18.4 ΔG0rxn, ΔH0rxn, ΔS0rxn, and Formation Reactions
PRE-HEALTH PREP
Calculating ΔG, ΔH, & ΔS with Formation Reactions and Absolute Entropy
General Chemistry 18.4
Standard thermodynamic quantities ΔG°, ΔH°, and ΔS° for a reaction are calculated using the products-minus-reactants formula, where ΔG° and ΔH° use formation values from thermodynamic tables while ΔS° uses absolute molar entropies grounded in the Third Law of Thermodynamics. A formation reaction is defined as one that produces exactly one mole of a single compound from elements in their standard states, requiring knowledge of the seven diatomic elements, carbon's graphite standard state, and the use of fractional coefficients. Combining ΔH° and ΔS° in the Gibbs free energy equation ΔG° = ΔH° − TΔS° requires careful unit conversion, as standard enthalpy of formation values are reported in kilojoules while absolute molar entropies are reported in joules per kelvin.
Why Entropy Uses Absolute Values (But ΔG and ΔH Don't)
Your textbook appendix contains a table of thermodynamic quantities — typically ΔG° of formation, ΔH° of formation, and standard molar entropies — and these three quantities will be your primary tools for calculating ΔG, ΔH, and ΔS in this lesson. Before diving into the calculations, it's worth understanding why entropy is handled differently from Gibbs free energy and enthalpy.
To calculate the entropy change of a reaction, you simply use absolute entropy values directly: products minus reactants. Think of it like tracking your net worth at a casino. If you walk in with $100 and walk out with $500, your net worth increased by $400 — computed as final state minus initial state: $500 − $100 = +$400. In the same way, because we know the exact amount of entropy each substance possesses, subtracting the reactants' total entropy from the products' total entropy gives the entropy change of the reaction directly.
The situation is fundamentally different for Gibbs free energy and enthalpy. We can never measure the absolute amount of G or H a substance contains — only changes in those quantities. Imagine being told someone lost 50 pounds on one diet, gained back 60, lost 30 on another, and so on. Without knowing their starting weight, you can never determine their current weight. Gibbs free energy and enthalpy are in exactly this position: without an absolute reference point, we cannot assign them absolute values.
The Third Law of Thermodynamics Provides an Entropy Reference Point
Entropy escapes this problem because of the Third Law of Thermodynamics: a perfect crystal at absolute zero (0 K) has zero entropy. This single reference point allows us to calculate the absolute entropy of any substance under any conditions by integrating the heat capacity from 0 K upward. Because that reference point exists, every substance in your appendix table carries an absolute entropy value — not merely a relative one.
Key Distinction: Absolute molar entropies are never zero at 25 °C (298 K), even for pure elements in their standard states. Zero entropy is only achieved at 0 K for a perfect crystal. This is why the entropy column in your thermodynamic data table always contains nonzero values, unlike the ΔH° and ΔG° of formation columns where elements in their standard state are defined as zero.
How to Use the Products-Minus-Reactants Formula
For all three thermodynamic quantities — ΔG°, ΔH°, and ΔS° — the calculation follows the same fundamental pattern. Whether you are calculating Gibbs free energy, the standard enthalpy of formation, or standard molar entropy changes, the formula is:
The General Formula
ΔX°rxn = Σ n·ΔX°f(products) − Σ m·ΔX°f(reactants)
where n and m are the stoichiometric coefficients of the products and reactants, respectively, and X represents G, H, or S.
For ΔG and ΔH, the values in the table are formation values — they represent the free energy or enthalpy change when one mole of that compound is formed from its elements in their standard states. For ΔS, the values are absolute molar entropies, but the products-minus-reactants arithmetic works identically.
Critical Rule — Elements in Their Standard States: The ΔG° and ΔH° of formation for any element in its standard state are zero by definition. You will not always find these values listed in a table — you are expected to know they are zero. The absolute entropy of an element in its standard state, however, is not zero (only 0 K gives zero entropy), so the entropy column will always show a nonzero value.
