I have read on Wikipedia that the flammable limits of gaseous ammonia are between 15–25% and I have also read that the autoignition temperature is $651\rm~^\circ C$. And that the combustion of ammonia in air is $$\ce{4NH3 + 3O2 -> 2N2 + 6H2O}$$ My understanding of this is that with a 20% air to $\ce{NH3}$ that I could hold a flame. Outside of that range, I cannot hold a flame.

My question is if I am at 30% air and above $651\rm~^\circ C$ will Ammonia still combust/react with Air or will it stay $\ce{NH3}$?

My follow up question would be that when if I sprayed ammonia into a heated chamber (above $650\rm~^\circ C$) would flames/the exothermic reaction only happen when the ammonia vapor and the air reach that 15–25% range as the vapor propagates into the air and not react outside that range?


Well you're mixing concepts which is leading to your confusion.

In the section named "Combustion" the Wikipedia text correctly states (per reference) that ammonia burns with difficulty when mixed with 16-25% of air. As you noted, the combustion of ammonia in air is: $$\ce{4NH3 + 3O2 -> 2N2 + 6H2O}$$

  • In the "Properties" section of the same Wikipedia article there is a subheading named "Solvent properties" which has the weird statement: "Ammonia does not burn readily or sustain combustion, except under narrow fuel-to-air mixtures of $15–25\%$ air." The statement is weird both because of the placement of the sentence as well as the wording. It should be that the fuel-to-air mixture is $16-25\%$ of ammonia, not $15–25\%$ air.

The gist is that you'd mix gaseous ammonia and air in range of $16-25\%$ and blow it through something like a Bunsen burner, which once lit, would allow the gas mixture to continue to burn. But if you just blew such a mixture through a burner, then the gas mixture would not autoignite at room temperature.

Now as to the autoignition temperature of $651^\circ \rm{C}$, ammonia the Wikipedia section "Laboratory use of anhydrous ammonia (gas or liquid)" notes that in the range of $16-25\%$ air that an explosive mixture can result. So at or above $651^\circ \rm{C}$ and outside the limits of $16-25\%$ air, the reaction would still take place but you wouldn't get an explosion. The reaction would proceed until most of the ammonia or the oxygen is used up.

  • With an explosion you can't really account for the temperature in all points of the explosion so there isn't really any well defined equilibrium.
  • Without an explosion there would some well defined equilibrium constant for the reaction which would be temperature dependent. (We're assuming that the container of gas is heated or cooled to maintain whatever the desired equilibrium temperature is.) Thus the kinetic energy of the gas molecules at the given temperature is providing the activation energy necessary for the reaction to occur. The equilbrium would be given by the equation of the partial pressures: $$\ce{K_{eq} = \dfrac{{p_{\ce{N^2}}}^2 {p_{\ce{H2O}}}^6}{{p_{\ce{NH3}}}^4{p_{\ce{O2}}}^3}}$$
  • $\begingroup$ @ MaxW is there a way to calculate the time it would take for this reaction to take? I do not know what an equilibrium constant is or how it is used. I assume it is for calculating the time of equations. I also assume the equation would be dependent on pressure as well as temperature. $\endgroup$
    – Pford
    Mar 1 '16 at 19:59
  • $\begingroup$ How fast the reaction occurs at a fixed temperature (assuming no explosion...) is a matter of chemical kinetics which is another level of complexity separate from the equilibrium. // If the gas were to start to burn, then as the gas burns it gets hotter which causes it to burn faster. The next result is an explosion. $\endgroup$
    – MaxW
    Mar 1 '16 at 23:45

Chemical reaction would proceed in an air/ammonia mix, even outside the concentration range at which the reaction is self-sustaining, particularly if there were a catalyst to help. It is just that at the specified concentration range, when it reaches $651^{\circ}\mathrm{C}$ at one point in the mixture, the reaction proceeds fast enough to heat the rest of the mixture to that ignition temperature.

This is similar to an Otto cycle spark-ignition engine: with the correct mixture, the flame front propagates from the spark plug throughout the cylinder.

In the Ostwald process, a $1:10$ ration of $\ce{NH3}$ and air can sustain the reaction at the surface of a catalyst.

In a similar fashion, hot copper can catalyze the oxidation of acetone, methanol or ethanol, providing a surface where the reactants can combine fast enough to keep the surface hot.


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