Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Problem of the Week Archive => Topic started by: Borek on April 16, 2012, 09:15:31 AM
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When the gaseous mixture of hydrogen and tetramethyllead (Pb(CH3)4) is heated, tetramethyllead decomposes, producing free methyl radicals, which further react with hydrogen, creating methane and hydrogen radicals. Methyl radicals can react with antimony producing a volatile methylides and "dissolving" the element.
The mixture of hydrogen and tetramethyllead was pumped at 14 m/s through a thin quartz tube, heated at a point P. Inside of the tube contained two identical antimony mirrors deposited at distances 22 and 37 cm from the point P in the direction of the gas flow. Mirrors disappeared in 45 and 150 seconds respectively. What was the half life for the methyl radicals in the experiment conditions?
Edit: modified the problem to make it solvable, see the following discussion.
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To determine half life I believe we need to know the order of the reaction and the rate constant. I have no clue how to go about doing that :(
It's been too long since I've done anything with kinetics.
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free methyl radicals, which further react with hydrogen, creating methane and hydrogen radicals
CH3· + H2 -> CH4 + H·
There is a large excess of hydrogen.
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I think we can say that rate = k[H2][CH3.]. Since the H2 concentration is large we can lump it in with k giving us
r = k[CH3.]
Thus first order half-life is ln(2)/k.
Need to find k correct?
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>:(
I am leaving it open for another week, but I am also starting another one.
Did you know I paid $5 for a book I have not seen since eighties to find this single question?
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Assuming the reaction of methyl radicals with the mirrors has the form v = k[CH3•], the rate at which the mirrors disappear is directly proportional to the concentration of methyl radicals.
Therefore, if we represent the concentration of methyl radicals as a function M(x) then M(22) = A/45 and M(37) = A/150, where x is the distance from point P and A is a constant.
By Stewie's argument, we expect the concentration of methyl radicals to decrease exponentially as a fucntion of x, M(x) = C exp(-x/L), where C and L are constants. Plugging the values above into the equation, we can solve for L, and find that L = 12.5 cm
Finding the half-life requires knowing the velocity of the flow F. The half-life would be given by F ln(2)/L.
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Finding the half-life requires knowing the velocity of the flow F.
You are given two times and two distances.
Not without a reason.
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Oh my, I am a complete idiot.
I don't know how I managed to do it, but the saved version I worked on before posting, and the version I posted are different :-[ Somehow I deleted several crucial words. As the version I have here is is grammatically idiotic, I guess I wanted to edit it and forgot to paste, or something like that :'(
The mixture of hydrogen and tetramethyllead was pumped through a thin quartz tube, heated at a point P.
should read
The mixture of hydrogen and tetramethyllead was pumped at 14 m/s through a thin quartz tube, heated at a point P.
I feel so ashamed I am thinking about banning myself.
You can't always look intelligent.
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.bpp.com.pl%2FIMG%2FIMG_1008.jpg&hash=7cd15673eac1a08812fa0519c33b2babb01affd1)
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Lets give it a try. ::) Here is what I propose:
* At time t=0, reactants are at the point P with an initial CH3. concentration named C0.
* After 22cm and a speed of 14m/s, time is t22=0.22/14=0.0157 s, and the concentration is C22=C0 e-kt22 (equation 1)
* After 37cm and a speed of 14m/s, time is t37=0.37/14=0.0264 s, and the concentration is C37=C0 e-kt37 (equation 2)
* The speed at which antimony is consumed is supposed to be proportional to the concentration of methyl radicals. So, if the second mirror takes 150/45=3.333 more times to disappear than the first one, it's because the concentration is 3.333 times less: C37=C22/3.333 so C22/C37=3.333 (equation 3)
* From (equation 1)/(equation 2), we get: C22/C37=ek(t37-t22) (equation 4)
* From (equation 3) and (equation 4), we get ek(t37-t22)=3.333. Knowing the numerical values for t22 and t37, we can calculate k=1/(t37-t22) ln(3.333) :rarrow: k=112.51 s-1
* For half life time t1/2, we have C1/2=C0/2 and C1/2=C0 e-kt1/2. When combined, it gives 1/2=e-kt1/2 :rarrow: t1/2=-ln(1/2)/k
The half life time is t1/2=0.00616 s
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Everyone makes mistakes Borek. I also made the mistake of omitting a crucial value from a homework problem I once wrote for an actual class (thank goodness for students who worked on the homework early and informed me of the omission). Of course, my previous post also has an error:
The half-life would be given by F ln(2)/L.
Which should read L ln(2)/F. Plugging in 14 m/s for F, I get t1/2 = 0.0062 s which agrees with cth's answer.
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Yes, around 6 ms is a correct answer.
This was the original method used to determine half life of the free radicals back in 1929: F.A. Paneth, W. Hofeditz, Berichte der Deutschen Chemischen Gesellschaft, 62, 1344 (1929).