Hess's Law

Energy Relations in Chemistry: Thermochemistry

Hess's Law

Hess's Law

It is often possible to calculate DH for a reaction from listed DH values of other reactions (i.e. you can avoid having to do an experiment)

Enthalpy is a state function

It depends only upon the initial and final state of the reactants/products and not on the specific pathway taken to get from the reactants to the products
Whether one can arrive at the products via either a single step or multi-step mechanism is unimportant as far as the enthalpy of reaction is concerned - they should be equal

Consider the combustion reaction of methane to form CO₂ and liquid H₂O

CH₄(g) + 2O₂(g) -> CO₂(g) + 2H₂O(l)

This reaction can be thought of as occurring in two steps:

In the first step methane is combusted to produce water vapor:

CH₄(g) + 2O₂(g) -> CO₂(g) + 2H₂O(g)

In the second step water vapor condenses from the gas phase to the liquid phase:

2H₂O(g) -> 2H₂O(l)

Each of these reactions is associated with a specific enthalpy change:

CH₄(g) + 2O₂(g) -> CO₂(g) + 2H₂O(g) DH = -802 kJ

2H₂O(g) -> 2H₂O(l) DH = -88 Kj

(Note: under conditions of standard temperature and pressure the liquid state of water is the normal state. Thus, the gas would be expected to condense. This is an exothermic process under these conditions. In a related process, it should get warmer when it rains)

Combining these equations yields the following:

CH₄(g)+2O₂(g)+~~2H₂O(g)~~ -> CO₂(g)+~~2H₂O(g)~~+2H₂O(l)

DH = (-802) kJ + (-88) kJ= -890 kJ

Hess's Law
if a reaction is carried out in a series of steps, DH for the reaction will be equal to the sum of the enthalpy changes for the individual steps

the overall enthalpy change for the process is independent of the number of steps or the particular nature of the path by which the reaction is carried out.

Thus we can use information tabulated for a relatively small number of reactions to calculate DH for a large number of different reactions

Determining DH for the reaction

is difficult because some CO₂ is also typically produced.

However, complete oxidation of either C or CO to yield CO₂ is experimentally pretty easy to do:

We can invert reaction number 2 (making it endothermic) and have CO(g) as a product. (This describes the decomposition of CO₂ to produce CO and O₂)

Thus, we now have two equations with known enthalpies of reaction: the first describes the combustion of carbon and oxygen to produce CO₂ and the second describes how CO₂ can be decomposed to produce carbon monoxide (and oxygen). We can combine these together to describe the production of carbon monoxide from the combustion of carbon and oxygen:

The overall reaction, going from left-hand side reactant(s) to the right-hand side product(s) would be:

We can algebraically subtract the one-half O₂ from both sides to yield the following equation with the associated overall enthalpy:

Another way to look at the method of combining reactions would be as follows:

plus

gives:

DH = (-393.5 kJ) + (283.0 kJ) = -110.5 kJ

canceling out identical compounds from the left and right hand sides of this reaction gives

Carbon occurs in two forms: graphite and diamond. The enthalpy of combustion of graphite is -393.5 kJ, and that of diamond is -395.4 kJ

C(graphite) + O₂(g) -> CO₂(g) DH = -393.5 kJ

C(diamond) + O₂(g) -> CO₂(g) DH = -395.4 kJ

Calculate DH for the conversion of graphite to diamond

What we want is DH for the reaction:

C(graphite) -> C(diamond)

C(graphite) + ~~O₂(g)~~ -> ~~CO₂(g)~~ DH = -393.5 kJ

~~CO₂(g)~~ -> C(diamond) + ~~O₂(g)~~ DH = +395.4 kJ

C(graphite) -> C(diamond) DH = +1.9 kJ

We can never expect to obtain more or less energy from a chemical reaction by changing the method of carrying out the reaction ("conservation of energy").

Another way of saying this is that the particular pathway chosen to arrive at the same reactants yields the same DH for the overall reaction.

Consider the previous example of the combustion of methane to produce gaseous H₂O and then the condensation of the gaseous H₂O to the liquid state. How is this represented in an Enthalpy Diagram?

The key features are the following:

1. Each line represents a set of reactants or products for a balanced chemical reaction. When going from one line to another, the atoms must balance. For example, if we were to ask what is the enthalpy associated with the condensation of water we would have (from the above data):

The CO₂(g) on both sides will cancel to yield:

2. The relative distance of each line must reflect the relative enthalpy difference (DH) between the reactants/products. If the enthalpy change in going from reactants to products is negative, then the line for the products must be below the reactants. Furthermore, the length of the distance must be proportional. For example, the distance reflecting the enthalpy associated with the condensation of water (DH = -88 kJ)is only about 10% as long as the distance between the reactants and products for the combustion of methane to CO₂ and liquid water (DH = -890 kJ)

1996 Michael Blaber