Why can we add/substract/cross out chemical equations for Hess law?

Question

Suppose I take the model example:

(1) $\ce{C + 1/2 O2 -> CO}$

(2) $\ce{CO + 1/2 O2 -> CO2}$

Adding both together:

(3) $\ce{C + CO + 1/2 O2 + 1/2 O2 -> CO + CO2}$

(=> $\ce{C + 1/2 O2 + 1/2 O2 -> CO2}$)

Now why is it generally ok to cross out CO on both sites? Clearly it's balanced with or without and it's not part of the reaction either but let's assume for a moment that this reaction (3) containing the reactants and products from (1) and (2) suddenly forms other products:

$\ce{C + CO + 1/2 O2 + 1/2 O2 -> C2O3}$

(Only hypothetically, there are certainly some reactions that then form something other than merely additively the same products), why can Hess law still be true than? The formation must clearly have another enthalpy.

Or do I always have to imagine with such manipulations that they are pure mathematical steps to the actual reaction of interest without the various reaction paths actually reacting in such ways?

Answer 1

From your example, you were trying to use Hess's law to find the enthalpy change of the reaction $\ce{C + O2 -> CO2}$ from a two-step process:

$(1) \ \ce{C + \frac{1}{2}O2 -> CO}$

$(2) \ \ce{CO + \frac{1}{2}O2 -> CO2}$

As you said, adding these two processes combine to produce the overall reaction we are interested in as the $\ce{CO}$ on both sides would cancel. Now, what you are saying is that, why can't we just not cancel the two $\ce{CO}$ and then combine the $\ce{CO}$ and $\ce{CO2}$ to make a different hypothetical molecule $\ce{C2O3}$.

Sure, you can do that, and obviously, the overall reaction would have a different enthalpy, but the $\ce{C2O3}$ never appeared in our steps. You can easily fix this by adding an extra step as below:

$(1) \ \ce{C + \frac{1}{2}O2 -> CO}$

$(2) \ \ce{CO + \frac{1}{2}O2 -> CO2}$

$(3) \ \ce{CO + CO2 -> C2O3}$

Now if you combine them, you'd get what you had,

$\ce{C + CO + \frac{1}{2}O2 + \frac{1}{2}O2 -> C2O3}$

or $\ce{C + CO + O2 -> C2O3}$

However, we had to consider an additional step for that overall process which would be the source of the difference in enthalpy that you are pointing to.

Answer 2

Enumerated equations of chemical reactions, regardless of if they reflect really ongoing processes or not,, are just accountant inventories of atoms counts on both sides. They say in your example that numbers of atoms of carbon and oxygen on both sides are equal.

If there are the same terms on both sides, they can be subtracted as in any other mathematical equation.

If atoms counts on both sides are equal for 2 reactions, summing the sides is the equivalent operations and counts for both sides will be equal too.

Not there you cannot imply from the above reaction equations that CO and CO2 form C2O3. Such knowledge is outside of mathematical formalism of these equations.

Answer 3

Fundamentally, this is precisely because of Hess's law, which says that (as long as the reactants and the products are both in consistent standard state) the change in enthalpy depends only on the changes between the initial reactants and final products of the reaction, not on how the reaction actually happens (or even whether it can actually happen).

In particular, since a "reaction" like $\ce{CO -> CO}$ where nothing actually changes has $\Delta H = 0$, it follows from Hess's law that adding such a (non-)reaction to another reaction cannot change its $\Delta H$. So the reactions $$\ce{C + O2 -> CO2}$$ and $$\ce{C + CO + O2 -> CO + CO2}$$ must have the same $\Delta H$, since they only differ by the addition of one $\ce{CO}$ molecule to both the reactants and the products. According to Hess's law, it doesn't matter whether the $\ce{CO}$ actually participates in the reaction somehow or not: one molecule of $\ce{CO}$ goes in and one molecule of $\ce{CO}$ comes out, and that's all that matters. Indeed, for all we know, that $\ce{CO}$ might be sitting in a sealed bottle on a shelf somewhere and might never even come anywhere near the other reactants!

It is, of course, extremely convenient that Hess's law happens to hold, since if it didn't, we might not even be able to define the concept of a standard enthalpy of reaction without having to consider details like the specific pathway the reaction takes and what other chemical species might be present and how they might affect the kinetics of the reaction. But it does, and indeed that's exactly why reaction enthalpy is such a useful concept in chemical thermodynamics.