Worked Example: How to Calculate ΔG°, ΔH°, and ΔS° Step by Step
Consider the following reaction, with thermodynamic data provided in the table below:
Reaction: PCl₃(g) + 3 HCl(g) → Cl₂(g) + PH₃(g) (not balanced as written in the lesson — coefficients below are from the source data)
The reaction used in the calculations is: PCl₃(g) + 3 HCl(g) → 3 Cl₂(g) + PH₃(g)
The thermodynamic data used are:
| Species | ΔG°f (kJ/mol) | ΔH°f (kJ/mol) | S° (J/mol·K) |
|---|---|---|---|
| Cl₂(g) | 0 | 0 | 223.0 |
| PH₃(g) | +5.4 | +13.4 | 210.2 |
| PCl₃(g) | −288.1 | −269.6 | 311.7 |
| HCl(g) | −92.3 | −95.3 | 186.7 |
Calculating ΔG° of Reaction
Using the ΔG° of formation column and applying products minus reactants:
Calculation: ΔG°rxn
ΔG°rxn = [3(0) + 5.4] − [(−288.1) + 3(−92.3)]
ΔG°rxn = 5.4 − (−288.1 − 276.9)
ΔG°rxn = 5.4 − (−565.0)
ΔG°rxn = +570.4 kJ
Sign Trap: Students frequently make sign errors at this step. When you subtract a negative sum, you end up adding. Always enter the full expression into your calculator with explicit parentheses around the reactant sum — let the calculator handle the arithmetic rather than trying to manage the signs mentally.
Calculating ΔH° of Reaction
Repeat the same procedure using the standard enthalpy of formation column:
Calculation: ΔH°rxn
ΔH°rxn = [3(0) + 13.4] − [(−269.6) + 3(−95.3)]
ΔH°rxn = 13.4 − (−269.6 − 285.9)
ΔH°rxn = 13.4 − (−555.5)
ΔH°rxn = +568.9 kJ
Calculating ΔS° of Reaction
Now use the absolute molar entropy column. Note that Cl₂(g) now contributes 223.0 J/mol·K rather than zero — absolute entropies of elements in their standard states are never zero.
Calculation: ΔS°rxn
ΔS°rxn = [3(223.0) + 210.2] − [311.7 + 3(186.7)]
ΔS°rxn = [669.0 + 210.2] − [311.7 + 560.1]
ΔS°rxn = 879.2 − 871.8
ΔS°rxn = +7.4 J/K
Units: ΔG° and ΔH° from formation values are reported in kilojoules (kJ), while ΔS° from absolute molar entropies is reported in joules per kelvin (J/K). These are different units — keep this distinction in mind whenever you combine ΔH and ΔS in the same equation.
Units Warning: Calculating ΔG° via the Gibbs Free Energy Equation
Occasionally, an exam will provide ΔH° of formation values and absolute molar entropies but omit the ΔG° of formation column — and still ask you to calculate ΔG° at a specific temperature (commonly 298 K). The approach is to calculate ΔH° and ΔS° from the table as shown above, then plug those values into the Gibbs free energy equation:
Gibbs Free Energy Equation
ΔG° = ΔH° − TΔS°
where T is the temperature in Kelvin.
The critical trap here is a unit mismatch. Your ΔH° will be in kilojoules, while your ΔS° will be in joules per kelvin. You must convert one to match the other before substituting into the Gibbs free energy equation.
Unit Conversion Strategy: If the answer choices on a multiple-choice exam are in kilojoules, convert ΔS° from J/K to kJ/K before plugging in. For the example above, ΔS° = 7.4 J/K becomes 0.0074 kJ/K. If the answers are in joules, convert ΔH° instead. Mixing units in this equation is one of the most common errors on thermodynamics exams.
What Is a Formation Reaction?
A formation reaction (also called a standard molar formation reaction) is defined by two conditions:
- It produces exactly one mole of a single compound as the sole product.
