Sciencemadness Discussion Board

Exothermic redox reactions

DalisAndy - 4-12-2015 at 18:19

Just a general question. Thermite like reaction behave very similar to all other reactions correct? The reactivity series stills applies right? I'm just trying to figure out how to use lots of lithium I got out of a battery. I figure why not make pure calcium and silicon; possibly other metals

j_sum1 - 4-12-2015 at 19:24

Wow. you have asked a big question. Short answer is that reduction potentiak tables can give an indicatiin but realky aren't the tool you want. The figures are fir aqueous solutions at 1M and standard laboratory conditions. For thermites it is better to go for heats of formation.

Ca is not something you are likely to get. Si will work but you need to do some tricks. Search for some recipes. MrHomeScientist has some good thermites on his youtube channel. There is an exotic thermites thread and also a useful table of thermodynamic information on a range of thermites downloadanle somewhere. I may be able to give references later but am on my phone at present. And of course, blogfast is a specialist on these and can provide useful expert help once you get to the stage of asking specific questions.

In short, thermites are lots of fun.

careysub - 4-12-2015 at 20:09

You can get calcium, but it isn't cheap. The best source I know is Onyxmet.com.

They have calcium for:
Calcium metal (dendritic) 99.87% 100.0 g $26.64

Galliumsource has a price almost as good, but you have to buy a lot more:
Calcium metal 99.0% 250.0 g $69.00

DalisAndy - 4-12-2015 at 20:10

Oh just my luck. Maybe I can convince blogfast I'm not chemically illiterate. Thank you anyway

deltaH - 4-12-2015 at 21:13

Quote: Originally posted by DalisAndy  
Just a general question. Thermite like reaction behave very similar to all other reactions correct? The reactivity series stills applies right? I'm just trying to figure out how to use lots of lithium I got out of a battery. I figure why not make pure calcium and silicon; possibly other metals


Tables of standard electrode potentials are a type of thermodynamic evaluation, BUT for solutions in water AND under standard conditions of temperature and pressure.

What you want to do in this case is work out the dG_reaction. I'll walk you through it:

First write the chemical equation, e.g.

2Li + CaO <=> Li2O + Ca

I use an equilibrium symbol because I do not yet know which direction is favoured.

Now I need dG_formation_standard for Li2O and CaO only, because the dGf's of elements are 0 by definition.

For simple and common substances, this can be gotten from:

http://webbook.nist.gov/chemistry/

For example...

For Li2O, select "formula", then type in "Li2O" and then select the "condensed phase" checkbox under thermodynamic data. Scroll down on the next page for the results, specifically:

ΔfH°solid -598.73 kJ/mol
S°solid 37.85 J/mol*K

From these two value we can calculate the standard Gibbs free energy of formation by using the simple relation:

ΔfG° = ΔfH° - T.S°

Since we're using standard conditions, T = 298K

so for Li2O:

dGf_std_f_Li2O = -598.73kJ/mol - (298K)(37.85J/mol.K)/(1000J/kJ) = -610 kJ/mol

Similarly, for CaO, we can look up the values and work them out and get:

dGf_std_f_CaO = -673kJ/mol

Now, dG_reaction_std = SUM[(Stoichiometric coefficients component i)*(dG_std_f component i)

where the stoichiometric coefficients are positive for products and negative for reagents and the value is gotten from the stoichiometry from the balanced equation.

So for our case, this is simply:

dG_reaction_standard = -610 -(-673) kJ/mol = +63 kJ/mol

So under standard conditions, the reverse reaction is very favoured, i.e.

Li2O + Ca => 2Li + CaO

So bad luck, sorry :(



[Edited on 5-12-2015 by deltaH]

deltaH - 5-12-2015 at 04:01

I was curious about the thermodynamics of the reaction:

3CaO + 2Al <=> Al2O3 + Ca

dGf standard Al2O3 is -1691kJ/mol

So dG_reaction_standard as I've written the equation is +328kJ/mol

So that's unfortunately also NOT proceeding in a forward direction :mad:

Even for magnesium:

Mg + CaO <=> MgO + Ca

dG_reaction_standard = -609kJ/mol - (-646kJ/mol) = +37kJ/mol

There's good reason for using calcium metal to reduce strongly reducing metals (like thorium :D).

Also for

Ti + 2CaO <=> Ca + TiO2

dGf_standard_TiO2 = -959.08 kJ/mol
So dGf_reaction_standard = +333kJ/mol

Which means you can even reduce titanium dioxide to titanium metal with calcium metal.

[Edited on 5-12-2015 by deltaH]