Chemistry · Thermodynamics · NEET
There are two clear cases. First, when the reaction has only solids and liquids (no gases), because their volume change is so tiny that Δn_g RT is almost zero. Second, when gases are present but the total moles of gas on the product side equals the total moles of gas on the reactant side, so Δn_g = 0. In both cases the term Δn_g RT vanishes and ΔH = ΔU.
The full relation is ΔH = ΔU + Δn_g RT. Here Δn_g means (moles of gaseous products) − (moles of gaseous reactants). If this difference is 0, then Δn_g RT = 0, so ΔH = ΔU + 0 = ΔU. Example: N2(g) + O2(g) → 2NO(g). Reactant gas moles = 2, product gas moles = 2, so Δn_g = 0 and ΔH = ΔU.
Yes, for practical NEET purposes. Solids and liquids barely change volume when they react, so the pΔV work is negligible and Δn_g is taken as 0. So for reactions involving only solids and liquids, we say ΔH ≈ ΔU. Only count GAS moles for Δn_g; ignore solids and liquids completely.
No. Count ONLY the gas molecules. Ignore every solid (s) and liquid (l). For CaCO3(s) → CaO(s) + CO2(g): the only gas is CO2 on the product side, so Δn_g = 1 − 0 = 1. Here Δn_g ≠ 0, so ΔH is NOT equal to ΔU.
Only if Δn_g = 0. If Δn_g is not zero, you must convert: ΔH = ΔU + Δn_g RT. This is a very common NEET trap. A question gives you ΔU for a gas reaction and expects you to add Δn_g RT before using it. Never assume ΔH = ΔU unless you have checked Δn_g.
The condition Δn_g = 0 does not depend on temperature. But the SIZE of the difference does: the extra term is Δn_g RT, so higher T makes the gap bigger when Δn_g is not zero. When Δn_g = 0, T does not matter because the whole term is zero.
For the reaction 2A(g) + B(g) → 2D(g), ΔU° = −10 kJ mol⁻¹ and ΔS° = −44 J K⁻¹ at 298 K. Identify the correct ΔG° and the spontaneity. (Given R = 8.31 J mol⁻¹ K⁻¹)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
H2(g) + I2(g) → 2HI(g). Reactant gas moles = 2, product gas moles = 2, so Δn_g = 0 and ΔH = ΔU exactly.
CaCO3(s) → CaO(s) + CO2(g). Here Δn_g = 1 − 0 = 1, so ΔH = ΔU + RT, which is larger than ΔU.
Δn_g = (total moles of gaseous products) − (total moles of gaseous reactants). Count only gases.
Their volumes hardly change during a reaction, so the pressure-volume work pΔV is nearly zero. That work is exactly the difference between ΔH and ΔU.
Yes. NEET regularly gives ΔU and asks for ΔH (or ΔG), testing whether you remember to add Δn_g RT. Knowing when they are equal saves you from a very common mistake.