- All reactants are elements in their standard states — no compounds are permitted on the reactant side.
The standard enthalpy of formation (ΔH°f) and standard Gibbs free energy of formation (ΔG°f) for a compound are simply the ΔH° or ΔG° of that compound's formation reaction. This construction is deliberately designed so that the products-minus-reactants formula recovers the correct ΔG° or ΔH° for any overall reaction — you do not need to understand the derivation to apply the method, but recognizing formation reactions is a separate exam skill you will also need.
Consider HCl(g) as an example. Its formation reaction produces exactly one mole of HCl(g), and the only reactants permitted are the elements hydrogen and chlorine in their standard states.
Why Elements in Their Standard States Have Zero ΔG° and ΔH° of Formation: A formation reaction for an element in its standard state would have identical reactants and products — the "reaction" does nothing. A reaction that accomplishes nothing has ΔG = 0 and ΔH = 0 by definition. This is why you will always fill in zero for those entries in any thermodynamic table.
Standard States and the 7 Diatomic Elements You Must Know
Writing a formation reaction correctly requires knowing the standard state of every element involved. The standard state of an element is its most stable form at 25 °C (298 K) and 1 atm pressure. In practice, most elements fall into one of three categories:
- Solids: The vast majority of metals and many nonmetals (e.g., sodium, iron, phosphorus, sulfur, carbon).
- Liquids: Only 2 elements are liquid at standard conditions — Hg(l) (mercury) and Br₂(l) (bromine). Some tables also list gallium, cesium, and francium as near-room-temperature liquids, but mercury and bromine are the ones you need to know.
- Gases: A handful of nonmetals, including the noble gases and the seven diatomic elements.
The 7 Diatomic Elements
Seven elements exist as diatomic molecules in their standard states. You must know all 7 — writing a formation reaction with H instead of H₂, for instance, would make the reaction invalid.
The 7 Diatomic Elements: H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂
Memory Aids:
- Six of the seven — N, O, F, Cl, Br, I — trace the shape of a 7 on the periodic table. Hydrogen is the seventh.
- Mnemonic: "Never Have Fear Of Ice Cold Beer" (N, H, F, O, I, Cl, Br).
State Matters: An element must be in its correct state to qualify for a formation reaction. Cl₂(g) is the standard state for chlorine — Cl(g), Cl₂(l), and Cl₂(s) are all non-standard and would carry nonzero ΔG° and ΔH° of formation values.
Special Case: Carbon's Standard State Is Graphite
Carbon deserves special attention because it exists in multiple solid allotropes: diamond, graphite, and buckminsterfullerene (C₆₀). Simply writing "C(s)" is technically ambiguous. The thermodynamically stable form at standard conditions is graphite — it has the lowest free energy of the three — so the standard state of carbon is specifically graphite.
Exam Precision: If a formation reaction shows C(diamond) as a reactant, it is not a formation reaction — diamond is not the standard state. Write C(graphite) whenever carbon appears as a reactant in a formation reaction. Some professors will accept "C(s)" without specification, but writing "C(graphite)" is always unambiguously correct.
Sulfur also has multiple solid allotropes (rhombic and monoclinic), and the standard state is rhombic sulfur. This is less commonly tested than carbon, but worth noting.
Fractions Are Required in Formation Reactions
Because a formation reaction must produce exactly one mole of a single product, you will frequently end up with fractional stoichiometric coefficients on the reactant side — and that is perfectly acceptable.
For HCl(g): hydrogen and chlorine are both diatomic gases in their standard states. Balancing the formation of exactly 1 mol of HCl requires:
Formation Reaction for HCl(g)
½ H₂(g) + ½ Cl₂(g) → HCl(g)
ΔG°f = −92.3 kJ/mol ΔH°f = −95.3 kJ/mol
Do Not Double Through: It is tempting to multiply through by 2 to eliminate the fractions, giving H₂(g) + Cl₂(g) → 2 HCl(g). That is a valid balanced equation, but it is not a formation reaction because it produces two moles of product. Fractions (or their decimal equivalents, such as 0.5) are mandatory whenever diatomics are involved. The ΔG° and ΔH° of formation values in tables always correspond to the one-mole-of-product reaction.
