Chemistry · Thermodynamics · NEET
Enthalpy (H) is the total heat energy stored in a substance at a given pressure. You cannot measure H directly, but you can measure how much it changes during a reaction. That change is called ΔH (delta H). When a reaction happens at constant pressure, ΔH is exactly the heat absorbed or released. This is why enthalpy is called the 'heat content' of a system.
Internal energy U is all the energy inside the system (kinetic + potential of particles). Enthalpy adds one extra term: H = U + pV. The pV part accounts for the work needed to make space for the system against the outside pressure. At constant volume the heat equals ΔU (q_v = ΔU). At constant pressure the heat equals ΔH (q_p = ΔH). NEET loves testing which one applies to which condition.
By the first law, ΔU = q + w, and pressure-volume work is w = -pΔV. So q = ΔU + pΔV. At constant pressure, ΔH = ΔU + pΔV. Comparing the two, q_p = ΔH. In short, when pressure stays constant, all the heat you add or remove shows up as a change in enthalpy. Most lab reactions are open to the air, so pressure is constant and ΔH is the useful heat quantity.
Yes. Enthalpy is a state function because it is built only from state functions: U, p and V. This means ΔH depends only on the initial and final states, not on the path taken. That is why you can add reactions in Hess's law and why ΔH is fixed for a given reaction. Heat (q) by itself is a path function, but at constant pressure it happens to equal the state function ΔH.
A negative ΔH means heat is released, so the reaction is exothermic (products have lower enthalpy). A positive ΔH means heat is absorbed, so the reaction is endothermic (products have higher enthalpy). For example, combustion has ΔH < 0. If you reverse a reaction, the sign of ΔH flips but the size stays the same. NEET often tests this reversal rule.
No. Just like internal energy, the absolute value of H cannot be measured because we cannot count all the energy inside a system. We can only measure the change ΔH between two states, usually with a calorimeter. That is why every thermodynamics problem works with ΔH (a difference), never with a single H value.
Which amongst the following options is the correct relation between change in enthalpy and change in internal energy?
For 2H₂(g) + O₂(g) → 2H₂O(g), Δ_r H° = −483.64 kJ. What is the enthalpy change for the decomposition of 1 mol of water vapour into H₂ and O₂?
For the liquid–vapour equilibrium Liquid ⇌ Vapour, which relation is correct? (ΔH_v = enthalpy of vaporisation)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Only at constant pressure. When pressure is held constant, q_p = ΔH. At constant volume the heat equals ΔU instead. So ΔH equals heat only under the constant-pressure condition.
Enthalpy and ΔH are energy quantities, so the SI unit is the joule (J). In reactions we usually use kilojoules per mole (kJ mol⁻¹) because we compare energy per mole of substance.
Most reactions happen in open containers at constant atmospheric pressure, not at constant volume. Under constant pressure the measured heat equals ΔH, so enthalpy is the more convenient quantity for real lab chemistry.
Not always. A negative ΔH (exothermic) favours spontaneity, but the entropy term also matters. Spontaneity is decided by Gibbs energy ΔG = ΔH − TΔS, not by ΔH alone.
They are linked by ΔH = ΔU + Δn_g RT, where Δn_g is the change in moles of gas. If there is no change in gas moles (Δn_g = 0), then ΔH = ΔU.
