5.3 Hess's Law and Enthalpies of Formation
PRE-HEALTH PREP
Hess's Law vs. Enthalpy of Formation
How to Calculate ΔH
General Chemistry 5.3
Lesson Overview
Hess's Law and the enthalpy of formation method are two approaches for calculating the enthalpy change (ΔH) of a chemical reaction from published thermodynamic data, both grounded in the principle that enthalpy is a state function. Hess's Law involves manipulating known reactions—reversing, scaling, and combining them—until intermediates cancel and the target reaction emerges, while the enthalpy of formation method applies a straightforward products-minus-reactants formula using tabulated ΔH°f values. Coverage also includes the definition and identification of formation reactions, which require exactly one mole of a single product formed from elements in their standard states.
Table of Contents
- ΔH as a State Function
- Hess's Law
- The 5-Step Core Method
- Example 1: 3 NO → N₂O + NO₂
- Exam Strategy: The Quick Method
- Example 2: C₂H₄ + 6 F₂ → 2 CF₄ + 4 HF
- Example 3: Propane Combustion
- Enthalpy of Formation
- The Enthalpy of Formation Formula
- What Is a Formation Reaction?
- Writing Formation Reactions
- Identifying Formation Reactions: 3-Step Filter
- Why the Enthalpy of Formation Method Works
- Key Takeaways
ΔH as a State Function
In the previous lesson on calorimetry, you learned how to measure the heat released by a reaction in the lab and use it to calculate ΔH. This lesson takes a completely different approach: no lab required. Instead, you will learn two methods for calculating ΔH by looking up published thermodynamic data — Hess's Law and the enthalpy of formation method.
Both methods rest on a single foundational principle: ΔH is a state function. A state function depends only on the initial and final states of a system, not on the path taken between them. This means that as long as your reactants and products are the same, it does not matter whether ΔH was measured directly in a calorimeter or calculated by combining a set of intermediate reactions. The answer will be identical either way.
Key Principle: Because ΔH is a state function, any set of reactions that sum to your target reaction will have ΔH values that sum to the target ΔH. This is the logical foundation for both Hess's Law and the enthalpy of formation formula covered in this lesson.
Hess's Law
Hess's Law states that if a series of reactions can be combined to give an overall target reaction, the ΔH values of those individual reactions sum to give the ΔH of the target reaction. In practice, you are given a set of reactions with known ΔH values and must manipulate them — reversing, scaling, or both — until they add up exactly to the reaction you care about.
Fair warning: Hess's Law problems require methodical bookkeeping and most students find them tedious. But there is a clear, systematic procedure that makes them manageable — it is not guesswork. Once you internalize the method, you will also learn a faster shortcut suitable for exam conditions.
The 5-Step Core Method for Hess's Law
Before diving into examples, understand what you can do to any provided reaction without violating Hess's Law:
- Reverse the reaction: flip the direction and change the sign of ΔH.
- Multiply by a scalar: multiply all coefficients and ΔH by the same factor.
- Combine both: reverse and scale simultaneously.
Three provided reactions might look like a small dataset, but each one implicitly represents an infinite family of scaled and reversed variants. Your job is to select exactly the right variant of each provided reaction so that when you add them all together, every intermediate species cancels and you are left with the target reaction.
Follow this 5-step procedure for every Hess's Law problem:
- 1Identify the first species in the target reaction. If it appears in more than one provided reaction, skip it — it will be handled last.
- 2Move to the next species in the target reaction. If it appears in only one provided reaction, decide whether to reverse, scale, or both to place the correct number of moles on the correct side. Lock that reaction in — no other provided reaction contains this species, so it will never change.
- 3Continue through each remaining species the same way.
- 4Return to any species that appeared in multiple reactions. With all other reactions already locked in, determine what the remaining reaction must contribute and adjust accordingly.
- 5Add all ΔH values together to get the target ΔH.
Example 1: 3 NO(g) → N₂O(g) + NO₂(g)
The target reaction is:
Three reactions with known ΔH values are provided as thermodynamic data. The worked solution below demonstrates the full Hess's Law method.
| THERMODYNAMIC DATA | |
|---|---|
| 2NO2(g) ➔ 2NO(g) + O2(g) | ΔH° = +113kJ |
| 2N2(g) + O2(g) ➔ 2N2O(g) | ΔH° = +163kJ |
| N2(g) + O2(g) ➔ 2NO(g) | ΔH° = +181kJ |
Setting Up the Problem
The first species in the target is NO(g), but NO appears in two of the three provided reactions. Per the 5-step procedure, skip it for now.
