Q1(e). Dimerization and temperature

Answer in brief

For exothermic dimerization, dln K/dT=ΔᵣH°/(RT²)<0, so the dimerization constant decreases on heating.

Key points

• Write the association equilibrium as 2M⇌ M₂, with K=a_(M₂)/a_M².
• The van' t Hoff relation is dln K/dT=ΔᵣH°/(RT²). Formation of a stabilising bond or intermolecular association normally releases heat, giving ΔᵣH°<0.
• Therefore increasing temperature favours dissociation and reduces K. If the enthalpy is approximately constant, ln (K₂/K₁)=-ΔᵣH°(1/T₂-1/T₁)/R, which is negative for T₂>T₁.
• This sign is conditional on exothermic dimerization; dimerization alone is not a mathematical guarantee that its enthalpy must be negative.