0
$\begingroup$

Someone asked me a question, that which of the following can be decomposed chemically:

  1. Ammonia
  2. Iron
  3. Neon
  4. Hydrogen
  5. Fluorine

My Analysis:

At first attempt I answered Ammonia but changed it to 'I can't answer'. My analysis of the question is as follows

  1. If Fluorine (the name implies fluorine gas) is decomposed, but it will require to 242kJ/mol (bond dissociation enthalpy). So, its quite difficult that we will be able to decompose it chemically.
  2. If Hydrogen (the name implies Hydrogen gas) is decomposed, but it will require to 435.88kJ/mol (bond dissociation enthalpy). So, its even more difficult that we will be able to decompose it chemically.
  3. If If Ammonia (the name implies Ammonia gas) is decomposed, but it will require to 389kJ/mol (bond dissociation enthalpy). So, its significantly difficult that we will be able to decompose it chemically.
  4. Its useless to say that we will be decomposing Iron or Neon, because they are already in elemental state.

Why I tempted to answer Ammonia: I thought about Haber's Process, and remembered that the enthalpy change of the main reaction is negative (N2+3H2=NH3), about -46.1kJ/mol; this implies that the enthalpy of the just opposite reaction is positive (Enthalpy change of NH3=N2+3H2). This made me to conclude that the reaction is not feasible, thermodynamically and therefore ammonia cannot be decomposed easily. Therefore I changed my answer to 'I can't Answer'

My Question: I asked my friend my what is the answer, he said Ammonia. I was surprised but still confused. I have written all of my thoughts above, but neither I am able to justify my answer, nor is my friend able to understand my logic. I want your discussion and a final review of the question understanding '

$\endgroup$
  • $\begingroup$ It's exactly as "useless" in case of iron as fluorine or hydrogen. You simply don't say than most stable form of element decomposes. $\endgroup$ – Mithoron May 28 '16 at 22:28
  • $\begingroup$ I can't understand what are you trying to convey. $\endgroup$ – Shivanshu Gupta May 30 '16 at 10:52
  • $\begingroup$ "Its useless to say that we will be decomposing Iron or Neon, because they are already in elemental state." Hydrogen and fluorine also are. You should read your own text. $\endgroup$ – Mithoron May 30 '16 at 19:45
  • $\begingroup$ That means I have wrongly interpreted them as gases, they all are in elemental state, except ammonia, which is a compound and hence can be decomposed chemically. The answer is therefore ammonia. $\endgroup$ – Shivanshu Gupta May 31 '16 at 11:44
1
$\begingroup$

It really depends on what the definition of "decompose" is in this case. If you take the definition of "decompose" to be:

... to break down or cause to break down into component elements or simpler constituents

then the only option would be ammonia because the others are elements whereas ammonia is a compound which can be broken down into elemental Hydrogen and Nitrogen by the Haber Process. The energetics is not of concern in this case.

$\endgroup$
  • $\begingroup$ Thank you! It's simply about compounds and elements. Compounds can be broken into elements while there's no such a thing into which element can broken. Am I right? $\endgroup$ – Shivanshu Gupta May 30 '16 at 13:59
  • 2
    $\begingroup$ Well, elements could be broken down into smaller elements by nuclear fission but that's not at the chemical level anymore. $\endgroup$ – IT Tsoi May 30 '16 at 14:03
  • $\begingroup$ If it's not chemical level, then what name chemists use for it. Is it 'Nuclear level'. $\endgroup$ – Shivanshu Gupta May 30 '16 at 14:06
  • $\begingroup$ Yep - a nuclear reaction. $\endgroup$ – IT Tsoi May 30 '16 at 14:07
  • $\begingroup$ One last question from 'IT Tsoi', could you comment on my question language and grammar and way I have asked it. I want to improve the way I ask question because for the last few days I have been criticised for it. $\endgroup$ – Shivanshu Gupta May 30 '16 at 14:28

Not the answer you're looking for? Browse other questions tagged or ask your own question.