Move on to N₂O(g). It appears in exactly one provided reaction:
The target requires 1 mole of N₂O on the product side. This reaction has N₂O on the correct side but with 2 moles. Cutting the reaction in half fixes both issues:
Reaction A (½ × original):
N₂(g) + ½ O₂(g) → N₂O(g) ΔH = ½ × 163 = +81.5 kJ
Because N₂O appears in no other provided reaction, this is locked in and will never change.
Handling NO₂
Next, locate NO₂(g). It appears in only one provided reaction:
The target needs 1 mole of NO₂ on the product side. Here, NO₂ is on the reactant side and there are 2 moles — so reverse the reaction and cut it in half:
Reaction B (reversed, ½ × original):
NO(g) + ½ O₂(g) → NO₂(g) ΔH = −½ × 113 = −56.5 kJ
Notice that reversing a reaction changes the sign of ΔH, and scaling it multiplies the magnitude. Both operations apply simultaneously.
Returning to NO
The target needs 3 moles of NO on the reactant side. Reaction B already contributes 1 mole of NO on the reactant side. The only remaining provided reaction must supply the additional 2 moles.
The third provided reaction is:
It has 2 moles of NO, but on the product side. Reversing it puts NO on the reactant side as needed:
Reaction C (reversed):
2 NO(g) → N₂(g) + O₂(g) ΔH = −181 kJ
Verification and Final Answer
Adding Reactions A, B, and C together and canceling species that appear on both sides:
Combining all three reactions:
Reactants: N₂ + ½ O₂ | NO + ½ O₂ | 2 NO → total: N₂ + O₂ + 3 NO
Products: N₂O + NO₂ | N₂ + O₂
Canceling N₂ and O₂ from both sides leaves:
ΔH = +81.5 + (−56.5) + (−181) = −156 kJ
Exam Strategy: The Quick Method for Hess's Law
Writing out every step is essential when learning the method. Once you are proficient, however, you can dramatically speed up the process on an exam by calculating ΔH directly without writing out the full reaction algebra. Here is how that looks for Example 1:
- NO(g): appears in two reactions — skip it.
- N₂O(g): appears in one reaction, on the correct side, but at 2 moles. Cut in half: ΔH contribution = ½ × 163.
- NO₂(g): appears in one reaction, on the wrong side, at 2 moles. Reverse and halve: ΔH contribution = −½ × 113.
- Back to NO(g): after steps 2 and 3, one mole of NO is already on the reactant side. The target needs 3 moles total, so the third reaction must provide 2 more — but NO is on the product side there, so reverse it: ΔH contribution = −181.
Exam Tip: The quick method is reliable only when you have used every provided reaction. If any reaction remains unused, you have made an error somewhere. Under time pressure, compute the answer first, select the matching choice, and verify the full reaction algebra afterward if time permits.
Example 2: C₂H₄(g) + 6 F₂(g) → 2 CF₄(g) + 4 HF(g)
The target reaction is:
| THERMODYNAMIC DATA | |
|---|---|
| H2(g) + F2(g) ➔ 2HF(g) | ΔH° = -537kJ |
| C(s) + 2F2(g) ➔ CF4(g) | ΔH° = -680kJ |
| 2C(s) + 2H2(g) ➔ C2H4(g) | ΔH° = +52.3kJ |
Quick Method
- C₂H₄(g): appears in one reaction (the third), on the product side. Need 1 mole on the reactant side — reverse the whole reaction: ΔH = −52.3 kJ.
- F₂(g): appears in two reactions — skip it.
- CF₄(g): appears in one reaction (the second), on the product side, at 1 mole. Need 2 moles — double the reaction: ΔH = 2 × (−680) kJ.
- HF(g): appears in one reaction (the first), on the product side, at 2 moles. Need 4 moles — double the reaction: ΔH = 2 × (−537) kJ.
With all three reactions used, the F₂ bookkeeping resolves automatically: doubling the first reaction gives 2 mol F₂, and doubling the second gives 4 mol F₂, totaling exactly the 6 mol needed on the reactant side.
Verification
Reaction A (reversed): C₂H₄(g) → 2 C(s) + 2 H₂(g) ΔH = −52.3 kJ
Reaction B (× 2): 2 C(s) + 4 F₂(g) → 2 CF₄(g) ΔH = 2(−680) kJ
Reaction C (× 2): 2 H₂(g) + 2 F₂(g) → 4 HF(g) ΔH = 2(−537) kJ
Summing: C₂H₄ + 2 C(s) + 6 F₂ + 2 H₂ → 2 C(s) + 2 H₂ + 2 CF₄ + 4 HF
Canceling 2 C(s) and 2 H₂ from both sides leaves:
ΔH = −52.3 + (−1360) + (−1074) = −2486.3 kJ
Example 3: Propane Combustion — C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l)
This example is slightly more challenging because it introduces a phase change for water. The target reaction is:
| THERMODYNAMIC DATA | |
|---|---|
| 2C3H8(g) + 7O2(g) ➔ 6CO(g) + 8H2O(g) | ΔH° = -2390kJ |
| CO(g) + ½O2(g) ➔ CO2(g) | ΔH° = -283kJ |
| H2O(l) ➔ H2O(g) | ΔH° = +40.7kJ |
Quick Method
- C₃H₈(g): appears in one reaction (the first), on the reactant side, at 2 moles. Need 1 mole — cut in half: ΔH = ½ × (−2390) kJ.