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Ps. Of course, as others have already noted, a hypothetical reaction like $\ce{C + CO + O2 -> C2O3}$ would have a different $\Delta H$ than $\ce{C + CO + O2 -> CO + CO2}$, since it has different products: $\ce{C2O3}$ instead of $\ce{CO + CO2}$. There is no contradiction here, since the reaction $\ce{CO + CO2 -> C2O3}$ has a non-zero $\Delta H$ that exactly accounts for the difference.

Answer 4

Now why is it generally ok to cross out CO on both sites?

Every chemical reaction tells you about the amount of reactant used up, and the amount of product made. To make things easier, we will consider a situation where exactly one mole reacts, e.g. a stoichiometric coefficient of "1" in front of a reactant means we use up one mole, and a stoichiometric coefficient of "1" in front of a product means we make one mole.

To understand why we can cancel CO on both sides, imagine we have a thousand moles of all species present (reactants and products). Each reaction will change the amounts in our stockpile. Because there is so much, we will never run out.

The first reaction adds a mole of CO to our stockpile. The second reaction uses up a mole of CO in our stockpile. The net effect is no change in CO. It does not matter whether reaction (1) happens first, second, or at the same time as reaction (2).

Or do I always have to imagine with such manipulations that they are pure mathematical steps to the actual reaction of interest without the various reaction paths actually reacting in such ways?

If the reactions are real, substracting or adding reactions is real, as well. When you add them in a one-to-one fashion, there is usually a physical reason. For example, if you start out with no CO at all, the second reaction can only use up as much CO (or less) as the first reaction produces.

Why can we add/substract/cross out chemical equations for Hess law?

When you substract a reaction, you are saying that it runs in the reverse direction. In this case, your stockpile of products will diminish, and your stockpile of reactants will increase. Again, this might actual happen, or it might be artificial.

For Hess' law, it does not matter. The quantity you calculate with Hess' law, enthalpy, is a state function. Any path from A to B will result in the same enthalpy change, no matter whether real or fictitious. If you can measure the reaction from B to A, for example, you can also give the enthalpy for the reaction from A to B, even if you can't do the experiment.

UPDATE

[OP in comments] "If the reactions are real, substracting or adding reactions is real, as well." - That is a little unclear to me, although I can have two real reactions, but if I add them together, it can also happen that instead of the products merely being additive, another products are formed because, for example, a reactant from reaction a now suddenly prefers to bond with a reactant from reaction b. A+B->C and D+E->F, instead of A+B+D+E->C+F it is A+B+D+E->I+J, where A+D->I and B+E->J. That shouldn't be simply additive anymore because it doesn't produce the same products as the 1. and 2. reactions alone.

That is true in general. If I don't mix reactant, I don't get a reaction. If I add a third thing (maybe an acid or a base, for example), I might get a different product. Hess's law just says that once you describe a certain set of reactions with some reaction equations, the enthalpies add up in the same way that these component reactions do.

Answer 5

Hess' Law is inferred from the nature of Enthalpy: It is a state function. This is derived through its equation, ΔH=ΔU(or ΔE depending on how you represent the internal energy)+PΔV.

P, Pressure, and V, volume, are both state functions: their values are only based on the present state (for example you could use the present P/V/n/T values, assuming ideal gas, or using the Van der Waals equation to find P and V respectively). Similarly, change in internal energy is also a state function, since its value is irrespective of the pathway taken to achieve the change in internal energy (q+w is always constant).

There's an example in Atkins' Chemical Principles that illustrate this fact: Consider this: If you're rolling a ball, A, down a hill, it ends up bumping into another ball, B, making B roll up a part of the hill, gaining GPE, and A itself slows to rest at the bottom of the hill.

In this example, A does work w on B by allowing it to gain KE which is transferred into GPE as it climbs the hill, and A loses energy as heat q to the hill surface by frictional heating as it moves down the hill. Therefore, the total loss of Internal Energy of A, ΔU, is equal to q+w. If A had lost transferred more heat instead of work, it means the process' pathway has changed. However, we know that A must have lost an amount of energy equal to its initial GPE, which is constant.

Therefore, the value of Internal energy is irrespective of the relative pathway it takes to achieve it. As we get here, you might now have a clue about why this was brought up, because it's beginning to sound a lot like the statement of Hess' Law: The total enthalpy change of a reaction is irrespective of the path taken.

We know that ΔH=ΔU+PΔV. We also know that U, P and V are all state functions. Therefore, since enthalpy is the sum of these functions, we can conclude that enthalpy is also a state function.

Therefore, we only have to consider the net outcomes of a reaction when calculating enthalpy: the net reactants leading to the net desired products, which justifies the cancelling of chemicals on both sides.