As in the previous page about the enthalpy of formation $\Delta H_f^\circ$ for reactions of formation changes in enthalpy are important in more complex reactions such as those with more than one product or compounds on the side of reactants. This shall be treated here.
Enthalpy analysis for a common chemical reaction
This section was adapted from Castelland GW, Physical chemistry.
Now, with the reaction of formation theory as background we may deal with more complex and traditional chemical reactions. Then the same question as before is posed: what is the amount of energy required for a given reaction to occur?
In order to answer properly to the above question, some math development has to be done. Let us then consider a reaction for which the change of enthalpy is unknown.
Fe$_2$O$_3$(s) + 3H$_2$(g) $\rightarrow$ 2Fe(s) + 3H$_2$O(l) Eq. (01)
For the above reaction, each reactant and product have its ownn enthalpy so that substraction between those of the products and those of reactants gives the change in enthalpy $\Delta H$. This can be written schematically as follows,
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| Fig. 1 A sketch for the change of enthalpy in a reaction of two reactants and two products. |
In other words, since, for given temperature and pressure, all compounds have a given mole enthalpy $\bar{H}$ (enthalpy per unit of mole [J/mol]) we may express symbolically the enthalpies of each reactant and product in the reaction Eq. (01) as follows,
- molar enthalpy of Fe$_2$O$_3$(s) is represented as $\bar{H}$(Fe$_2$O$_3$, s)
- molar enthalpy of H$_2$(g) is represented as $\bar{H}$(H$_2$, g)
- molar enthalpy of Fe(s) is represented as $\bar{H}$(Fe, s)
- molar enthalpy of H$_2$O(l) is represented as $\bar{H}$(H$_2$O, l)
Thus, if we compare the reaction in Eq. (01) with the rough formula to determine the change in emthalpy $\Delta \bar{H}$ and make use of the above definitions for mole enthalpies $\bar{H}$ we obtain the following,
$\Delta \bar{H}=$ 2$\bar{H}$(Fe, s) + 3$\bar{H}$(H$_2$O, l) - $\bar{H}$(Fe$_2$O$_3$, s) - 3$\bar{H}$(H$_2$, g) Eq. (02)
Equation (02) could be very helpful if only all enthalpies were known.
However, nowadays there is enthalpy data available for several molecules. This is, $Delta H$ in Eq. (02) can be easily determined. On the other hand, there may be missing data for some compounds either because these are knew or because experimental determination is too complicated.
Let us do as if we did have few data at hand so that $\Delta H$ can not be estimated from Eq. (02). What do we do?
Algebra of reactions
A way of going arund Eq. (02) to find $\Delta H$ is by using what we know about the enthalpy of formation $\Delta H_f^\circ$ for reactions of formation. A closer look to Eq. (02) shows that if we were capable of finding $\bar{H}$(H$_2$O, l) and $bar{H}$(Fe$_2$O$_3$, s) were known our problem would be greatly reduced, which is at first a naive idea.
Then if we write the reactions of formation for H$_2$O and for Fe$_2$O$_3$ we have:
H$_2$(g) + $\frac{1}{2}$O$_2$ $\rightarrow$ H$_2$O(l) Eq. (03)
2Fe(s) + $\frac{3}{2}$O$_2$(g) $\rightarrow$ Fe$_2$O$_3$(s) Eq. (04)
respectively.
Also, we may think as before in terms of $\Delta \bar{H}$ for each of reactions in Eqs. (03-04). Then, if we use a notation similar to that in the previous section we may write as follows,
for reaction in Eq. (03)
- molar enthalpy of H$_2$(g) is represented as $\bar{H}$(H$_2$, g)
- molar enthalpy of O$_2$(g) is represented as $\bar{H}$(O$_2$, g)
- molar enthalpy of H$_2$O(l) is represented as $\bar{H}$(H$_2$O, l)
for reaction in Eq. (04)
Next, with the above notation the reaction in Eq. (03) may be rewritten in terms of the enthalpies, as shown schematically in Fig. 1. This procedure is barely the same followed previously to get Eq. (02). We obtain the following,
$\Delta \bar{H}$(H$_2$O, l) = $\bar{H}$(H$_2$O, l) - $\bar{H}$(H$_2$, g) - $\frac{1}{2}\bar{H}$(O$_2$, g) Eq. (05)$\Delta \bar{H}=$ 2$\bar{H}$(Fe, s) + 3$\Bigl[$$\Delta \bar{H}$(H$_2$O, l) + $\bar{H}$(H$_2$, g) + $\frac{1}{2}\bar{H}$(O$_2$, g) $\Bigr]$
Next, square brackets can be dropp if products are performed. We arrive to,
A further simplification in Eq. (10) can still be made if additions and subtractions are performed. Thus,
Understanding the change in enthalpy for a chemical reaction
The procedure to arrive at Eq. (10) was as clear as possible and all mole enthalpies $\bar{H}$ that were dropped during the process disappear due to the procedure itself, let us say. No magic trick was done here.
This lead us to some key conclusions:
- for a chemical reaction, as that shown in Eq. (01), the mole enthalpies $\bar{H}$ of the reactants and products, in the form of molecular species of elements in their stable form, has no effect on the mole enthalpy change $\Delta \bar{H}$;
- the change in mole enthalpy $\Delta \bar{H}$ for a chemical reaction, as that shown in Eq. (01), depends solley in the mole enthalpy changes of the compunds, as reactants and products.
Now, a side effect of the induced conclusions stated above is that if the elements, as reactants and/or products, have no incluence on the overall mole enthalpy change $\Delta \bar{H}$; then one may assign any value to the mole enthalpies of these reactants/products (any!).
If we observe the above so called side effect as if you did not know the value of $\bar{H}$ for the molecular species of elements in their stable form, appearing as reactants/products, Eq. (10) would be an unexpected life saver.
The fact, that those molecular species of elements, in their stable form, appearing as reactants/products, in a chemical reaction has no effect on $\Delta \bar{H}$ is the reason to find in data tables $\bar{H}_f=0$ for such elements. Of course, chemists could have set $\bar{H}_f$ in another arbitrary number but zero has mathematical advantages when simplifying.
Other stuff of interest
- Some basic concepts on thermodynamics
- Who was Clausius?
- Processes at constant volume and pressure
- Composition variables in mixtures
- The first law of thermodynamics and some concepts
- Partial and total derivatives (for thermodynamics)
- First law of thermodynamics for chemical reactions
- Solving the Colebrook equation
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Ildebrando.
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