By John Brennan – Updated Aug 30, 2022

Knowing whether a reaction will occur on its own is a cornerstone of chemistry. A reaction that proceeds without external energy input is termed thermodynamically spontaneous. The key indicator of spontaneity is the standard Gibbs free energy change, ΔG°, which compares the free energy of products and reactants in their standard states. A negative ΔG° signals a spontaneous reaction as written; a positive value indicates that the reaction is non‑spontaneous under the conditions considered.

Step 1 – Write the Balanced Equation

Begin by writing a complete, balanced chemical equation for the reaction. If you need a refresher on how to do this, consult the introductory resource linked below. For example, the combustion of methane is written as:

CH₄ + 2 O₂ → CO₂ + 2 H₂O

Step 2 – Retrieve Thermodynamic Data

Open the NIST Chemical WebBook (a trusted, authoritative database). Search for each species in your equation to obtain the standard enthalpy of formation, ΔfH°, and the standard molar entropy, S°, values. These are typically listed in kJ mol⁻ⁱ and J mol⁻ⁱ K⁻ⁱ, respectively.

Step 3 – Calculate Standard Enthalpy of Reaction

Sum the ΔfH° values of all products, then sum those of all reactants. Subtract the reactants’ total from the products’ total to obtain the standard enthalpy change, ΔH°:

ΔH° = ΣΔfH°(products) – ΣΔfH°(reactants)

Example for methane combustion:

• ΔfH°(CH₄) = –74.5 kJ mol⁻ⁱ
• ΔfH°(CO₂) = –393.5 kJ mol⁻ⁱ
• ΔfH°(H₂O, l) = –285.8 kJ mol⁻ⁱ
• ΔfH°(O₂, g) = 0 (by definition)

Products: –393.5 + 2(–285.8) = –965.1 kJ mol⁻ⁱ
Reactants: –74.5 kJ mol⁻ⁱ
ΔH° = –965.1 – (–74.5) = –890.6 kJ mol⁻ⁱ

Step 4 – Calculate Standard Entropy Change

Sum the S° values of products and reactants separately, then subtract reactants from products to find ΔS°:

ΔS° = ΣS°(products) – ΣS°(reactants)

Example values:

• S°(CH₄) = 186.25 J mol⁻ⁱ K⁻ⁱ
• S°(CO₂) = 213.79 J mol⁻ⁱ K⁻ⁱ
• S°(H₂O, l) = 69.95 J mol⁻ⁱ K⁻ⁱ
• S°(O₂, g) = 205.15 J mol⁻ⁱ K⁻ⁱ

Reactants: 186.25 + 2(205.15) = 596.55 J mol⁻ⁱ K⁻ⁱ
Products: 2(69.95) + 213.79 = 353.69 J mol⁻ⁱ K⁻ⁱ
ΔS° = 353.69 – 596.55 = –242.86 J mol⁻ⁱ K⁻ⁱ

Step 5 – Convert Entropy Change to kJ mol⁻ⁱ

Multiply ΔS° by the absolute temperature (298.15 K for room temperature) and divide by 1000 to align units with ΔH°:

(–242.86 J mol⁻ⁱ K⁻ⁱ) × 298.15 K ÷ 1000 = –72.41 kJ mol⁻ⁱ

Step 6 – Compute Standard Gibbs Free Energy

Subtract the temperature‑scaled entropy term from the enthalpy term:

ΔG° = ΔH° – TΔS° = (–890.6 kJ mol⁻ⁱ) – (–72.41 kJ mol⁻ⁱ) = –818.2 kJ mol⁻ⁱ

A negative ΔG° confirms that the methane combustion reaction is thermodynamically spontaneous at 298.15 K.

What You’ll Need

• Pencil and paper (or a digital spreadsheet)
• Scientific calculator or computational software
• Access to reliable thermodynamic tables (e.g., NIST Chemical WebBook)

References

• Atkins, P., et al. Chemical Principles: The Quest for Insight. 2008.
• Vollhardt, P., et al. Organic Chemistry, Structure and Function. 2011.