Question about enthalpy of reaction


by gokuls
Tags: enthalpy change, thermochemistry
Borek
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#19
Dec28-12, 02:35 AM
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Quote Quote by gokuls View Post
How is it that in both cases we deal with 1 mol worth of reactions? In the first equation you wrote, it would be 1 mol worth of reactions because you produce 1 mol of water, but in the second equation, you must have 2 mols of reactions, since you produce 2 mol of water, n'est pas?
No, it depends on what you consider a "reaction". That's why it is so important to remember it is "reaction as WRITTEN".
morrobay
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#20
Dec29-12, 01:39 AM
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Quote Quote by gokuls View Post
Say there were multiple reactants and products, then which one would it be referring to?
As written . This example is right out of the book:
N2 (g) + 3H2 (g) --> 2NH3 (g)
This reaction involves splitting one mole of N2 triple bond requiring 945 kJ/m
And 3 moles of H2 single bond, requiring 3(436) kJ
And the formation of 6 moles of N-H bonds releasing 6(391) kJ
The enthalpy change is 2253 kJ reactants - 2346 kJ products = - 93 kJ
That is twice the molar enthalpy of one mole of ammonia formation
Studiot
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#21
Dec29-12, 04:18 AM
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As written . This example is right out of the book:...........................
...............That is twice the molar enthalpy of one mole of ammonia formation
And is entirely consistent with my convention and what Borek has been saying.

Take the first named substance

The enthalpy of this reaction is 93kJ per mole of nitrogen

If you want the figure per mole of ammonia you have to divide this figure by 2 and if you want it per mole of hydrogen you have to divide it by 3.

This is where the older use of gram-mole makes life easier than sitting up late at night counting molecules.


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