2.4: Hess’ law

Measuring the enthalpy change for every possible reaction experimentally would be time-consuming. In addition to the investment of time and equipment, there are also potential chemical complications. What if a reaction does not proceed? Or what if multiple reactions happen concurrently, or the reaction does not proceed reasonably quickly? These concerns would make it more challenging to measure ΔH_rxn experimentally.

To get around these issues, we can exploit the fact that enthalpy is a state function (i.e. path independent) and thus can use any reaction path between the reactants and products to calculate the overall change in enthalpy. This concept is known as Hess' law, which states that the enthalpy change for a reaction will be the same whether it occurs in a single step or multiple steps. So, if our overall chemical reaction of interest can be expressed as a series of reaction steps, then the enthalpy change for the overall reaction is equal to the sum of the enthalpy changes for each step.

For example, consider the generic reaction A → C with an associated enthalpy change ΔH₁. If this reaction can alternately be written as the sum of the generic reactions A → B (ΔH₂) and B → C (ΔH₃), then our overall enthalpy change is equal to the sum of the enthalpy changes of the steps (ΔH₁ = ΔH₂ + ΔH₃), as show in the diagram below.

When applying Hess' law, two other properties of enthalpy are helpful:

1. If the stoichiometric coefficients of a reaction are multiplied by a constant, then the enthalpy change for that reaction is multiplied by the same constant.

   For example, if the enthalpy change for one mole of A reacting with two moles of B to give three moles of C is ΔH₁, then the enthalpy change for two moles of A reacting with four moles of B to give six moles of C is 2ΔH₁.

   

2. When a reaction is reversed, its associated enthalpy change has the same magnitude with the opposite sign.

   For example, if the enthalpy change for one mole of A reacting with two moles of B to give three moles of C is ΔH₁, then the enthalpy change for the reverse reaction of three moles of C forming one mole of A and two moles of B is -ΔH₁.

   

Let's apply Hess' law and the properties of enthalpy to determine the enthalpy change for the formation reaction of propane, C₃H₈, at 298 K:

Graphite does not readily react with hydrogen gas to form propane at 298 K, so we cannot measure the enthalpy change for this reaction directly. Instead, we can use the known enthalpy changes for the reactions below at 298 K to deduce our desired enthalpy change using Hess' law.

The first step is to manipulate the three reactions above by flipping and/or multiplying their coefficients by a constant so that they add together to give our desired overall reaction. Reaction (1) has C₃H_8(g) on the reactant side, but our desired reaction has it on the product side, so we will flip this reaction and, correspondingly, the sign of its enthalpy change. Reaction (2) has C_{(s, graphite)} on the reactant side, as desired, but we must multiply the coefficients and enthalpy by three to get the desired coefficient in the overall reaction. Likewise, we must multiply reaction (3) by four to get the requisite 4 H_2(g) in the overall reaction.

Notice that when these three reactions are arranged in this way, everything cancels out except the reactants and products in our desired propane formation reaction below. Thus, the enthalpy change for this reaction is simply the sum of the enthalpy changes for the steps we added together

This Hess' law method can be used to solve for the enthalpy change of any reaction.

Combining the method above with the formation reactions introduced in section 2.3, we can arrive at a second version of Hess' law. For any reaction we can imagine inserting an additional step where (1) the reactants are broken down into elemental "building blocks" in their standard states (i.e., the reverse of a formation reaction) and then (2) the products are built up from these same elemental building blocks (i.e., a formation reaction). Let's revisit the pair of formation reactions for methanol, where the first reaction forms methanol in the liquid phase, and the second forms methanol in the gas phase, both at P = 1 bar and T = 298 K:

We can use these two reactions to determine ΔH for the evaporation of methanol at T = 298 K and P = 1 bar:

We can flip the first formation reaction (i.e. make the reactants become products, and the products become reactants), which will change the sign of ΔH:

Taking the sum of these two reactions gives an alternate path from CH₃OH_(l) to CH₃OH_(g) (pay attention to stoichiometry to ensure that everything except the desired reactant(s) and product(s) cancel out).

By Hess' law, the total change in enthalpy for the evaporation of methanol is calculated by adding the enthalpy changes for these two steps:

Notice that we flipped the formation reactions of the reactants (and changed the sign of ΔH), and added it with the formation reactions of the products. We can do the same thing with a more complicated example.

The balanced reaction for the combustion of propane is:

We can write the reaction with an additional intermediate step

The intermediate step we added breaks down the reactants into the elements in their standard states, which are then reassembled to form the products. The overall difference in enthalpy between the initial state (reactants) and the final state (products) is not affected. Breaking down the reactants corresponds to the reverse of the formation reaction, and building up the products corresponds to the forward formation reaction. We can use this to write an equation to determine ΔH_rxn for any reaction:

where is the sum of the enthalpies of formation () for each product multiplied by its coefficient, m, in the balanced reaction and is the sum of the enthalpies of formation () for each reactant multiplied by its coefficient, n, in the balanced reaction.

We can use this summation approach to calculate the change in enthalpy for the combustion of propane

Using enthalpies of formation provided below:

the overall change in enthalpy is

The enthalpy of formation for O_{2 (g)} is 0 kJ/mol because it is a pure element in its standard state.

Enthalpy, in units of J or kJ, is an extensive property of the system. The amount of enthalpy (energy) will depend on the system size. ΔH_rxn is usually given in kJ to describe the reaction as written. Doubling the stoichiometric coefficients of the reaction will double the change in enthalpy, if ΔH_rxn is reported in kJ.

Tabulated values of enthalpy changes, such as enthalpies of formation, combustion, and phase transformations, are usually reported as molar enthalpies (an intensive property) with units of kJ/mol. Molar enthalpies (or changes in molar enthalpies) do not depend on system size.

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