Molar Enthalpies of Formation

1. Molar Enthalpies of Formation When comparing enthalpy changes for formation reactions of different compounds, we must choose a reference energy state. It is convenient to set the enthalpies of elements in their most stable form at SATP to be zero. As an arbitrary convention, for the sake of simplicity, all other enthalpies of compounds are measured relative to that reference energy state. A formation reaction always begins with elements, so any standard enthalpy of formation reactions are measured from the reference energy state of zero.

2. Thermal stability is the tendency of a compound to resist decomposition when heated. The lower (i.e. more negative) the value of a compound’s standard molar enthalpy of formation, the more stable it is. Δf Hm° = – 280.7 kJ/mol SnO Δf Hm° = – 577.6 kJ/mol SnO2 Tin(IV) oxide has a greater thermal stability than tin(II) oxide.

3. The standard enthalpy change of a reaction is the sum of the standard enthaplies of formation of the products minus the sum of the standard enthalpies of formation of the reactants. ΔrH° = ΣnΔfPHm° – ΣnΔfRHm°

4. – 985.2

5. Steps to success write the formula equations & corresponding standard enthalpy of formations. NOTE: data booklet gives you molar enthalpy, so you need to change it to just enthaply! Arrange your equations! Apply Your answer should be: ∆rH°= -65.4 kJ ΔrH° = ΣnΔfPHm° – ΣnΔfRHm°

7. Solution... ∆cH°= (1 mol CO2 x -393.5 kJ/mol CO2 + 2 mol H2O x -241.8kJ/mol H2O) – (1 mol CH4 x -74.6 kJ/mol + 2 mol O2 x 0 kJ/mol O2) = -877.1 kJ – (-74.6 kJ) =-802.5 kJ ∆cH°=n ∆cHm° ∆cHm°= -802.5 kJ/ 1mol = -802.5 kJ/mol CH4

8. Homework: • Read pgs. 510 – 513 • pgs. 514 – 515 Section 11.5 Questions #’s 1 – 9