Question

How can I understand Hess's law step by step?

Answer 1

Hess's law affirms that the energy and enthalpy change in every transformation does not depend by the different intermediate states that the transformation is carried through, but depend only by the initial and final states.

This is equivalent to say that energy and enthalpy are "state-functions".

This affirmation sounds a bit abstract, but it is very simple indeed. If you understand it, Hess's law will turn obvious.

In the following step by step explanation I won't distinguish between energy and enthalpy, those are quite similar, and the same explanation is good for both.

The state of a system tells us the set of conditions which describe "how is" the sistem in that state (which pressure, volume, kind and quantity of every substance in it, and so on), like the place of a man on the Earth surface is caracterized by a couple of coordinates (longitude and latitude).

The total amount of energy of a system depends by the state, that is, by the set of state variables, like the altitude of the man on the Earth crust depends by the place where he stands (coordinates).

Thus the energy change between initial and final state (e.g reactants to products) is like the altitude variation for going from place A to place B. You may change path, but the energy change, as well as the altitude variation, will depend only by the initial and final states (places), but not by the different intermediate states (places) of any different path. This explains the Hess's law.

I hope this explanation was useful.

Now let me make an example. If you burn two moles of methane (initial state) to form carbon dioxide (final state), the final state can be attained in two (and even more) ways:

A) directly:
`2CH_4(g) + 4O_2(g) → 2CO_2(g) + 4H_2O(l)`

B) in two steps:
`2CH_4(g) + 4O_2(g) → 2CO(g) + 4H_2O(l) + O_2(g)`
`2CO(g) + 4H_2O(l) + O_2(g) → 2CO_2(g) + 4H_2O(l)`

The energy of the initial state of the system can be calculated taking the formation enthalpy of 2 moles of methane, -150 kJ, whereas for the elementary substances, like gaseous biatomic oxygen, the formation enthalpy is zero. The formation enthalpies of many substances can be found here .Thus, the initial state has energy ("altitude") -150 kJ.

The energy of the final state of the same chemical system, two moles of carbon dioxide and four moles of liquid water is:
-393.5 kJ • 2 + (-286.0 • 4) = -1931 kJ.

Hess's law states that the energy variation is `E_f - E_i` = -1931 - (-150) = -1781 kJ, is independent by the path, A, B or whichever.

In the path B) the energy of the intermediate state of the system, made of 2 moles of carbon monoxyde, four moles of liquid water and one mole of dioxigen, is: -110.5 kJ • 2 + (-286.0 • 4) = -1365 kJ.

We can expect that for the second step, the combustion of two moles of CO(g), there will be released -1931 - (-1365) = -566 kJ. And indeed this amount exactly to the double of the energy released in the combustion of one mole of carbon monoxide, -283 kJ.

Then the energy change in the direct transformation A), - 1931 kJ, is equal to the overall energy change in the two step reaction B), -1365 + (-566) = -1931 kJ, as Hess's law does pretend.