# The thermal decomposition of NH4Cl [closed]

When you heat $$\ce{NH4Cl}$$ it decomposes into $$\ce{NH3 + HCl}$$ gases. My question is why does it decompose into ammonia and hydrogen chloride gases on the molecular level. Is it because in the $$\ce{N-(HCl)}$$ bond, The $$\ce{N-H}$$ bonds breaks easily since it requires only $$\pu{390 kJ/mol}$$ and $$\ce{H-Cl}$$ requires far more heat energy? Thus the $$\ce{N-H}$$ bonds breaks first?

• HCl isn't acidic in gas phase. – Mithoron Mar 21 '19 at 17:01
• I don’t see how that answers my question? I specifically said hydrogen chloride instead of hydrochloride acid... – Serapion Mar 21 '19 at 17:02
• What is the alternate set of products you think it might have decomposed into? – Karsten Theis Mar 21 '19 at 17:09
• That is simply not my question, I know it will decompose into ammonia and hydrogen chloride. But the question is on heating, how is the strength of bonds playing out? Is it that H-Cl bond is stronger and thus it will be more resistant? And in the N-H-Cl, the N-H requires less energy to break off? Consequently leaving HCl? – Serapion Mar 21 '19 at 17:16
• Look, the questions of "why is this" type are inherently flawed and can't be answered comprehensively. There are just too many alternatives. You'd better ask "why is this and not that". These often have a relatively short and meaningful answer. – Ivan Neretin Mar 21 '19 at 17:43

In the solid state of ammonium chloride, the nitrogen and hydrogen atoms are connected by covalent bonds, forming the ammonium ion. The bond length is about 98 pm. Each ammonium ion is surrounded by eight chloride ions (and in turn, each chloride ion is surrounded by eight ammonium ions). The distance between a chloride and the closest hydrogen atoms is about 237 pm. In contrast, the bond length in HCl is 127 pm.

In the reaction $$\ce{NH4Cl(s) -> NH3(g) + HCl(g)},$$

you lose the ionic interactions, and ammonium gets deprotonated while chloride gets protonated (lose an N-H bond and gain a H-Cl bond). The change in entropy for the reaction is positive, so it is favored at high temperature.

We might think that high temperature makes the H-Cl bond stronger and the N-H bond weaker. The bond dissociation energy for N-H is 390 kJ/mol while that of H-Cl is 432 kJ/mol. However, bond dissociation energies are not directly relevant because we are not transferring a hydrogen atom (making $$\ce{NH3+ + HCl-}$$). Instead, we are transferring a proton, so the pKa of ammonium and hydrogen chloride are more relevant but not directly either because this does not happen in aqueous solution.

The key is that the products are in the gas phase. If the products of the decomposition were ions and had opposite charges, they would not remain in the gas phase. In aqueous solution, you get $$\ce{NH4Cl(s) -> NH4+(aq) + Cl-(aq)}$$ because water is good at solvating ions. In the gas phase, the most stable combination of the atoms at hand is $$\ce{HCl + NH3}$$.

• From your answer, what I inferred was that the increase in entropy is the best justification for the products obtained. Is that what you were trying to convey? – Yusuf Hasan Mar 22 '19 at 3:06
• That's the rationale for gas phase products, even if the reaction is endothermic (losing all the ionic interactions). Which products are formed depends on the most stable species possible with that set of atoms. Unbound atoms have even higher entropy, and a plasma would be even better in terms of entropy, but its formation is highly endothermic. – Karsten Theis Mar 23 '19 at 14:32

Here we have a protolytic reaction like

H2O + HCl <=> H3O+ + Cl-

NH3 + HCl <=> NH4+ + Cl-

There no other products possible.

• Sure other things are possible equations. For instance $$\ce{H2O + HCl(aq) ->[\Delta] H2O ^ + HCl ^}$$ – MaxW Mar 21 '19 at 19:54
• This is only change from liquid to gaseous phase, no chemical reaction. – Nobby Mar 21 '19 at 20:47