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Name: John R. Carriveaux III
Status: N/A
Age: N/A
Location: N/A
Country: N/A
Date: 1999 


Question:
Hi! I'm a student in high school researching for a project to do in fortran. I've been looking into chemistry and I think I have an idea. I plan to make a program that accepts a chemical equation, and gives back various forms of inforemation about the reaction, such as the Enthalpy of reaction, the Entropy of reaction, and Gibbs free energy. However, I do not wish to use huge data files containing Enthalpies of formation, or force the user to enter multiple equations. So here is my question: Is it possible to calculate a theoretical value for the Enthalpies of formation?

Thank you very much for your assistance.



Replies:



Joe's right...but there is a third, easier way. As long as you restrict yourself to reactions where the only thing that happens is the formation and breaking of bonds, you can estimate heats of reaction from bond energies. All you need to do is to specify the Lewis Dot structure of all the reactants and products, thereby identifying which bonds are broken and which bonds are formed. Then, you look up in a relatively small table the dissociation energies of all bonds broken and all bonds formed. The enthalpy of reaction is then given approximately by the formula delta H(rxn) = {sum over broken bonds} DE - {sum over formed bonds} DE

Example: CH4 + Cl2 -> CH3Cl + HCl

In this reaction, a C-H bond and a Cl-Cl bond are broken, and a C-Cl and an H-Cl bond are formed.

Thus, delta H(rxn)
= DE(C-H) + DE(Cl-Cl) - DE(C-Cl) - DE(H-Cl)
= 411 + 240 - 327 - 428 = -104 kJ/mole.
Experimental value = -101 kJ/mole....not bad!!!

Now, try the decomposition of hydrazine, N2H4, to form N2 and H2...

N2H4 -> N2 + 2 H2

An N-N and 4 N-H bonds are broken, and an N:::N and two H-H bonds are formed. Using DE(N-N)=159, DE(N-H)=389, DE(N:::N)=946, DE(H-H)=435, you should find delta H(rxn) = -101 kJ/mole, which compares well to -95.4 kJ/mole from experiment.

-prf tppr



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