The Thermodynamic Difference Rule (TDR) for non-aqueous solvates. Part 1. Review of methodology, investigation and prediction of thermodynamic data for sulfur dioxide solvates, MpXq.nSO2, routes to expand the database and forecast of future science and technology

Abstract

This paper investigates methods of estimation of new as well as already known data for the standard enthalpy of formation, ΔfHo the standard free energy of formation, ΔfGo and the standard (absolute) entropy, S298o for SO2 solvates, MpXq.nSO2 at 298 K and validation of the latter solvates’ existing data. A new approach enabling extension of the existing database is presented which involves the use of additional thermodynamic data for hydrates MpXq·nH2O or, ammoniate salts MpXq·nNH3 if and when available. This supplements the normal TDR approach which uses thermodynamic data for the parent compounds, MpXq. In principle use of these procedures will extend to other solvates too. The whole of the thermochemical data for these and for other solvate materials can be thought of as a vast matrix of self-consistent cross-linked linear equations of the type displayed below. Several TDR equations are involved which take the analytical form:[ΔfGo(MpXq.nSO2,s) − ΔfGo(MpXq,s)]/kJ mol−1 = ϴGf(SO2, s − s) n = −299.9n (N = 2, R2 = 1.00)[ΔfHo(MpXq.nSO2,s) − ΔfHo(MpXq,s)]/kJ mol−1 = ϴHf(SO2, s − s)) n = −338.3n (N = 9, R2 = 0.999)[S298o (MpXq.nSO2,s) − S298o (MpXq,s)]/J K−1 mol−1 = ϴSo(SO2, s − s)) n = 106.9n (N = 2, R2 = 0.999)[ΔfGo(MpXq·nH2O,s) − ΔfGo(MpXq,s)]/kJ mol−1 = ϴGf(H2O, s − s)) n = −242.4n (N = 93, R2 = 0.998)[ΔfHo(MpXq·nH2O,s) − ΔfHo(MpXq,s)]/kJ mol−1 = ϴHf(H2O, s − s)) n = −298.6n (N = 342, R2 = 0.999)[S298o (MpXq·nH2O,s) − S298o (MpXq,s)]/J K−1 mol−1 = ϴ So(H2O, s − s) n = 40.9n (N = 83, R2 = 0.978)[ΔfGo(MpXq.n NH3,s) − ΔfGo(MpXq,s)]/kJ mol−1 = ϴGf(NH3, s − s) n = −21.0n (N = 4, R2 = 0.922)[ΔfHo(MpXq·nNH3,s) − ΔfHo(MpXq,s)]/kJ mol−1 = ϴHf(NH3, s − s)) n = −104.2n (N = 277, R2 = 0.930)[S298o (MpXq·nNH3,s) − S298o (MpXq,s)]/J K−1 mol−1 = ϴ So(NH3, s − s)) n = 64.1n (N = 9, R2 = 0.989)[ΔfGo(MpXq.nSO2, s)]/kJ mol−1 = [ΔfGo(MpXq.nH2O, s)] − 57.5 n[ΔfHo(MpXq.nSO2, s)]/kJ mol−1 = [ΔfHo(MpXq.nH2O, s)] − 39.7 n[S298o (MpXq.nSO2, s)]/J K−1 mol−1 = [S298o (MpXq.nH2O, s)] + 66.0 n[ΔfGo(MpXq.nSO2, s)]/kJ mol−1 = [ΔfGo(MpXq.nNH3, s)] − 278.9 n[ΔfHo(MpXq.nSO2, s)]/kJ mol−1 = [ΔfHo(MpXq.nNH3, s)] − 234.1 n[S298o (MpXq.nSO2, s)]/J K−1 mol−1 = [S298o (MpXq.nNH3, s)] + 42.8 n