Heat in reactions

[half-death]

Hello,
I'd like to know why some dissolutions are exothermic (NaOH + H2O, for example) even if the entropy increases (if I'm right, solid state is less chaotic than aquous state...)
Also, I'd like to find some examples of uses of exothermic and endothermic reactions in every-day life. The only one I can think of, is a cold compress (NH4NO3 + H2O)...
Thank you for your help!

[Fishteacher73]

All reactiuons are either endothermic or exothermic. One can form a more chemically stable solution given the current factors. Some just need a bit of energy to find that "happy spot" and are endothermic. Others release energy when breaking apart from the intial configuration. These are exothermic.

[UncleAl]

Different solutes have different mechanisms in water. You can substantially increase the entropy of the solution vs. pure water, or not. You have heats of solvation (Lewis acid-base interactions). You can disrupt hydrogen bonding. Dissolving sodium acetate in water gets hot; dissolving NaOH or KOH in water gets real hot. Slaking lime gets boiling hot and more. Dissolving ammonium nitrate or urea gets real cold.

[half-death]

I totally forgot to take the hydrogen bond into account! If I understand well, to predict if a dissolution is exothermic or endothermic, you have to look the electronegativity of the ions to see how they will react with water because of the hydrogen bond. Is that it?

[TeleMad]

One nitpick.

Exothermic reactions do not necessarily give off energy to the surroundings, and endothermic reactions do necessarily take in energy from the surroundings. For example, if a reaction is exothermic, but involves a large enough reduction in entropy, then the net affect is that the system took in energy from the surroundings.

When dealing with overall energy the terms are exergonic and endergonic, not exothermic and endothermic.

[half-death]

Originally Posted by TeleMad

if a reaction is exothermic, but involves a large enough reduction in entropy, then the net affect is that the system took in energy from the surroundings.

When the entropy decreases, doesn't energy has to be released? If so, the system won't take energy from the surronding...
Well, I guess... I'm really not sure..........

[TeleMad]

TeleMad: if a reaction is exothermic, but involves a large enough reduction in entropy, then the net affect is that the system took in energy from the surroundings.

Originally Posted by half-death

When the entropy decreases, doesn't energy has to be released? If so, the system won't take energy from the surronding...
Well, I guess... I'm really not sure..........

The standard equation for measuring changes in a system's free energy (or whether or not a process is spontaneous) is as follows:

[delta]G = [delta]H - T[delta]S

where [delta]G is the change in Gibb's free energy, [delta]H is the change in enthalpy, T is the temperature, and [delta]S is the change in entropy.

An exergonic process has a negative [delta]G and is spontaneous. An endergonic process has a postiive [delta]G and is nonspontaneous.

A process can be endergonic even if it is exothermic because the latter term takes only [delta]H into consideration, whereas the first term takes both the change in enthalpy and the change in entropy into account.

Here are some completely made up numbers, made up just to illustrate the point. Assume [delta]H = -5, T = 2, and [delta]S = -4. The process would be exothermic because [delta]H is negative. But if we calculate [delta]G we get [delta]G = -5 - [2(-4)] = +3, showing that it is endergonic. It is endergonic because the change in entropy was large enough to dominate the change in enthalpy, turning what would have otherwise been a negative [delta]G into a positive one.