[Corvettaholic]

Back when I was in high school, my teacher dumped some elemental (I think it was elemental, or close to it) sodium into a glass of water. The stuff reacted pretty quick and made a neat little show. I thought you need some kind of activation energy to set off a reaction, so why did it behave this way?

[Mitch]

You don't need any activation energy. Read Jdurg's articles to the right of your screen.

[lindeproctor]

Elemental sodium is reactive in sodium and water because the element Na is more stable in a -1 oxidation state. It reacts with air to form Na20, and with water to form NaOH.

[Borek]

Na is more stable in a -1 oxidation state

+1

[lutesium]

There's no need for an activation energy for such reaction! Sodium Metal is active enough


Lutesium...

[Borek]

You need activation energy as always. However, it is so low, that thermal energy of water and metal are enough.

[mnakhla]

its thermodynamically favored at STP..meaning it has a negative delta G ..a largely negative delta G

[Greenport]

Are you sure it was elemental Sodium and not Sodium oxide?

Na would react with air to form Na2O.

Na2O reacts with water to form NaOH. While the elemental would likely do the same (or at the very least forming NaH), I don't see a way to really keep it from reacting with the air since Sodium is so reactive. It's kind of like saying you have elemental Chlorine - again that's not likely -- it'd likely be a Chloride or another chlorine-containing compound.

[Borek]

It most likely was elemental sodium. While it is highly reactive it can be kept safely for a long time under naphta. Then you can cut it into pieces - if it is done in air, but in a timescale of several minutes, metal will get covered with oxide, but what you have is still mostly metallic sodium. Throwing it into water is a classic experiment.

[nj_bartel]

its thermodynamically favored at STP..meaning it has a negative delta G ..a largely negative delta G

nothing to do with reaction rate

[mnakhla]

doesnt largely negative delta G mean lots of energy release?

[Yggdrasil]

Not necessarily.  ΔG = ΔH - TΔS.  Even if you have an endothermic reaction (ΔH > 0), the reaction can still have a large negative ΔG if the reaction results in a large gain in entropy (ΔS>0).  For example, even though melting ice requires an input of heat energy, the process is spontaneous (ΔG < 0) at room temperature because the reaction results in a gain in entropy (i.e. the TΔS term outweights the ΔH term).

[nj_bartel]

and reaction rate has more to do with activation energy than dg