Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Dolphinsiu on March 24, 2007, 03:32:11 AM
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In my notes,
Reactivaties of 1', 2', 3' H
Cl
Cl2 l
CH3CH2CH3 --------> CH3CH2CH2Cl + CH3 - CH - CH3
hv 25C 45% 55%
Rate of reaction 45/6 = 7.5 55/2 = 27.5
per atom
CH3
l Cl2
CH3 - CH - CH3 ---------> (CH3)2 - CH - CH2 - Cl + (CH3)3 - C - Cl
hv 25C 63% 37%
Rate of reaction 63/9 = 7 37/1 = 37
per atom
What I want to ask is why we need to calculate the rate of reaction per atom?
Also how to determine the dividing number (e.g. 6 in 45/6) ? Thank you!
My teacher are not willing to tell me, I hope some kind people can help me!
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Dear Dolphinsiu,
Once again: Try to draw “A Picure”
Hints:
1.) How/Why you got the “6” and “2” in Eq. 1, and How/Why the “9” and “1” in Eq. 2?
(Sorry, Do you think, that Mother Nature reads first your paper and “?COUNTS?” ?)
2.) What’s the reason you count “6” and “2”? NO, - for Mother Nature to distinguish?
Distinguished by which criterion, - for HER?
3.) As your Eq. are correct enough to solve the problem (There is a little redundancy), so
every Part of the Eq. is important !!!
In this moment: What means the Term “hv” to you?
4.) What tells you the Rate of the Rates (?? = 7.5 / 27.5)?
Hint to the Hints:
A possible/good Start-Point for at least one Hint lies in the answer to “H2 + Br2” in
the “Analytical Chemistry Forum” (Keyword: induced).
Let’s have you THE First Try!
Good Luck!
ARGOS++
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I am so sorry to tell you I get no idea in count as the count varies and do not follows mother nature.
For the first eq, I will use 3 and 2
For the second eq, I will use 3 and 1
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I know how to count la. But why we need to count?
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Dear Dolphinsiu,
Please, tell me: Where have You dropped, - in the meantime - , the
other “6” (sex) “Hydrogens” of the second Reaction/Compound?
Is it not a “small” little toooo “radical” from you?
In secret, I hopped, that you have discovered, chem. Reactions have
something to do whit breaking bounds and building others instead.
But that means, that the whole thing depends on the Quality of the
bound to be broken, AND the Quality of the bound which has to be build.
In your case, it’s the bound between C and H, C-H.
So the Question in real is: How many bound of which Quality you have ??
(You may translate: Quality in this chem. sense as “Strength”!!)
About “counting”: What You and I call counting is for Mother Nature,
HER liked “Likelihood” !
With simple words: As long as “hv” is involved, AND you
replace EVERY from You now numbered “Hydrogens” ONCE with a Chlorine, —
How many different products (mono-"Chlorides"), AND How much of each you should get?
But: Why can’t you find your “counted” ratio (= Likelihood for Mother N.)?
Are you now able to answer the Question of my fourth Hint?
Hope it may help
Good Luck!
ARGOS++