Introduction 

Hess’s Law. Source: Boundless. CC-BY-SA 3.0.

Hess’ Law states that a reaction will have the same enthalpy change regardless of whether the reaction occurs in a single step or in multiple. This is beneficial in that we can use step reactions to find the enthalpy of a target reaction. Consider the image to the right. The energy change from A → D is the same as the sum of the changes from A → B, B → C, and C→ D. This concept is occasionally used in industry to allow a reaction that is highly endothermic or slow to occur through stepwise reactions that are easier (or cheaper) to manipulate.

When working a Hess’ Law problem, you need to follow two main rules:

1. If you reverse the direction of a reaction, the sign of the ∆H changes.
2. If you multiply or divide the coefficients of a reaction, you must also multiply or divide the ∆H value.

These rules are applied to a set of steps in solving Hess’ Law problems.

1. Evaluate the Target Reaction. This reaction is the one you want to solve for and will be what you manipulate the step reactions for.

2. Look over the Step Reactions. These are the reactions you want to eventually “add up” to get the target. You may wish to highlight or underline substances that are specifically in the Target Reaction. You may also wish you indicate substances that appear only one time throughout the Step Reactions. These will not be cancelled and will help you line up your reactions faster.

3. Line up the Step Reactions. You will need to rewrite the step reactions so that the reactants in the target reaction are listed on the left and products in the target reaction are on the right of the arrow. It is helpful if the arrows for all Step Reactions line up. Also be aware that some reactants or products may appear more than once and they will cancel (similar to spectator ions) when adding the reactions.

4. Manipulate the Step Reactions. Once your Step Reactions are lined up, you may need to multiply or divide the reactions to guarantee the correct number of each substance from the target reaction is present.

5. Add the Step Reactions. This will make sure that the sum of the step reactions will give the target reaction.

6. Add the Enthalpies. Once you have the ensured the Step Reactions add together to give the Target Reaction, Add the enthalpies of the reactions to give the enthalpy of the target reaction.

Consider the following example where we want to calculate the ΔH values for the following reaction given the subsequent steps and enthalpy values.

1. Target Reaction:

C(s graphite) → C(s diamond)            ΔH = ?

2. Step Reactions:

C(s graphite) → CO₂ (g) +O₂ (g)                  ΔH = -393.41 kJ

2C(s diamond) → 2CO₂ (g) + 2O₂ (g)          ΔH = -790.80 kJ

We know from the Target Reaction we want graphite on the left. It is on the left in the first Step Reaction, so we will probably leave this reaction alone and rewrite it as is. We also know that we want diamond on the right so we will need to reverse the second reaction to get diamond to the product side. We will need to reverse the sign of the second reaction. Lining up the Step Reactions gives:

C(s graphite)               →  CO₂ (g) +O₂ (g)       ΔH = -393.41 kJ

2CO₂ (g) + 2O₂ (g)      →  2C(s diamond)        ΔH = +790.80 kJ

We can tell that we will have too many diamond particles by the coefficients in the second Step Reaction. We will also not be able to cancel the carbon dioxide or oxygen because the coefficients are not the same in the first and second Target Reaction. Therefore we need to divide the WHOLE second reaction AND the enthalpy value by 2:

C(s graphite)                  → CO₂ (g) +O₂ (g)      ΔH = -393.41 kJ

1CO₂ (g) + 1O2 (g)    → 1C(s diamond)        ΔH = 395.4 kJ

Adding these reactions gives:

C(s graphite) + CO₂ (g) +O₂ (g) → CO₂ (g) +O₂ (g) + C(s diamond)

Notice there are equal numbers of carbon dioxide and oxygen on the left AND right of the equation. Any substance on both sides will cancel. This gives:

C(s graphite) → C(s diamond)

Which is our Target Reaction. Therefore we can add the adjusted enthalpies to get the enthalpy of the target reaction.

-393.41 kJ + 395.4 kJ =1.99 kJ