|Mar1-08, 10:32 PM||#1|
I totally can't figure out rate mechanisms for whatever reason.
A three-step mechanism has been suggest for the formations of carbonyl chloride:
Step 1 Cl2 -> 2 Cl(fast, equilibrium)
Step 2 Cl + CO -> COCl (fast equilibrium)
Step 3 COCl + Cl2 -> COCl2 + Cl (slow)
What is the molecularity of the rate determining step?
Here since it was multiple choice, I guess bimolecular & got it right, but have no reason why...
Here is another
When the concentration of A is doubled, the rate for the reaction: 2 A + B-> 2 C quadruples. When the concentration of B is doubled the rate remains the same. Which mechanism below is consistent with the experiment observations?
The answer is (it was multiple choice, but I'll just post the correct answer)
Step 1: 2A -> D (slow)
Step 2: B + D -> E (fast)
Step 3: E-> 2 C (fast)
So any ideas regarding how I can better understand the concepts so I don't get slammed for the exam?
|Mar1-08, 10:44 PM||#2|
1. Which is the rate determining "step"? The fast or slow step?
2. Well, from the 2nd sentence, we now know that B is 0 order.
That's really all I can help you with. I haven't studied this in a while, wait for Chemistree or someone else to respond.
|Mar1-08, 10:51 PM||#3|
Slow step is the rate determining step. Suppose though that I wasn't told which one was slow or fast. It seems like such a simple concept... just can't seem to grasp it from the textbook.
|Mar2-08, 02:20 AM||#4|
i think you will be told about it. the slow step is the one having the higher activation energy. it is difficult for the molecules to collide effectively to form products....
if you are not told which step i sthe rate determining step, i'm afraid you will have to use logic....
step 1 involves homolysis of Cl2. this is quite easy... you can use UV,....
step 2 involves addition of Cl to CO. CO is linear, the Cl can add from behind the O atom.
step 3 though, involves the addition of another Cl atom. this time the highly electronegative O and Cl repels it. the Cl must have enough energy to reach the carbon and from a covalent bond with it. hence this is a slow step.
i think the question will state which step is slow or fast.
|Mar2-08, 02:24 AM||#5|
for 1. step 3 involves 2 molecules, Cl and COCl..... it should be bimolecular.
for 2. with respect to A, the rate is 2nd order (rate quadruples when A doubles). and with respect to B, it is 0 order (no change to rate when B doubles).
then, the rate determining step involves only A.
step with A is slow. but step with B is fast.
hope it helps, i'm also doing kinetics at school right now.
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