Sunday, December 28, 2008

Answer to: A mind snapping quiz

Two weeks ago I posted a quiz about a reaction that I considered interesting. Its text is the following:

You have some chlorinated water. To the solution you add some sodium nitrate, but not in stoechiometrical amounts (the sodium nitrate is fewer than it is necessary to react entirely with the mixture). The reaction products shall be:

  1. Sodium hypochlorite and nitrogen dioxide
  2. Nitric acid, hydrochloric acid, hypochlorous acid, kitchen salt and sodium hypochlorite
  3. Sodium hypochlorite, hydrochloric acid and NO2
  4. Nitric acid, sodium hypochlorite, hydrochloric acid, kitchen salt and hypochlorous acid
  5. Kitchen salt, oxygen and hydrochloric acid

From the early beginning we can see that we have "chlorinated water". This means, that when we bubble chlorine into water the following reaction can be observed:

Cl2 + H2O → HClO + HCl

Next, we pour some NaNO3 into the solution. This means that the hypochlorous acid (HClO), and the hydrochloric acid will react to form nitric acid and sodium salts:

HClO + NaNO3NaClO + HNO3 HCl + NaNO3NaCl + HNO3

Because of the fact that the initial mixture is in excess, in the final solution will remain HClO and HCl too, but not even a drop of NaNO3.

To sum up the compounds of the final solution will be :

  • HClO
  • HCl
  • NaClO
  • NaCl
  • HNO3

So the correct response is the fourth variant

I have seen that only 5 persons have answered to my quiz (from which only one of them has responded correctly), although I had expected a lot more to react. I know that this may be explained by the Christmas period that we are passing through, and people are more implied in celebrating this event in family than looking on chemistry blogs :-), and also by the fact that my blog is pretty new and hasn't got many followers yet. Merry Christmas and a Happy New Year!

Friday, December 26, 2008

Hydrogen Peroxide and Ozone

Three days ago, while I was reading an article about Oxygenated Water (H2O2) and Ozone (O3) chemistry, I saw 3 different reactions between the same substances mentioned above, all three possible. The reactions were:
  1. 2H2O2 + 4O3 → 2H2O + 7O2
  2. H2O2 + O3 → H2O + 2O2
  3. H2O2 + 3O3 → H2O + 5O2
I let the book down for a minute and asked myself why: "Why all three reactions are possible?". Immediately, I realised that the first one is the sum of the last two; so, all I had to do is to prove the last 2 reactions.
  1. H2O2 + O3 → H2O + 2O2
    • RedOx Method
    • Firstly everything went wrong. I thought that the oxygen atoms in the H2O2 molecule have the oxidation state number -2, but after a little bit of documentation I had found that, due to it's allotropic state, oxygen in H2O2 has an N.O. of -1 .

      OK. This wasn't a problem any more. Then, i noticed that O atoms are both the oxidizers and the reducers in the reaction (Nothing new, of course. There are such reactions, in nature, where the same element is both reducer and oxidizer; for example H2SO4 + S → SO2 + H2). In conclusion the partial ionization reactions are:

      O-1(H2O2) —( -1e- )→ O0(O2) O0(O3) —( +2e- )→ O-2(H2δ+Oδ-)

      Reaction Possible.

    • Separate Decomposition Equations Method
    • In a more simple way, I could solve this by taking the reactions separately, making an equation system, multiplying each reaction by the coefficients resulted and finally, adding everything up.

      So we have:

      H2O2 → H2O + 1/2 O2

      O3 ↔ 3/2O2

      ————————————(+)

      H2O2 + O3 → H2O + 2O2 (→TRUE)

  2. H2O2 + 3O3 → H2O + 5O2
  3. We observe that this reaction is just the anterior one with one more O3 ↔ 3/2O2 added. That means that in an athmosphere of O3 in exces this is perfectly possible. (→TRUE)
From these two we deduce that the first one is also plausible. And that's it: All three are true!

Monday, December 22, 2008

An illusiory paradox

Because the Chemistry Olympiad is on 10 January and there is very little time left, I use the winter holidays for training myself for the Olympiad. It was 11 o'clock and I was solving problems from a collection of more complex exercises, when my attention fell on one entangled question. It said: In a closed recipient heated at high temperature, ŋ particles are introduced (ŋ is a natural number). In the same conditions are introduced ŋ particles more, and it is observed that the pressure remains the same. Explain the phenomena. The first thing that crossed my mind was the ideal gas equation PV = ƞRT. This meant that the pressure, and volume remained the same, so the temperature lowered. But immediately I saw the part of the text that stated that the conditions remained the same. I had no inkling for, let's say, about half an hour, although I was endeavoring to find the explanation. Finally I gave up. I looked at the answers part of the book and what I saw shocked me by it's unexpected simplicity We know that the pressure depends on the number of collisions of the recipient's wall with the particles. That means that if the pressure is constant , we have the same number of particles (in the same conditions). You may say now: "What? But we can't have the same number of particles because we added more. It's a paradox!". Yeah! You're half right, but you neglected a very important aspect of chemistry. Particles react with each other. Nobody told you that the particles we introduced the first and second time were of the same type. So here's the explanation: The first particles (A) reacted stoechiometrically (they were echimolecular) with the second set of particles (B) as following: A + B → AB. We had initially ŋ particles, we added ŋ particles, and in the end we had ŋ particles. The simplest reaction model of this type that crosses my mind is Mg + S → MgS.

Sunday, December 14, 2008

A mind snapping quiz

Hi. Sorry for not being active for such a long period but I had to give my exams. Today I shall propose you a chlorine reaction related quiz:
You have some chlorinated water. To the solution you add some sodium nitrate, but not in stoechiometrical ammounts(the sodium nitrate is fewer than it is necessary to react entirely with the mixture). The reaction products shall be:
sodium hypochlorite and nytrogen dioxide nitric acid, hydrochloric acid, hypochlorous acid, kitchen salt and sodium hypochlorite sodium hypochlorite, hydrochloric acid and NO2 nitric acid, sodium hypochlorite, hydrochloric acid, kitchen salt and hypochlorous acid kitchen salt, oxygen and hydrochloric acid
UPDATE: I have posted the answer followed by the explanation at the following post http://dailychem.blogspot.com/2008/12/answer-to-mind-snapping-quiz.html