Worked Example: Formation Reaction for Sodium Bicarbonate
Write the standard formation reaction for solid sodium bicarbonate, NaHCO₃(s).
Step 1 — Identify the product and its quantity. A formation reaction produces exactly one mole of the target compound as the only product:
→ NaHCO₃(s) (1 mol)
Step 2 — List the constituent elements. Sodium bicarbonate contains Na, H, C, and O.
Step 3 — Write each element in its standard state.
- Sodium: Na(s) — solid metal
- Hydrogen: H₂(g) — diatomic gas
- Carbon: C(graphite) — most stable solid allotrope
- Oxygen: O₂(g) — diatomic gas
Step 4 — Balance with fractional coefficients as needed. NaHCO₃ contains 1 Na, 1 H, 1 C, and 3 O atoms:
Formation Reaction for NaHCO₃(s)
Na(s) + ½ H₂(g) + C(graphite) + 3/2 O₂(g) → NaHCO₃(s)
Equivalently: Na(s) + 0.5 H₂(g) + C(graphite) + 1.5 O₂(g) → NaHCO₃(s)
Both representations are correct. The fractional coefficients cannot be avoided without violating the definition of a formation reaction.
How to Identify a Formation Reaction on an Exam: 3 Criteria
When an exam presents several reactions and asks which one is a formation reaction, use this 3-step elimination sequence:
- 1Check the product side. Any reaction that produces more than one type of product, or more than one mole of a single product, is not a formation reaction. Eliminate it.
- 2Check that all reactants are elements. If any reactant is a compound rather than a pure element, it cannot be a formation reaction. Eliminate it.
- 3Verify standard states. Elements must appear in their correct standard-state forms. Eliminate reactions showing C(diamond) instead of C(graphite), Cl(g) instead of Cl₂(g), H₂(l) instead of H₂(g), and so on.
Process of Elimination: On a four-option multiple-choice question, systematically eliminating wrong answers is faster and more reliable than trying to construct the correct formation reaction from scratch and then matching it to an option. Work through the 3 criteria above in order — you should arrive at exactly one surviving option.
Quick Reference — Common Disqualifiers
- Two or more products → not a formation reaction
- Coefficient > 1 on the product → not a formation reaction
- Any compound on the reactant side → not a formation reaction
- C(diamond) as reactant → not a formation reaction
- Cl(g) or H(g) (monatomic) as reactant → not a formation reaction
- Br₂(g) as reactant (bromine should be liquid) → not a formation reaction
8 Key Takeaways
- ΔG°, ΔH°, and ΔS° of a reaction are all calculated using the same products-minus-reactants formula: multiply each species' tabulated value by its stoichiometric coefficient, sum the products, then subtract the sum of the reactants.
- The ΔG° and ΔH° of formation for any element in its standard state are zero by definition; the absolute molar entropy of an element in its standard state is never zero.
- ΔG° and ΔH° values from formation tables are in kilojoules (kJ), while ΔS° values from absolute entropy tables are in joules per kelvin (J/K). Convert to consistent units before using the Gibbs free energy equation ΔG° = ΔH° − TΔS°.
- A formation reaction produces exactly one mole of a single compound from elements in their standard states only — no compounds allowed on the reactant side.
- The 7 diatomic elements (H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂) must appear in their diatomic form in formation reactions; using monatomic versions disqualifies the reaction.
- The standard state of carbon is graphite, not diamond. Writing C(diamond) in a formation reaction makes it invalid.
- Fractional coefficients are mandatory in formation reactions whenever diatomics are involved — you cannot multiply through to remove them without violating the one-mole-of-product requirement.
- To identify a formation reaction on an exam, eliminate options in order using 3 criteria: multiple products → compound reactants → non-standard states. The surviving option is your answer.