Chemistry · Thermodynamics · NEET
ΔfH° is the heat change when ONE mole of a compound forms from its elements, with every element in its most stable form at 25°C and 1 bar. Example: C(graphite) + O2(g) → CO2(g), ΔfH° = -393.5 kJ/mol. The word 'formation' always means 'built from elements'.
To 'form' an element like O2, you would write O2(g) → O2(g). Nothing changes, no bonds break or form, so the heat change is zero. NCERT states it directly: by convention, ΔfH° of an element in its reference state (most stable form) is taken as zero. It is a chosen reference point, like sea-level for measuring height.
Only for the MOST STABLE form (the reference state). For carbon, C(graphite) has ΔfH° = 0, but C(diamond) has ΔfH° = +1.90 kJ/mol, because graphite is more stable than diamond. Diamond is NOT the reference state, so its formation enthalpy is not zero.
It is the most stable state of that element at 25°C and 1 bar pressure. Examples: hydrogen is H2 gas, oxygen is O2 gas, carbon is C(graphite), sulphur is S(rhombic). Each of these has ΔfH° = 0. Wrong forms like atomic O or O3 (ozone) are not the reference state, so they are not zero.
Absolute enthalpy (H) of a substance cannot be measured, only changes (ΔH). So we pick a common zero level: elements in their stable form. Then every compound's enthalpy is measured relative to that. This lets us calculate any reaction enthalpy using ΔrH° = Σ ΔfH°(products) − Σ ΔfH°(reactants).
No. That is NOT a formation reaction of O2; it is breaking O2 into oxygen atoms (bond dissociation / atomization), which needs energy and is positive. Formation of O2 means O2 → O2 (nothing), which is zero. Do not confuse forming an element with breaking it into atoms.
The standard heat of formation (in kcal/mol) of Ba²⁺ is: [Given: ΔfH° of SO4²⁻(aq) = −216 kcal/mol; standard heat of crystallisation of BaSO4(s) = −4.5 kcal/mol; ΔfH° of BaSO4(s) = −349 kcal/mol]
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Only in its most stable form (reference state) at 25°C and 1 bar. H2, O2, N2, C(graphite), S(rhombic) all have ΔfH° = 0. Less stable forms like diamond, ozone, or atomic oxygen are not zero.
Graphite is the more stable form of carbon at standard conditions, so it is chosen as the reference and set to zero. Diamond is slightly higher in energy, so forming diamond from graphite absorbs a little heat, giving a positive value.
No. ΔfH°(O2, g) = 0 because O2 is the stable element form. Making O atoms (O2 → 2O) is bond breaking and is strongly positive (about +249 kJ/mol per O atom), not zero.
Because absolute enthalpy cannot be measured. We only measure changes. Choosing elements as the zero baseline gives every compound a consistent reference, so reaction enthalpies can be calculated by subtraction.
Yes. When you use ΔrH° = Σ ΔfH°(products) − Σ ΔfH°(reactants), any element in stable form contributes 0. Forgetting this (or wrongly giving diamond/ozone zero) is a common NEET mistake.
