Enthalpy and enthalpy changes
Chemical Thermodynamics · Chemistry
Study notes
For N₂ + 3H₂ → 2NH₃, ΔU = −87 kJ at 298 K. Find ΔH. Step 1: Δn_g = 2 − 4 = −2 mol gas. Step 2: Δn_g RT = (−2)(8.314)(298)/1000 = −4.96 kJ. Step 3: ΔH = ΔU + Δn_g RT = −87 − 4.96 = −92 kJ. Step 4: ΔH more negative — system shrinks, surroundings get extra PV work. For reactions with Δn_g = 0, ΔH = ΔU.