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 ∘C. And that the combustion of ammonia in air is 4NHX3+3OX2⟶2NX2+6HX2O
My question is if I am at 30% air and above 651 ∘C will Ammonia still combust/react with Air or will it stay NHX3?
My follow up question would be that when if I sprayed ammonia into a heated chamber (above 650 ∘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?
Answer
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: 4NHX3+3OX2⟶2NX2+6HX2O
- 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∘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∘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: KXeq=pNX22pHX2O6pNHX34pOX23
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