Chemistry · Thermodynamics · NEET
It is the heat released or absorbed when exactly 1 mole of a compound forms from its elements, with each element in its most stable natural form. 'Standard' means we fix the conditions: pressure 1 bar and (usually) temperature 298 K. Example: for CO2, it is the ΔH of C(graphite) + O2(g) → CO2(g), which is −393.5 kJ per mole. The '1 mole of product' part is a rule: the equation must make one mole of the compound, even if you need fractions like 1/2 O2.
Because forming an element from itself needs no change at all. ΔfH° is measured from elements in their most stable state, so an element in that stable state is already the starting point. O2(g), H2(g), N2(g), and C(graphite) are all the most stable forms, so their ΔfH° = 0 by definition. Careful trap: C(graphite) is zero, but C(diamond) is NOT zero (+1.9 kJ/mol), because diamond is not the most stable form. Same for O3 (ozone), which is not zero.
Use the formula ΔrH° = Σ ΔfH°(products) − Σ ΔfH°(reactants). Multiply each ΔfH° by the number of moles (the coefficient) in the balanced equation, add them for products, add them for reactants, then subtract. Always products first. If you flip the order you get the wrong sign, which is the most common mistake in NEET.
Enthalpy of formation (ΔfH°) is a special case: it is only for making 1 mole of ONE compound from its elements. Enthalpy of reaction (ΔrH°) is the heat change for ANY balanced reaction. You use the formation values as building blocks to calculate the reaction value. So formation is the ingredient, reaction is the final dish.
Because ΔfH° is defined 'per mole of compound formed'. If your equation makes 2 moles, the ΔH you get is twice the formation enthalpy. That is why formation equations often use fractions like 1/2 N2 + 3/2 H2 → NH3. NEET questions test this: if a reaction makes 2 mol of water and releases 483.64 kJ, then ΔfH° of 1 mole of water is half, −241.82 kJ/mol.
Usually yes. A negative ΔfH° means energy was released when the compound formed, so it sits at lower energy than its separate elements. Lower energy generally means more stable. A large negative value (like CO2 at −393.5) means a very stable compound. A positive ΔfH° (like for ozone) means the compound is at higher energy and is less stable relative to its elements.
Consider the reaction 2H2(g) + O2(g) → 2H2O(g), ΔrH° = −483.64 kJ. What is the enthalpy change for the decomposition of one mole of water?
The standard heat of formation, in kcal mol⁻¹, of Ba²⁺ is: [Given: ΔfH°(SO4²⁻, aq) = −216 kcal/mol; standard heat of crystallisation of BaSO4(s) = −4.5 kcal/mol; ΔfH°(BaSO4, s) = −349 kcal/mol]
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Pressure of 1 bar and a specified temperature, usually 298 K (25 °C). Each element must be in its most stable physical state at these conditions, for example carbon as graphite and oxygen as O2 gas.
No. Most compounds have negative ΔfH° because forming them releases energy. But some, like ozone (O3) and nitric oxide (NO), have positive ΔfH° because they store energy and are less stable than their elements.
Always products minus reactants: ΔrH° = Σ ΔfH°(products) − Σ ΔfH°(reactants). Reversing this order flips the sign and is the top cause of wrong answers in NEET.
The formation formula is really a shortcut version of Hess's Law. Hess's Law lets you add and subtract any reactions to find an unknown ΔH, while the formation formula is the special ready-made version that uses tabulated ΔfH° values.
Yes. Multiply each compound's ΔfH° by its coefficient in the balanced equation before adding. For example, if the equation has 2H2O, use 2 × ΔfH°(H2O).
