Thermodynamics Formulas — Chemistry Mnemonic
Target: mnemonic for thermodynamics formulas enthalpy entropy Gibbs
Why is this hard to memorize?
Chemical thermodynamics predicts whether a reaction is spontaneous using three state functions: enthalpy (H, heat content), entropy (S, disorder), and Gibbs energy (G = H − TS). The master equation is ΔG = ΔH − TΔS: if ΔG < 0, the reaction is spontaneous. NEET and JEE test: Hess's law (enthalpy is path-independent), bond energy calculations, standard enthalpy of formation, entropy change, and predicting spontaneity at different temperatures. Calculation-heavy chapter — formula recall is critical.
Classic mnemonics you should know
"ΔG = ΔH − TΔS. Negative ΔG = spontaneous. "Go Hiking Through Snow" = G, H, T, S"
ΔH < 0 (exothermic) and ΔS > 0 (disorder increases) → always spontaneous at all T (ΔG always negative). ΔH > 0 and ΔS < 0 → never spontaneous. Mixed signs → depends on temperature. At equilibrium: ΔG = 0 → T = ΔH/ΔS.
"ΔH is path-independent. ΔH(reaction) = Σ ΔHf(products) − Σ ΔHf(reactants)"
Enthalpy change depends only on initial and final states, not the path. You can add enthalpy changes of intermediate steps to get the overall ΔH. This lets you calculate ΔH for reactions that can't be measured directly.
"ΔH = Σ(bonds broken in reactants) − Σ(bonds formed in products). "Breaking costs, forming pays""
Breaking bonds requires energy (positive). Forming bonds releases energy (negative). ΔH = energy needed to break all reactant bonds − energy released forming all product bonds. If more energy is released than consumed → exothermic (ΔH < 0).
The complete list
- ΔG = ΔH − TΔS (Gibbs equation)
- ΔG < 0 → spontaneous
- ΔG = 0 → equilibrium
- Hess's law: ΔH is path-independent
- ΔH = ΣΔHf(products) − ΣΔHf(reactants)
- Bond energy: ΔH = bonds broken − bonds formed
- ΔG° = −RT ln K
- Cp − Cv = R (for ideal gas)
Frequently asked questions
How do I predict if a reaction is spontaneous at a given temperature?
Use ΔG = ΔH − TΔS. Four cases: (1) ΔH<0, ΔS>0 → always spontaneous. (2) ΔH>0, ΔS<0 → never spontaneous. (3) ΔH<0, ΔS<0 → spontaneous at LOW T (below T = ΔH/ΔS). (4) ΔH>0, ΔS>0 → spontaneous at HIGH T (above T = ΔH/ΔS). Exam shortcut: find T(crossover) = ΔH/ΔS, compare with given temperature.
How do I use bond energies to calculate ΔH?
Step 1: Draw structural formulas of reactants and products. Step 2: List ALL bonds in reactants (to be broken) and their bond energies. Step 3: List ALL bonds in products (to be formed). Step 4: ΔH = Σ(bond energies of bonds broken) − Σ(bond energies of bonds formed). Important: this gives an approximate ΔH because bond energies are average values, not specific to the molecule.
What is the relationship between ΔG° and the equilibrium constant K?
ΔG° = −RT ln K. If K > 1: ΔG° < 0 (products favoured at equilibrium). If K < 1: ΔG° > 0 (reactants favoured). If K = 1: ΔG° = 0. At 25°C: ΔG° = −(8.314)(298) ln K = −5706 log K (in J/mol). This equation connects thermodynamics to equilibrium — very important for JEE numericals.
Explore more Chemistry mnemonics
We have 100+ memory tricks for Chemistry — periodic table groups, reactions, exceptions, and more.
Browse all Chemistry mnemonics