3 minutes solid tips on D&F Block – Colour of Ions by Sanjit Sir for Science #Chemistry Board Exam Easy way to remember Colour of Ions and prepare for board exam.
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IV, Chemical Kinetics. V, Surface Chemistry. VI, Isolation of Elements, 19. VII, p-Block Elements. VIII, d- and f-Block Elements. IX, Coordination Compounds. X, Haloalkanes and Haloarenes, 28. XI, Alcohols, Phenols and Ethers. XII, Aldehydes, Ketones and Carboxylic Acids. XIII, Organic Compounds containing Nitrogen.You can only do this if you know the mechanism of a reaction. If something is an elementary step, then you can get the order by inspection. For example, the reaction:
A + B + C D
is not necessary first order in each component for an overall reaction order of three (i.e. rate = k[A][B][C]). We need to consider the reaction mechanism.
If A and B react slowly to form an unobserved intermediate, I, which then quickly reacts with C to give our product, then the concentration of C doesn’t affect the rate (i.e. the reaction of A + B is the rate-limiting step).
A + B I (slow)
I + C D (fast)
The bottleneck in the production of D (the rate limiting step) is the reaction of A + B! As soon as I is formed, it reacts with C to make D very quickly as long as we have some amount of C present. If we explored this experimentally, we’d find that if we held [A] and [B] constant, the rate would be the same regardless of the [C].
A Real World(tm) analogy here. To get a new car we have to work, save and spend $$. Our first step is working (A) and saving (B) to get the money we need (I). In fact, we can’t even do the second step of the reaction, writing a check to the dealer (C) even though this takes very little time. Clearly, it is not the transaction itself that determines how soon we car, it is the first step, working and saving.
Now that we have a series of elementary steps, we can determine the reaction order. In the first step A and B react (again, an elementary step of our reaction), so our rate = k[A][B], a second order reaction. The total reaction is also second order overall because the second step of our mechanism has no effect on the rate (consider it to be instantaneous if you wish).
Intermediates can’t appear in your overall rate law.
Intermediates are defined as species that appear in the mechanism of a reaction, but not in the overall balanced equation. Let’s write out a mechanism and then add up the products and reactants to come up with the overall balanced reaction:
NO(g) + N2O(g) N2(g) + NO2(g)
2 NO2(g) 2 NO(g) + O2(g)
If we multiply the first equation by two and then add up the products and reactants, we’d see that NO2 is present in equal amounts on both sides and cancels from our final balanced equation. NO2 is therefore an intermediate
2 NO(g) + 2 N2O(g) + 2 NO2(g) 2 N2(g) + 2 NO2(g) + 2 NO(g) + O2(g)
So our net reaction is:
2 NO(g) + 2 N2O(g) 2 N2(g) + 2 NO(g) + O2(g)
Obviously, we can not explain the observed rate with respect to the concentration of an unobserved species!
Finally, note that NO is a catalyst in this reaction — it is not consumed in the reaction and is present both at the beginning and end!