was the method used for the measurement.

[curiouscat]

Then it is a matter of poorly worded problem (or my English fails me). Statement

Measurements of the effusing gas show that it contains 33 times as much H2 as CH3OH.

suggests molar ratio 33:1, not 33:4. Doesn't matter what was the method used for the measurement.

I disagree with you @Borek; I thought the English was ok . Composition of the effusing gas need not equal composition of the internal gas; I think that's the mistake you are making.  The molar ratio can be 33:1 just outside the  hole and 33:4 inside the flask.

Does that make sense? Do I have you convinced? It's a tricky question, no doubt.

[curiouscat]

Curious, your way does give the right answer. The answer key says K = 0.2

However, I don't understand the meaning/purpose of dividing 33/4?

Think of the hole as, say, a semi-permeable / selective membrane: The composition of the gas effusing outside is not equal to the gas inside. Hence you apply the correction based on Grahams law.

[Borek]

Composition of the effusing gas need not equal composition of the internal gas; I think that's the mistake you are making.

Good point. Tricky, but correct.

[Sophia7X]

The molar ratio can be 33:1 just outside the  hole and 33:4 inside the flask.

Ahhh, I get you!