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My textbook gives me the following formula for calculating the enthalpy change of any reaction:

$$\Delta H_\mathrm{r}^\circ=\sum(\Delta H_\mathrm{c}^\circ)_\text{reactants}-\sum(\Delta H_\mathrm{c}^\circ)_\text{products}\label{b}\tag{0}$$

where $\Delta H_\mathrm{c}^\circ$ is the standard enthalpy of combustion. Note that the enthalpy of combustion for same elements must be to the same product (i.e. all the elements of nitrogen in whichever compound they are, should be combusted to the same nitrogenous oxide)

My textbook simply assume it as a "fact". However, I have yet not found an intuitive or a mathematical way to prove it.

For starting the proof, I would write this balanced equation involving an arbitrary hydrocarbon:

$$\ce{C_xH_y + H_2 -> C_xH_{y+2}}\label{a}\tag{1}$$

The next step is obviously to write their standard combustion reactions, like this one for the first reactant: $$\ce{C_xH_y + $\left(x + \frac{y}{4}\right)$O2 -> xCO2 + \frac{y}{2}H2O}$$

and then another two for the other two compounds, and then show that, by Hess's Law, they can be remodeled into the original equation by simple arithmetic.

However, there are certainly a lot of problems with this approach:

  1. What if reaction $\ref{a}$ had more than one reactants, or more than one products? How do I write a proof for the general reaction involving $m$ reactants and $n$ products?
  2. What if the reactant wasn't a hydrocarbon but rather an arbitrary organic compound like $\ce{C_xH_yS_zN_pO_q}$ with many more different elements. How would we write a proof then?

I find that in trying to approach a proof for a general reaction, I have made the situation extremely complex. I wonder if an intuitive or a mathematical proof of such a complex equation is even possible. And if it is not, then how can we even assume equation $\ref{b}$ as true? Or, is it that the reaction $\ref{b}$ does not apply to all organic reactions, but only a subset of them?


PS: While I was experimenting with different equations, I stumbled upon $\ce{N2 + 3H2 -> 2NH3}$. Here, if you write out the individual combustion equations and balance them out, you'll get $\Delta H_\mathrm{r}^\circ=(\Delta H_\mathrm{c}^\circ)_{\ce{N2}} + 3(\Delta H_\mathrm{c}^\circ)_{\ce{H2}}-2(\Delta H_\mathrm{c}^\circ)_{\ce{NH3}}$ as expected from the general formula. However, the interesting thing to note here is that it does not matter whether you combust nitrogen to nitrous oxide, dinitrogenpentoxide, or perhaps any other nitrogenous oxide. In the end, it will all balance out and give you the answer you're looking for. (I call them self-balancing equations.)


Source: KS Verma; Physical Chemistry for JEE (Advanced): Part 1; Chapter - Thermodynamics Illustration 6.56III

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  • $\begingroup$ I think the heat of reaction should be net "Heat of Formation" ($\ce{$\Delta $H_f} $) of products minus that of reactants. $\endgroup$ – Soumik Das Mar 22 '18 at 11:30
  • $\begingroup$ @SoumikDas of course you're right. I never contradicted that. In fact, there's also a third way: $\Delta H_r = \text{Bond energy of reactants} - \text{Bond energy of products}$ but I am not asking for its proof in my question, am I? :) $\endgroup$ – Gaurang Tandon Mar 22 '18 at 11:32
  • $\begingroup$ I know the formula seems unconventional, but, apart from that, if there is any other problem in the question, please let me know. Thank you.|| For the people who VTCed, (1) could you please explain what exactly is "unclear"here? (2) if you think the question is broad, why do you think so? Is that because the proof for the formula is long, or that the proof does not exist? In either case, it'd be more helpful if you could please show how the formula is unproveable/difficult to prove, instead of just implicitly assuming that it is. Thank you. $\endgroup$ – Gaurang Tandon Mar 23 '18 at 1:01
  • $\begingroup$ the simple intuitive way to get the result is to recognise that, if it were not true, there would be some cycle of reactions that could break the law of conservation of energy. $\endgroup$ – matt_black Mar 23 '18 at 11:35
  • $\begingroup$ @matt_black Oh that's interesting! But, how does it explain whether $\Delta H_\mathrm{r}^\circ=\sum(\Delta H_\mathrm{c}^\circ)_\text{reactants}-\sum(\Delta H_\mathrm{c}^\circ)_\text{products}$ or $\Delta H_\mathrm{r}^\circ=\sum(\Delta H_\mathrm{c}^\circ)_\text{products}-\sum(\Delta H_\mathrm{c}^\circ)_\text{reactants}$? (the order of the terms) $\endgroup$ – Gaurang Tandon Mar 23 '18 at 12:10
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Second approach, but very abroad assumptions. Assume a reaction

$$\ce{reactants -> products} \tag{1} \label{rxn1}$$

where $\text{A}$ and $\text{B}$ are not necessarily one compound each. Let's analyze element $x$ that appears in the reactants as $x_{a}^{\alpha}$, and appears in products as $x_{b}^{\beta}$, where $\alpha$ and $\beta$ are the oxidation states, and $a,b$ are the number of atoms of the element in each case.

