Question

10.56 Using the data given below (all values are in kJ/mol) and Born-Haber cycles for the formation of both the hydride and the chloride of sodium, calculate the standard heats of formation in each case. Which type of compound is more stable? Briefly discuss the principal reasons for the differences in stabilities as measured by the standard heat of formation. Would you expect differences in entropy of formation to change your conclusions? Un- E AH EA sub 808 Nal 107.3 107.3 -73 495.9 218.2 -787 NaC 495.9 121.7 -349

Answer 1

Answer: - For Nahi ДНЕ Nahlg) Nalo) + 1/2 H2 Al sub AH diss. lattice energylu). Nalg) Hig) TE TEA Natcg) + Hcg). - EA- 57 Atsub + I ETAH ist EATU. = (107.34 495.9 + 218.2-73-808). KT/mol = - 59.6 kJ/mol . For Nace. AHF Nacig) Nals) + 4/lyg) All sub Allaiss Nacg) cecgo IEL LEA Natig) + Cicg) Lattice energy cu) Alf = Alsub + TE + A Maisste Atun 3+ 495.9 +121.7 - 349-787 KJ/mol -- 411.1 kJ/mol.

Boren - Haber cycle applies Hess's law to calculate enthalpy of formation in other words the lattice e of formation of foeemalion sul Lattice enthalpymeasures the th strength of fouces between cons in an ionic solid. During formation, AH f = - veie exothermic. reaction occurs. But since exothermicity of Nace is higher than Nath ; so Nace is more Seable over Nat. Nace would require more amount of energy to break the bond and dissociate into constituent ions so it has higher stability In both cases a solid and a gaseous component combines to generate a gaseous conic lattice. This considering the differences en entowpy of foumation is not significant AHE d distance between the centre of ions. ; as radius becomes larger ad increases and secaction becomes less exothermie [lx-Yucet