Chemistry · Thermodynamics · NEET
Start with the definition of enthalpy: H = U + pV. For a change at constant pressure, ΔH = ΔU + Δ(pV). For ideal gases, pV = n_g RT, so at constant temperature Δ(pV) = RT·Δn_g = Δn_g RT. Substitute this in: ΔH = ΔU + Δn_g RT. That is the whole derivation. It is short, so learn each line: (1) H = U + pV, (2) ΔH = ΔU + Δ(pV), (3) Δ(pV) = Δn_g RT, (4) ΔH = ΔU + Δn_g RT.
The g stands for gas. Only gases change volume enough to do pressure-volume work that matters here. So Δn_g = (moles of gaseous products) − (moles of gaseous reactants). You count only the gas species in the balanced equation. Solids and liquids take up almost no volume, so we leave them out. If you wrote Δn (all moles) you would get the wrong answer.
From the ideal gas law: pV = nRT. For the gases in a reaction, the amount of gas is n_g moles, so pV = n_g RT. When the reaction runs at constant temperature T and pressure p, the change is Δ(pV) = Δn_g·RT, because R and T stay constant and only the number of gas moles changes. This is the key swap that turns the general relation ΔH = ΔU + Δ(pV) into the gas form ΔH = ΔU + Δn_g RT.
Two things: (1) constant temperature so that T is fixed when we write Δ(pV) = Δn_g RT, and (2) the gases behave like ideal gases so pV = n_g RT holds. It is used for reactions at constant temperature and pressure. If a reaction has no gases or Δn_g = 0, then ΔH = ΔU.
Count gas moles on each side. Products: 2 (NH3 is gas). Reactants: 1 (N2) + 3 (H2) = 4. So Δn_g = 2 − 4 = −2. It is negative because gas moles go down. Then ΔH = ΔU + (−2)RT, meaning ΔH is smaller than ΔU here. Always use the balanced equation and count only (g) species.
Match R to the unit you want for energy. Use R = 8.314 J K⁻¹ mol⁻¹ when ΔH and ΔU are in joules, or R = 2 cal K⁻¹ mol⁻¹ when they are in calories. T is always in kelvin. A common NEET slip is forgetting to convert Δn_g RT (which comes out in J) to kJ before adding to a ΔU given in kJ.
Which amongst the following options is the correct relation between change in enthalpy and change in internal energy?
For 2A(g) + B(g) → 2D(g), ΔU° = −10 kJ mol⁻¹ and ΔS° = −44 J K⁻¹ at 298 K. Find ΔG° and the spontaneity (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.
Yes, and in that case Δn_g = 0 because there are no gases. So the term Δn_g RT becomes zero and ΔH = ΔU. Solids and liquids have almost no volume change, so their enthalpy and internal energy changes are practically equal.
Pressure-volume work is done mainly by gases expanding or shrinking. Solids and liquids barely change volume, so they contribute nothing to pV. Using only gas moles (n_g) gives the correct change in pV for the reaction.
Δn_g counts only gaseous moles (products minus reactants), while Δn would count all moles including solids and liquids. For this formula you must use Δn_g, the gas moles only.
When Δn_g is positive, meaning gas moles increase in the reaction. Then Δn_g RT is positive and ΔH > ΔU. When gas moles decrease, Δn_g is negative and ΔH < ΔU.
Yes. T must be in kelvin because the ideal gas law pV = n_g RT uses absolute temperature. Convert Celsius to kelvin by adding 273.
