HS Chemistry - Heat in Reactions

Hess's Law

Overview of The Page

This page will cover:

What is Hess's Law?
How can Hess's Law be used?

Hess's Law builds off of the 1st Law of Thermodynamics. It states that the amount of energy involved in a reaction must always be equivalent, regardless of its direction. If the direction of the reaction would change, then the direction of the energy would also change, but the amount of energy would remain the same.

The following reaction $from the previous subpage$ :

H₂ + Cl₂ → 2HCl

Has an enthalpy change of -184 kJ. That is, 184 kJ of heat energy is released, or evolved, in this reaction.

It makes sense that if we repeated this reaction for another set of Hydrogen and Chlorine molecules, that reaction would also release 184 kJ of heat energy. After all, the reactant particles are of the same type and present in the same amount, and so are the products. If these don't change, then neither will the energy of the reaction.

It also follows that if we reversed the reaction:

2HCl → H₂ + Cl₂

The direction of the energy will also reverse. If forming 2 molecules of Hydrochloric acid from 1 molecule of Hydrogen gas and 1 molecule of Chlorine gas releases energy, then reversing that process should absorb energy.

From this, we can tell that the enthalpy change of:

2HCl → H₂ + Cl₂

Is +184 kJ of heat energy. 184 kJ of heat energy is absorbed in this reaction.

That amount of energy the same as the amount of energy that was released in the reaction:

H₂ + Cl₂ → 2HCl

The only difference is that in H₂ + Cl₂ → 2HCl, 184 kJ of heat energy was released $shown by the \- sign in the enthalpy change$ and energy was lost from the system $to the surroundings$ , while in 2HCl → H₂ + Cl₂, 184 kJ of heat energy was absorbed $shown by the \+ sign in the enthalpy change$ and energy was absorbed into the system $from the surroundings$ .

This can be shown using the potential energy diagram for one of these reactions. If we draw dashed lines at the energy levels of the reactants $in this case, H~2~ \+ Cl~2~$ and the products $in this case, 2HCl$ :

Potential Energy Diagram

The H₂ + Cl₂ will always have the same energy level, and so will the 2HCl. Therefore, the difference between their energy levels will also remain the same. The amount of energy involved in the reaction will therefore also remain the same.

If the reaction H₂ + Cl₂ → 2HCl has an enthalpy change of ∆H, then the reaction 2HCl → H₂ + Cl₂ has an enthalpy change of -∆H, or ∆H in the reverse direction.

Thus, the amount of energy involved in the reverse reaction is the same as the amount of energy involved in the original reaction, with the only difference being whether the energy is absorbed or released $evolved$ .

Manipulating Reactions

In accordance with Hess's law, reactions can also be manipulated, provided that all factors are appropriately balanced.

What does this mean?

Let's take the reaction:

H₂ + Cl₂ → 2HCl

∆H = -184 kJ

This can be broken up into three reactions:

H₂ → H + H
Cl₂ → Cl + Cl
2H + 2Cl → 2HCl

The first two absorb energy, as the atoms in the molecule are separated. But the third one releases energy, as atoms are combined into molecules.

Let's calculate the enthalpy change for each of the three reactions. For the first two, since the molecule is being split into separate atoms, we'll use the bond energies $taken from a table that's not shown here$ :

1) H₂ → H + H
∆H_r1 = +436 kJ

2) Cl₂ → Cl + Cl
∆H_r2 = +242 kJ

And for the third one, since atoms are being combined into a molecule, we'll use the heat of formation $also taken from a table that’s not shown here$ :

3) 2H + 2Cl → 2HCl
∆H_r3 = 2 × -431 = -862 kJ

The first two absorb energy, but the third one releases energy $shown by the \+ and \- signs on the enthalpy change$ . Adding all the enthalpy changes together gives the total enthalpy change of the reaction: 436 + 242 - 862 = -184 kJ.

But let's look more closely at how that was done; every equation's enthalpy change of reaction was added together, and so were their reactants and products.