- O₂(g): appears in two reactions — skip it.
- CO₂(g): appears in one reaction (the second), on the product side, at 1 mole. Need 3 moles — triple the reaction: ΔH = 3 × (−283) kJ.
- H₂O(l): appears in one reaction (the third), on the reactant side. Need 4 moles on the product side — reverse and multiply by 4: ΔH = −4 × (40.7) kJ.
Verification
Reaction A (× ½): C₃H₈(g) + 7/2 O₂(g) → 3 CO(g) + 4 H₂O(g) ΔH = ½(−2390) kJ
Reaction B (× 3): 3 CO(g) + 3/2 O₂(g) → 3 CO₂(g) ΔH = 3(−283) kJ
Reaction C (reversed, × 4): 4 H₂O(g) → 4 H₂O(l) ΔH = −4(40.7) kJ
Intermediate species 3 CO(g) and 4 H₂O(g) cancel between reactions A/B and A/C respectively. The O₂ tallies as: 7/2 from reaction A plus 3/2 from reaction B = 5 mol total on the reactant side.
ΔH = −2206.8 kJ
Tip — Tracking the "skipped" species: When you skip O₂ because it appears in multiple reactions, you can still verify it at the end. In this example, halving reaction A gives 3.5 mol O₂ on the reactant side, and tripling reaction B adds another 1.5 mol — totaling exactly 5 mol O₂ as required. This cross-check can confirm your work, but it is easier to handle it after all other species are resolved.
Enthalpy of Formation: The Faster Alternative
Now for the good news. The enthalpy of formation method accomplishes the same goal as Hess's Law — calculating ΔH of a reaction from tabulated data — but with a fraction of the effort. You do not need to manipulate or combine reactions at all. You simply look up standard enthalpy of formation (ΔH°f) values and plug them into a formula.
Think of it like your digestive system: you use it every day without consciously directing each enzymatic step. You can apply the enthalpy of formation formula effectively without immediately understanding why it works. That said, understanding the underlying logic — covered at the end of this section — will solidify your conceptual grasp of thermochemistry.
The Enthalpy of Formation Formula
The standard enthalpy of reaction is calculated using the products-minus-reactants formula:
where n and m are the stoichiometric coefficients (moles) of each product and reactant, respectively. The standard enthalpy of formation (ΔH°f) for each species is looked up in an appendix or data table.
Example: 3 NO(g) → N₂O(g) + NO₂(g)
Given ΔH°f values:
| ΔHf° (kJ/mol) | |
|---|---|
| NO(g) | 90.4 |
| N2O(g) | 81.6 |
| NO2(g) | 33.8 |
Products: (1 mol × 81.6 kJ/mol) + (1 mol × 33.8 kJ/mol) = 115.4 kJ
Reactants: 3 mol × 90.4 kJ/mol = 271.2 kJ
ΔH°rxn = 115.4 − 271.2 = −155.8 kJ
This matches the Hess's Law result of −156 kJ to within rounding — exactly as expected for a state function.
Common Mistake — Sign error with negative reactant sums: After calculating the sum of reactant enthalpies, students sometimes forget to subtract a negative value. If Σ(m · ΔH°f reactants) is negative, subtracting it means adding a positive number. Carry the sign carefully through the arithmetic.
Know which method to use: If the problem gives you standard enthalpies of formation and asks for ΔHrxn, use the products-minus-reactants formula. If it gives you a set of complete reactions with known ΔH values, use Hess's Law. The data provided tells you immediately which tool to reach for.
What Is a Formation Reaction? 2 Defining Criteria
A standard enthalpy of formation (ΔH°f) is the enthalpy change for a specific type of reaction called a formation reaction. Every formation reaction satisfies exactly two criteria:
- It forms exactly one mole of a single product. No more, no fewer. Coefficients greater than one on the product side are not permitted.
- All reactants are elements in their standard states. No compounds may appear on the reactant side — only pure elements in the physical form they adopt at 25 °C and 1 atm.
When you look up the enthalpy of formation of a compound in a reference table, you are looking up ΔH for exactly this type of reaction written for that compound.
