The trick is to NOT leave out the water and its ionic components. Ionic solution may be described most accurately by a chemical reaction like that below:
H[sub]2[/sub]O + NaCl <–> H[sup]+[/sup]+OH[sup]-[/sup] + Na[sup]+[/sup]+Cl[sup]-[/sup] <–> H[sup]+[/sup]+Cl[sup]-[/sup] + Na[sup]+[/sup]+OH[sup]-[/sup] <–> HCl + NaOH
All these reactions are going on simultaneously. None of them are separable until one of the original reactants is removed: i.e., by boiling off the water.
It isn’t, but that’s not what I commented on. You made the claim that evaporation was self sustaining. I pointed out that it is not.
There is no such thing as a self sustaining enothermic process. The energy to keep the process going must come from someplace else.
It makes no difference. I agree with chaoticbear that it is a bit of both. Either way, it obeys the same thermodynamic equations.
I need to see a cite for this. The energy for evaporation comes from the fire, not from anything in the beer. So why would it make the beer colder? I’m also reasonably sure that Mythbusters busted it. See here
Lattice energy is useless here since you are talking about an aqueous solution not gaseous ions. Instead, you need to look at solubility constants. Of course the solubility constants will be for pure water, but since you have a mixture of salts these things will be meaningless.
I think this is a very complicated problem and am not certain that enough information is present to determine what precipitates out first. Most likely, Chronos is right that you get a mixture of each, since as one ion precipitates out, the others become more concentrated as the solution evaporates. There is likely to be a slight preference for certain ions to be paired. I could guess that potassium and bromine would like each other best due to similarity in the size of the vacant and occupied orbitals.
Keep in mind that a salt crystal isn’t really a bunch of independent NaCl molecules, each Na+ is bonded to six Cl- and each Cl- is bonded to six Na+. If you percipitate two salts together, you will have a dissordered mess. I’m not certain what all the effects will be, but a lot will hve to do with the rate the solution is evaporated.
I’m not sure that electronegativity figures in here. The biggest electronegativity difference is between K and Cl, so KCl would have the strongest bond, but electronegativity has to do with atoms not ions. Chloride doesn’t have any electronegativity whatsoever since it is already negatively charged.
Ok, it’s spontaneous. I don’t mind splitting hairs, it’s what I do for a living, but if we’re going to split that one, and I guess we should, that’s what I meant.
It’s still not endothermic. It’s just that the amount of heat generated by the combustion reaction is not enough to maintain a stable flame. There are a lot of instances where a flame goes out because there is not adequate heat generated, but the combustion process itself is still endothermic.
What happens when a solid burns is that the heat of combustion causes the solid material to thermally decompose. The decomposition process requires heat, and the resulting products become gas-phase. The gas-phase decomposition products then become the fuel for the flame, and are further decomposed at high temperature, react with oxygen, and form CO2, water, and what are usually called combustion by-products (soot, various hydrocarbons, etc.). If the flame doesn’t generate enough heat to drive the first step, additional heat - in this case the hot plate - can provide the necessary heat to decompose the surface material, generate the gas-phase fuel, and sustain the flame. The flame itself is still exothermic, but it’s not generating enough heat to be stable on its own. It’s similar to the case where a gas grill runs out of fuel - the flame is still exothermic, but there isn’t enough fuel going into the flame to maintain temperatures high enough for the flame to be stable, and it goes out. There is too much air, not enough fuel, and the heat generated isn’t adequate to keep the reactions going.
I can’t answer Sunspace’s question - I don’t know. I suspect what is being asked is whether there are reactions that generate flames, but not at temperatures that are hot enough to burn flesh or other materials we’d like to have not burn. There may be, but I don’t know what the reactants would be. The argument against something like that happening is that the energy level has to be high enough for the reactants to release light energy, and that level may be too high for a cold flame.
Definitely not endothermic but still impressive and fun.
A friend and I used to make hydrogen balloons using a sodium hydroxide and water solution and adding aluminum foil. Mix this in a narrow neck bottle and fit a balloon over the neck. When it fills with hydrogen tie off the neck of the balloon and you have a nice toy.
Some balloons are just not filled enough and have a more-or-less neutral buoyancy. We would touch these with a kitchen match and the resulting fireball was impressive. If we kept doing it, the fire would eventually crinkle the hair on the backs of our hands. Safety glasses are a REAL good idea when doing this.
Regards
Testy
Well I’ve accidentally spilled flaming sambuca on my hands, during a drinking game at a party. As it was in a darkened room, my hands were covered with a dramatic - and delicious! - blue flame. The effect was impressing my friends so I let it burn for several seconds before clapping the flames out.
In truth I wasn’t feeling much pain at the time… but there were no burns in the morning. If I remember correctly, the flames were warm, even hot - though perhaps it was actually the liquid that was hot - but in no way enough to be painful.
So I’m sure something like this was the origin of the juvenile legend. Young chemistry students saw it happen and surmised that the lack of burns was due to an endothermic reaction. But it was a wild and wrong guess.
And after all, if it were really an endothermic reaction it would have actually felt cold.
Acetone burning in your hand is two reactions. One of them is endothermic.
The fire is not endothermic in sum, but it’s endothermic where it counts.
Here’s another one
Eh, maybe it’s time for a warning/disclaimer, for any among us who may be sincerely foolish.
Don’t assume that these stunts are always safe, every time!
You have the right to be a damn fool, but we’d rather you didn’t hurt yourself doing it.
In many of these experiments, showy displays of fire didn’t hurt anyone, because the flame was brief and the fuel was quickly spent. It doesn’t always happen so neatly.
For example, an old friend of mine used gasoline to start a brushpile on fire. In the time it takes to say whump, he was engulfed in flame, and the fire was gone. He still had his hair, his eyebrows, and his clothes. His bare legs, though, had 2nd degree burns.
That fart-lighting thing? Don’t do it in the nude, and do NOT do it while wearing synthetic fabrics! Just as a screen over a Bunsen burner stops the flame from going further, your cotton jeans will stop the flame from burning the hair around your anus. Synthetic fabrics, rather than protecting you, will melt to your skin unless everything goes just right.
Amen to that, brother! I am constantly advising students not to do any of the things they see on Youtube. Dry ice bomb, The Works bomb, whatever. Don’t do it. You WILL get hurt.
I have to constantly evaluate my demonstrations the ensure that:
A) The dangerous ones cannot be imitated because the students will not have the chemicals or apparatus and
B) There is some instructional purpose.
Are you sure that’s a benefit?
This classic thread speaks of the logical conundrum that is endothermic fire 