For a substance, the standard enthalpies of formation for its liquid and gaseous states are -280 kJ mol ⁻¹ and -235 kJ mol ⁻¹ respectively. The absolute standard molar entropies of the liquid and gas are 120 J K ⁻¹ mol ⁻¹ and 210 J K ⁻¹ mol ⁻¹ respectively. Assuming that enthalpy and entropy values are independent of temperature, the normal boiling point of the substance is ______ K.

Step-by-step solution

The vaporisation process is represented as: Substance (l) Substance (g) First, calculate the standard enthalpy of vaporisation: H_ vap = _ f H^ ( g ) - _ f H^ ( l ) H_ vap = -235 - (-280) = 45 kJ mol ⁻¹ = 45000 J mol ⁻¹ Next, calculate the standard entropy of vaporisation: S_ vap = S^ ( g ) - S^ ( l ) S_ vap = 210 - 120 = 90 J K ⁻¹ mol ⁻¹ At the normal boiling point, the liquid and vapour are in equilibrium, so G = 0 . G = H_ vap - T_ b S_ vap = 0 T_ b = H_ vap S_ vap T_ b = 45000 90 = 500 K Answer: 500