Chemistry · Thermodynamics · NEET
Hess's Law says: the total heat (enthalpy) change of a reaction is the same, no matter if the reaction goes in one step or many small steps. So if you cannot measure a reaction's ΔH directly, you can build that reaction by adding up other reactions whose ΔH you already know. The total ΔH will match.
It works because enthalpy (H) is a state function. A state function depends only on the current state (the starting point and the ending point), not on the path taken to get there. So the enthalpy change ΔH only cares about reactants and products. Whether you take a short road or a long road, the change is the same. This is why for NEET you can safely add, reverse, or scale reactions.
Follow three rules. (1) If you reverse a reaction (swap reactants and products), change the sign of ΔH (plus becomes minus, minus becomes plus). (2) If you multiply a reaction by a number (like 2 or 1/2), multiply its ΔH by the same number. (3) Add the adjusted reactions so that the extra species cancel out and you are left with the target reaction. Then add the adjusted ΔH values to get the answer.
The sign of ΔH flips. If the forward reaction releases 200 kJ (ΔH = −200 kJ), the reverse reaction absorbs 200 kJ (ΔH = +200 kJ). This is a very common NEET trap: they give you a formation reaction and ask about decomposition, which is just the reverse. Remember to flip the sign.
You multiply ΔH by the same number. If forming 2 moles of water releases 483.64 kJ, then forming 1 mole releases half of that, 241.82 kJ. Enthalpy is an extensive property, meaning it depends on the amount of substance. So doubling the amounts doubles ΔH; halving the amounts halves ΔH.
Yes. ΔrH° = [sum of ΔfH° of products] − [sum of ΔfH° of reactants]. Multiply each ΔfH° by its number of moles in the balanced equation. This shortcut is just Hess's Law applied through formation reactions, so it always agrees with adding reactions step by step. Note: ΔfH° of any element in its most stable form is zero.
They are not really different. The formation-enthalpy formula (products minus reactants) is one special use of Hess's Law. Hess's Law is the bigger idea: you can combine ANY set of known reactions, not only formation reactions. Use the formation formula when you are given ΔfH° values; use step-by-step adding when you are given random reactions.
Consider the following 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° of SO₄²⁻(aq) = −216 kcal mol⁻¹, standard heat of crystallisation of BaSO₄(s) = −4.5 kcal mol⁻¹, ΔfH° of BaSO₄(s) = −349 kcal mol⁻¹]
Try the real previous-year questions from this chapter — each with the answer and a full solution.
No. Because it comes from the idea of a state function, the same logic works for any state function, such as internal energy (ΔU) and entropy (ΔS). But in NEET Chemistry, you will mostly use it for enthalpy (ΔH).
Only when needed to match your target. First write the target reaction. Then adjust each given reaction (reverse or multiply) so the unwanted species cancel and the correct species end up on the right side. If a reaction is already in the right direction and amount, leave it alone.
Reverse a reaction → change the sign of ΔH. Multiply a reaction by n → multiply ΔH by n. These two rules solve almost every NEET Hess's Law numerical.
By definition, the standard enthalpy of formation of an element in its most stable form (like O₂ gas, or solid carbon as graphite) is taken as zero. This gives a common reference point so all ΔfH° values can be compared.
Enthalpy problems, including Hess's Law, appear almost every year in the Thermodynamics chapter. Learning to reverse, scale and add reactions is one of the highest-value skills for scoring in Physical Chemistry.
