ΔS from standard entropies

Worked example

Use the standard entropies S(N₂) = 191.8 J K⁻¹ mol⁻¹, S(H₂) = 130.8 J K⁻¹ mol⁻¹ and S(NH₃) = 192.4 J K⁻¹ mol⁻¹ to find ΔS for the Haber process, N₂ + 3H₂ → 2NH₃.

1. 1. Weighted sum of 2·S(NH₃) − 1·S(N₂) − 3·S(H₂) → ΔS

   = −199.4 J K⁻¹ mol⁻¹

Answer: ΔS = −199 J K⁻¹ mol⁻¹

Method

ΔS = ΣS(products) − ΣS(reactants) = 2S(NH₃) − S(N₂) − 3S(H₂). It is negative because four moles of gas become two.

Common mistakes

• Multiply each entropy by its coefficient in the equation: 2 NH₃ and 3 H₂.
• Unlike ΔfH, the entropy of an element is not zero: N₂ and H₂ must be included.

More thermodynamics questions

• ΔG from ΔH and ΔS
• K from ΔG°
• ΔG° from K
• K from ΔH° and ΔS°
• van 't Hoff: K at a new temperature
• Moles of an ideal gas
• ΔH at another temperature (Kirchhoff)
• Boiling point at reduced pressure
• Entropy change on heating
• Hess's law from enthalpies of formation
• ΔH from a bomb-calorimeter ΔU
• Enthalpy of combustion from a calorimetry experiment
• Enthalpy of neutralisation
• ΔfH from enthalpies of combustion
• ΔH from mean bond enthalpies
• An unknown bond enthalpy from ΔH
• Born–Haber cycle: lattice enthalpy of NaCl
• Born–Haber cycle: lattice enthalpy of MgCl₂
• Temperature at which a reaction becomes feasible
• ΔG from enthalpies of formation and entropies
• Enthalpy of solution from a thermodynamic cycle
• Methanol synthesis: K from tabulated data