# Hottest region of Bunsen flame

Since region 1 is closer to the source, I presume it to be the hottest as complete combustion takes place there. Also, this is the part where the gas mixture(responsible for flame) reacts with oxygen first(as soon as valve is open). So, the reaction should take place much quicker and more heat should be released. But in my book, the answer is given as region 2. Where am I wrong?

I know there exists as question similar to this in our community : Bunsen burner and hottest part but the answers don’t answer my question.

• There is more than one reaction happening ;) – Stian Yttervik Mar 22 '18 at 15:51
• Of all things, complete combustion is not what happens in region 1. – Ivan Neretin Mar 22 '18 at 16:01
• No, that's irrelevant. Burn pure methane, and the pattern will be the same. – Ivan Neretin Mar 22 '18 at 16:02
• @IvanNeretin Why not? We have an open valve and availability of oxygen. – MollyCooL Mar 22 '18 at 16:10
• From Dave’s and MaxW’s comments and answers, my take is that gases near region 1 are just coming out of tube and so they are relatively more cool. The rate of reaction is less than the mixture velocity, so the statement “ No unreacted mixture is present in region 1 “ is false. It is because of this mixture (which still has to react), that region 1 is not enitrely combusted and is relatively more cool than point 2(where there is no unreacted gas mixture). Correct me if I’m wrong and thank you very much for your time. – MollyCooL Mar 22 '18 at 18:01

The question itself is poorly written as given.

The diagram should have "regions" and "points."

So what seems to be labeled "Region 2" and "Region 4" I would name as "Point 2" and "Point 3" respectively.

So the innermost conical area would be "Region 1", the next conical area would be "Region 2" and the third conical area would be "Region 3."

Region 1 is where the mostly unburned gas and oxygen mixture are pushing above the lip of the Bunsen Burner. Region 1 exists because the gas coming out of the tube is cool. If the air port is open and the gas flow is too low then the gas will start to burn down the tube and you'll get a "strike back" where the flame is either (1) blown out or (2) burns at the jet. If the gas flow is too great you can blow the burning region off contact with the upper tube. If you increase the gas flow even more then you can in fact blow the flame out.

Region 2 would be a reducing region within the flame. This region is hot and burns the fuel and oxygen coming out of the tube.

Region 3 would be a oxidizing region of the flame. Here oxygen from "outside air" (oxygen which didn't come up the tube) is migrating into the flame to burn the excess fuel which is not bunt in region 2.

You'd use the reducing and oxidizing regions when doing bead tests for identification.

Point 2 would be the hottest part of the flame as shown by the composite image of a paperclip in the flame from the YouTube video.

• Oh yes, sorry about the labelling and use of language. Same doubt! Sorry if it is stupid, but why should there be unreacted mixture when the reaction occurs immediately and continues in the same trend as more gas is supplied? – MollyCooL Mar 22 '18 at 16:31
• @MollyCool - I'm not sure exactly what your asking. I think you're asking about Region 1. Region 1 exists because the gas coming out of the tube is cool. If the gas flow is too low the gas will start to burn in the tube and you'll get a sort of "backfire" which then blow the flame out. If the gas flow is too great you can blow the burning region off contact with the upper tube. If you increase the gas flow even more you can in fact blow the flame out. // Does that answer your question? – MaxW Mar 22 '18 at 16:37
• Yes I am referring to region 1 but you said unreacted mixture exists in that region. I’m confused on why there is unreacted mixture when flame is continuously burning. Entire amount of the gas mixture burns to produce the flame and it is being continuously supplied to keep it lit. So as the gas flows through tube and reaches region 1 by this time it should have reacted to produce flame right? – MollyCooL Mar 22 '18 at 16:43
• No, for Region 1 the gas is mostly unburned. Region 1 is cool because it is just coming out of the tube which is cool. Look at the Meker–Fisher burner. Without the metal grid, which is cooled by the cold gas coming up the tube, the very wide top would allow the flame to go down the tube. The wire grid serves as a boundary between the hot gases burning in the flame and the cool mixture coming up the tube. So there is essentially no "region 1" for this type burner, only regions 2 (reducing) and 3 (oxidizing) as shown. – MaxW Mar 22 '18 at 17:46
• FWIW I think the original question would have been clearer if they had simply used markers A, B, C and D and not used the word "region" at all. – MrWhite Mar 23 '18 at 22:17

Region 1 is much cooler because it contains a mixture of gases that it unreacted.

The gases in the tube are traveling too fast for the flame to enter the tube.

The inner cone is this unreacted gas mixture as it exists the tube.

• Oh! But aren’t the reactions too quick That the entirety of the gases react and there is nothing like “unreacted”? – MollyCooL Mar 22 '18 at 16:21
• @MollyCooL We know that the reaction does not propagate as quickly as the gases move within the tube, because the flame does not enter the tube. The mixture in the tube already includes air, so it would react within the tube if the reactions propagated faster than the mixture velocity. For more detail see: nvlpubs.nist.gov/nistpubs/jres/60/jresv60n6p535_A1b.pdf – DavePhD Mar 22 '18 at 16:39
• I get that the reaction rate is slower than mixture velocity in the tube. Is this true even at the inner cone (region 1) where the gas is technically out of the tube? – MollyCooL Mar 22 '18 at 18:15
• @MollyCooL yes, because it can not slow instantaneously. It slows as it interacts with the surrounding, stationary, air. The center of the gas stream moves the fastest. – DavePhD Mar 22 '18 at 18:30