If we add the energy changes of the equations together, then we need to add the material changes together as well to keep the equation balanced. What we do to one side $the energy$ , we must do to the other as well $the matter$ :

Matter Changes	Enthalpy Changes
H₂ → H + H	∆H_r1
+	+
Cl₂ → Cl + Cl	∆H_r2
+	+
2H + 2Cl → 2HCl	∆H_r3

This comes out to:

H₂ + Cl₂ + 2H + 2Cl → 2H + 2Cl + 2HCl

∆H_r = ∆H_r1 + ∆H_r2 + ∆H_r3

And since like terms cancel out:

H₂ + Cl₂ + ~~2H + 2Cl~~ → ~~2H + 2Cl~~ + 2HCl

∆H_r = ∆H_r1 + ∆H_r2 + ∆H_r3 = 436 + 242 + $\-862$ = -184 kJ

Thus, we're left with:

H₂ + Cl₂ → 2HCl ∆H_r = -184 kJ

This is useful because not only can it easily be applied to much larger chemical equations as well, but we can also use it to find the enthalpy heat of reaction for more complex reactions.

Here's another unknown chemical reaction:

Matter Changes	Enthalpy Changes
4Cr + 3O₂ → 2Cr₂O₃	∆H_r = ?

With two known chemical reactions provided:

Matter Changes	Enthalpy Changes
Cr₂O₃ → 2Cr + 3O	∆H_r1 = +1872 kJ
O₂ → 2O	∆H_r2 = +496 kJ

Using the bond energies to calculate the enthalpy heat of reaction for this one will be tricky, as the product molecule is made up of 5 atoms, and we don't know how they connect. We can't tell which atoms are bonded to which $is the 1st Oxygen atom connected to only a Cr atom, or is it connected to both Cr atoms, or is it connected to the other 2 Oxygen atoms but none of the Cr atoms, etc. We don’t know$ and therefore we don't know what the intramolecular bonds are. It would be very difficult to calculate the enthalpy change of reaction by breaking up the reactants into separate atoms and then calculating using bond energies without knowing what intramolecular bonds are being broken apart/formed.

But we can use Hess's Law to solve this reaction. We can try manipulating the second and third reaction, and then adding them together to get the third reaction. All we need to do is apply the same changes to the enthalpy heats, and then we can add them to get our answer.

First, both the second and third reaction have lone Oxygen atoms on the product side. Since the original reaction doesn't have lone Oxygen atoms, we want ours to cancel each other out as like terms, which means they need to be on opposite sides and have equal coefficients. We'll reverse one of the reactions:

1) 2Cr + 3O → Cr₂O₃
∆H_r1 = -1872 kJ

Since we reversed the direction of the reaction, we needed to reverse the sign of the enthalpy change as well. The number doesn't change, but since heat is going in the other direction, the sign flips.

The third reaction could have been reversed instead of the second reaction, but since the original reaction has Cr₂O₃ on the product side, reversing this reaction will save us a step later on. But either way is correct.

Now, let's write out our updated second equation and our third equation:

2) 2Cr + 3O → Cr₂O₃
∆H_r1 = -1872 kJ

3) O₂ → 2O
∆H_r2 = +496 kJ

Now the top equation has lone Oxygen atoms on the reactant side, and the bottom equation has them on the product side. Now all that's needed before they can cancel each other is for the lone Oxygen atoms to have the same coefficients.

2 × $2Cr \+ 3O$ → 2 × Cr₂O₃ ∆H_r1 = 2 × -1872 kJ

3 × O₂ → 3 × 2O ∆H_r2 = 3 × +496 kJ

And the coefficients can be expanded:

4Cr + 6O → 2Cr₂O₃ ∆H_r1 = -3744 kJ

3O₂ → 6O ∆H_r2 = +1488 kJ

Now, since the lone Oxygen atoms are on opposite sides and have the same coefficient, we can add the two equations together and cancel the lone Oxygen atoms out:

4Cr + 6O + 3O₂ → 2Cr₂O₃ + 6O ∆H_r = ∆H_r1 + ∆H_r2

4Cr + 6O + 3O₂ → 2Cr₂O₃ + 6O ∆H_r = -3744 + 1488 kJ

4Cr + 3O₂ → 2Cr₂O₃ ∆H_r = -2256 kJ

Now we have our original reaction, and we've found the enthalpy change of reaction for it as well, all using Hess's Law.