Chemistry · Thermodynamics · NEET
No, but formation enthalpy IS one type of reaction enthalpy. Every formation reaction is a reaction, so Δf H is a special case of Δr H. But most reactions are NOT formation reactions. A reaction only counts as a formation when 1 mole of ONE compound is made straight from its elements in their most stable form (like C graphite, H2 gas, O2 gas). So: all formation enthalpies are reaction enthalpies, but not all reaction enthalpies are formation enthalpies.
Because CaCO3 here is made from other COMPOUNDS (CaO and CO2), not from its elements. NCERT gives Δr H = -178.3 kJ/mol for this reaction. For it to be a formation enthalpy, CaCO3 would have to be made from Ca(s) + C(graphite) + O2(g). So this is just a reaction enthalpy, written Δr H, not Δf H. This exact example is in the NCERT Chemistry book, so NEET can test it directly.
No. The reaction is correct (HBr is made from its elements H2 and Br2), but it makes 2 MOLES of HBr, not 1 mole. Formation enthalpy is always PER ONE MOLE of the compound. To get Δf H of HBr, you divide by 2: -72.8 / 2 = -36.4 kJ/mol. This 'per 1 mole' rule is the trap NEET loves. NCERT uses this exact example.
Three conditions must ALL be true: (1) exactly 1 mole of ONE compound is formed as the product; (2) the reactants are all ELEMENTS, not compounds; (3) each element is in its reference (most stable) state at standard conditions, like O2 gas, C graphite, Br2 liquid. If any one condition fails, it is only Δr H, not Δf H.
Because Δf H values are TABULATED (given in data tables), and you use them to calculate any Δr H using the formula Δr H = Σ Δf H(products) - Σ Δf H(reactants). If you wrongly treat a reaction enthalpy as a formation enthalpy (or forget the 'per 1 mole' rule), your Hess's law calculation gives the wrong sign or value. NEET 2025 tested this directly with a BaSO4 crystallisation problem.
Yes. For any element in its most stable (reference) form, Δf H = 0, because making an element 'from itself' involves no change. Example: Δf H of O2(g), C(graphite), and H2(g) is 0. This is why the formula Δr H = ΣΔf H(products) - ΣΔf H(reactants) drops out all pure elements.
The standard heat of formation, in kcal/mol, of Ba2+ is: [Given: standard heat of formation of SO4^2- ion (aq) = -216 kcal/mol; standard heat of crystallisation of BaSO4(s) = -4.5 kcal/mol; standard heat of formation of BaSO4(s) = -349 kcal/mol]
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Enthalpy of formation is written Δf H (the 'f' means formation). Reaction enthalpy is written Δr H (the 'r' means reaction). When measured at standard conditions (1 bar pressure), a small circle is added: Δf H° and Δr H°.
Yes. If forming the compound from its elements absorbs heat (endothermic), Δf H is positive. Example: many oxides of nitrogen have positive Δf H. Exothermic formations (like water, -285.8 kJ/mol) have negative Δf H.
Use Δr H = Σ Δf H(products) - Σ Δf H(reactants), multiplying each Δf H by its number of moles in the balanced equation. Elements in their stable form have Δf H = 0, so they drop out.
Yes, that is the key point. Δf H is defined per exactly 1 mole of the compound. If your equation makes 2 moles, the reaction enthalpy is twice the formation enthalpy, so you must divide by 2 to get Δf H.