Then, combustion of $x$ can be written as:

$$\ce{ $x_{a}^{\alpha}$ + $\frac{ay}{2c}$ O2 -> $\frac{a}{c}$ $x_{c}^{\gamma}$O_{y}}$$

The enthalpy of this reaction is

$$\Delta H_\mathrm{c}(x_a^\alpha) = \frac{a}{c}\Delta H_\mathrm{f}\left(x_{c}^{\gamma}O_{y}\right) -\Delta H_\mathrm{f}(x_a^\alpha) $$

Analogously, and assuming the product of combustion is the same:

$$\Delta H_\mathrm{c}(x_b^\beta) = \frac{b}{c}\Delta H_\mathrm{f}\left(x_{c}^{\gamma}O_{y}\right) -\Delta H_\mathrm{f}(x_b^\beta) $$

If the stoichiometric coefficient of $x$ in the reaction \ref{rxn1} is $m$, the stoichiometric coefficient in the products will be $\frac{am}{b}$ The difference of enthalpies of combustion (taking into account stoichiometric coefficients of \ref{rxn1}), is:

$$ \begin{align} \Delta H_\text{rxn} &= m\Delta H_\mathrm{c}(x_a^\alpha) - \frac{am}{b}\Delta H_\mathrm{c}(x_b^\beta)\\ &= \frac{am}{c}\Delta H_\mathrm{f}\left(x_{c}^{\gamma}O_{y}\right) - m\Delta H _\mathrm{f}(x_a^\alpha) - \frac{am}{b}\left(\frac{b}{c}\Delta H_\mathrm{f}\left(x_{c}^{\gamma}O_{y}\right) -\Delta H_\mathrm{f}(x_b^\beta) \right)\\ &= \frac{am}{b}\Delta H_\mathrm{f}(x_b^\beta) - m\Delta H_\mathrm{f}(x_a^\alpha) \end{align} $$

Which is the definition we all know. Considering this for all elements in the reaction, then the formula is true. Now, taking this with a grain of salt, since I haven't found a way to prove it for cases where the combustion product is not an oxide (I got a feeling it can be done with charge balance, that's why I was considering the oxidation states, but I haven't gotten to a proof for that case, I left them in case someone else has an idea based on that).

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This is an application of Hess' Law which states that the overall enthalpy change of a reaction is independent of the route taken, which essentially comes from the conservation of energy.

It enables the enthalpy change of a reaction, which does not take place directly, to be easily measured.

In many cases, the reactant and product can be burnt and the enthalpy change measured by calorimetry. Because the reactant and product are made up of the same elements, the products of combustion will be the same.

Enthalpies of formation are often difficult to measure directly so you can go through the following steps:

  1. Write down the equation you are interested in.

  2. Underneath, write down the combustion products, which will be the same.

  3. Apply Hess' Law

In general you get:

Hess 1

You can see that, from Hess' Law, the energy change of the green route must be the same as the energy change of the red route, since the arrows start and finish at the same place.

The enthalpy of formation of methane is a good example since carbon does not react with hydrogen directly to give methane. However carbon, hydrogen and methane can be burnt easily and the enthalpy changes measured.

The Hess Cycle looks like this:

Hess Cycle

Applying Hess' Law:

$\rm \Delta H_f+(-890)=-394+2(-288)$

$\rm \Delta H_f-890=-960$

$\rm \Delta H_f=-960+890$

This where your formula comes from.

$\rm \underline{\Delta H_f=-70\,kJ/mol}$

A graphical method is often better than trying to remember and apply formulae.

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Maybe not an answer, but I think the idea may go like this. It is sort of a Born-Haber like cycle. For the reaction of $X$ compound giving products $Y$, we take each element $x_i$ of $X$ and oxidize them to a "high" oxidation state intermediate $I$ (combustion of $X$). Then, we reduce $I$ to the oxidation state $y_i$ of that element in product $Y$ (reverse reaction of the combustion of $Y$). Since $I$ is the same for both reactions, as long as both combustion (or reverse combustion) reactions are balanced by mass, the difference between them will be the same as the difference between $X$ and $Y$, which is the enthalpy of the reaction. Maybe the figure explains it a bit better.

enter image description here

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  • $\begingroup$ (not my downvote) I think the figure contradicts what you're suggesting. You say that "we reduce $I$ ... in product $Y$" But, in the figure you've shown combustion of $I$ to $Y$. Isn't combustion always an oxidation instead? $\endgroup$ – Gaurang Tandon Mar 23 '18 at 3:18
  • $\begingroup$ Yes, but we're doing the opposite to get to $Y$ (that's why it's the difference of heat of combustions, not the sum). $\endgroup$ – ralk912 Mar 23 '18 at 3:19
  • $\begingroup$ That's interesting. But how do you know that the moles of the intermediates will cancel out in the end though? $\endgroup$ – Gaurang Tandon Mar 23 '18 at 3:53
  • $\begingroup$ @GaurangTandon see my new answer, it doesn't cover all cases, but I think it covers a few at least. $\endgroup$ – ralk912 Mar 23 '18 at 4:21

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