Standard States of Common Elements
Knowing the standard state of each element is essential for writing formation reactions correctly. Many periodic tables color-code this information: solids in black, liquids (mercury and bromine) in blue, and gases in red. The diatomic elements — H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂ — appear as diatomic molecules in their standard states.
| STANDARD STATES | |
|---|---|
| Gases | He, Ne, Ar, Kr, Xe, Rn, H2, N2, O2, F2, Cl2 |
| Liquids | Br2, Hg |
| Solids | Everything Else |
| Special | C(graphite), S8 |
Special case — Carbon: Carbon exists in multiple solid allotropes (different structural forms of the same element in the same phase). The standard state of carbon is graphite, not diamond, because graphite is the lower-energy form. Formation reactions involving carbon must use solid graphite as the reactant, not generic "C(s)".
Writing Formation Reactions
Example: Formation reaction for NaCl(s)
Identify the elements: sodium (Na) and chlorine (Cl).
Standard states: Na is a solid metal; Cl is a diatomic gas, Cl₂(g).
Write one mole of product and balance:
Note that you cannot double through to eliminate the fraction — doing so would produce 2 mol NaCl, violating the one-mole-of-product requirement. The fractional coefficient is mandatory here.
Example: Formation reaction for NaHCO₃(s) (sodium bicarbonate)
Elements present: Na, H, C, O.
Standard states: Na(s), H₂(g), C(s, graphite), O₂(g).
Unbalanced skeleton:
Balanced formation reaction:
Again, doubling to clear fractions is not allowed — it would place 2 mol of NaHCO₃ on the product side. Fractional coefficients on the reactant side are perfectly acceptable in formation reactions.
Identifying Formation Reactions on Exams: 3-Step Filter
A common exam question presents several reactions and asks which one is a valid formation reaction. Use this 3-step filter to eliminate wrong answers quickly:
- Check the products first. Does the reaction form exactly one product at exactly one mole? If there are multiple products, or more than one mole of the product, it is not a formation reaction.
- Check the reactants for compounds. If any reactant is a compound rather than a pure element, it is not a formation reaction.
- Verify standard states. All elements must be present in their standard-state form (correct phase and allotrope).
Quick filter: Eliminate immediately any reaction with more than one product or any compound on the reactant side. What remains can be checked for standard states. This approach is much faster than analyzing every reaction from scratch.
Why the Enthalpy of Formation Method Works
The enthalpy of formation method is not a separate principle — it is a specific, elegant application of Hess's Law. To see why, consider the following imaginary 2-step pathway for any reaction A → B:
- Decompose each reactant into its constituent elements in their standard states. This is the reverse of each reactant's formation reaction, so each step contributes −ΔH°f(reactant).
- Recombine those elements to form each product. This is precisely the formation reaction for each product, contributing +ΔH°f(product).
This 2-step detour passes through the same initial state (your reactants) and arrives at the same final state (your products). Because ΔH is a state function, the overall ΔH is identical to that of the direct reaction — regardless of how contrived the intermediate pathway is.
Analogy: Imagine you need to cut a small hole in a wall for an air-conditioning unit. One approach would be to demolish the entire room down to rubble and rebuild it with the hole already in place. That is a roundabout path, but if you end up with the same room plus a hole, the net change is identical. The enthalpy of formation method does exactly this on a molecular level: it "demolishes" reactants into elements, then "rebuilds" products from those elements.
This is why the standard enthalpy of reaction formula takes the form products minus reactants: you are summing the formation enthalpies of everything you are building (products) and subtracting the formation enthalpies of everything you are "unbuilding" (reactants, via the reverse process).
Key Takeaways
- ΔH is a state function: it depends only on the initial and final states, not the path. This principle justifies both Hess's Law and the enthalpy of formation method.
- Hess's Law lets you calculate ΔH by combining known reactions. You may reverse any reaction (changing the sign of ΔH) or multiply it by a scalar (scaling ΔH by the same factor) to make intermediates cancel and yield your target reaction.
- The 5-step systematic procedure for Hess's Law: process each species in the target reaction one at a time, skipping any that appear in more than one provided reaction until all others are resolved.
- The enthalpy of formation formula — ΔH°rxn = Σ(n · ΔH°f products) − Σ(m · ΔH°f reactants) — is the faster approach when standard enthalpies of formation are tabulated. Multiply each ΔH°f by the stoichiometric coefficient, sum the products, sum the reactants, and subtract.
- A formation reaction forms exactly one mole of a single product from elements in their standard states. No compounds are permitted on the reactant side, and fractional coefficients on the reactant side are acceptable.
- Carbon's standard state is graphite, not diamond. Use C(s, graphite) in any formation reaction involving carbon as a reactant.
- The enthalpy of formation method is a special case of Hess's Law: it imagines decomposing reactants into elements and reforming them into products, and works because ΔH is path-independent.