Step 6: Thermochemical stoichiometry

Yes stoichiometry can even be used with thermodynamics. It shows the movement of energy through out a reaction. There is two types of energy that can used. There is enthalpy (heat), or free energy. Entropy (randomness) can also be used. Free energy is just the combination of entropy and enthalpy.

There are two ways you find these. There are tables that  list all three. These can be in many different units, so make note of it. The energy differs for every substance so you have to look them all up. The other way to find it is using the equation G=H-S*K.
G- free energy
H- enthalpy
S- entropy
K- temperature in Kelvin

Using this method you still need two of the three, so I think you should use a table for all three unless you want to practice your algebra. 

The important thing to remember here (and relates most to stoichiometry) is that its measured in units of energy per mole. That lets use do stoichiometry with it. The units I'm using are joules/mole or J/mole.

For example I'll use a reaction that very relevant to you. The combustion of sugar using enthalpy.

You have 30g. of sugar. How much energy can you get from it.

C6H12O6 + 6 O2 --> 6 H2O + 9 CO2
C6H12O6 + 6 O2 - 1250 kJ/mol --> 6 H2O + 6 CO2 + 209 kJ
      30g.                              l                                                     l
         l                                  l                                                     l
    30/180                           l                                                     l
         l                                  l                                                     l
    .167 moles----  .167*1250-----------------------------------209 kJ

You get 209 kJ of energy.

The first thing you may have noticed is that I changed the sign from a - to a +. That is because the number in the equation is the standard enthalpy of formation or the energy needed to create it. Since we are destroying it we get that energy back. All you have to really know is switch the signs. The second thing you make have noticed is that I didn't include the energy from the oxygen. That is because oxygen is an element and, all elements have a standard enthalpy of formation of 0.

This part may be wrong so correct me if I am. Its been awhile since I have used it, but I'm pretty sure this work.
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&quot;6.022&times;1023&quot; 6.022&times;10 <em><strong>^</strong></em> 23
Yeah didn't notice that. When I wrote it up I used an exponent, but it did show up I guess. Thank you :)
In the NaCl example, how did you determine the molecular weight of NaCl to be 28?
Uh oops thank you :) I was going through it fast and used the number instead of weight.
Ok, so 58?
Yep, that one should be right unless my arithmetic is way off. If you are wondering about any molecules just look on a periodic table for the weights of each element and add them up.

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