What are the products of the reaction between fluorine and ammonia? Is it the same as with chlorine and ammonia, which depend upon which reactant is taken in excess?
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4 Answers
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Fluorine reacts with ammonia to give dinitrogen and hydrogen fluoride as the major product.
The Formation of Dinitrogen Tetrafluoride in the Reaction of Fluorine and Ammonia - J. Am. Chem. Soc. 1959, 81 (23), 6338-6339 - states that dinitrogentetrafluoride accounts for 6% of the yield while the major reaction is:
$\ce{3F2 + 2NH3 → N2 + 6HF}$
The same article quotes in its first paragraph that the result remained the same irrespective of the concentration of the reactants. (In the case of chlorine it matters which reactant is taken in excess)
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$\begingroup$ I wonder if the N2F2 produced is just the product of NF3 degredation... $\endgroup$– J. LSFeb 15, 2015 at 14:12
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Actually, there are three types of reaction occuring between ammonia and fluorine which are given here. They all depend on reaction conditions and concentration of reactants. All the reactions are dangerous and produce volatile compounds. Reaction should be done under expert supervision. Please do not try this if you new in chemistry field.
- $\ce{2NH3 + 3F2 ->[\Delta] 6HF + N2}$ (mentioned)
- $$\ce{2NH3 + 6F2 → 6HF + 2NF3}$$
Ammonia react with fluorine to produce hydrogen fluoride and nitrogen(III) fluoride. The reaction takes place at the low temperature.
- $$\ce{4NH3 + 3F2 → 3NH4F + NF3}$$
Ammonia react with fluorine to produce ammonium fluoride and nitrogen(III) fluoride. Ammonia should be aqueous. In this reaction, the catalyst can be metallic copper. Impurities: tetrafluorohydrazine $\ce{N2F4}$.
answered Jul 13, 2016 at 3:58
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The reaction is this $$\ce{5 F2 {(g)} + 2 NH3 {(g)} -> N2F4 {(g)} + 6 HF {(g)}}$$ And it doesn’t depend upon which reactant is taken in excess; I believe it’s because fluorine is just too reactive.
answered Feb 15, 2015 at 12:06
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3$\begingroup$ "And it doesn’t depend upon which reactant is taken in excess." I find that hard to believe. Can you provide a source? As worded your statement implies that reactive 1 mol of F2 with 100 mol of ammonia will still form N2F4. $\endgroup$– Curt F.May 15, 2015 at 12:19
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$\begingroup$ N2F4 is a minor product , major product is always N2 $\endgroup$ Jan 11, 2020 at 14:02
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When fluorine reacts with an excess of ammonia it forms $\ce{NH4F}$. When fluorine reacts with a limited supply of ammonia it forms $\ce{HF}$. Both these reactions yield nitrogen ($\ce{N2}$) gas.
