The Born-Haber cycle for KCl is evaluated with the following data : ƒH for KCl = – 436.7 kJ mol–1 ; subH for K = 89.2 kJ mol–1 ; ionizationH for K = 419 kJ mol–1 ; electron gainH for Cl(g) = –348.6 kJ

Question

The Born-Haber cycle for KCl is evaluated with the following data : ƒH for KCl = – 436.7 kJ mol–1 ; subH for K = 89.2 kJ mol–1 ; ionizationH for K = 419 kJ mol–1 ; electron gainH for Cl(g) = –348.6 kJ mol–1 ; bondH for Cl2 = 243.0 kJ mol–1 The magnitude of lattice enthalpy of KCl in kJ mol–1 is…………. (Nearest integer) KCl ds fy, ckWuZ gkcj pØ dk ewY;kadu fuEufyf[kr vk¡dM+ks ds vk/kkj ij fd;k x;k gS& KCl ds fy,ƒH = – 436.7 kJ mol–1 ; K ds fy, subH = 89.2 kJ mol–1 ; K ds fy, ionizationH = 419 kJ mol–1 ; Cl(g) ds fy, electron gainH = –348.6 kJ mol–1 ; Cl2 ds fy, bondH = 243.0 kJ mol–1 KCl dh tkyd ,UFkSYih dk ifjek.k kJ mol–1 esa gS…………. (fudVre iw.kkZd esa)

Prashant Kotian · Accepted Answer

Correct answer: .. K(s) + ½Cl2(g)  KCl(s); ƒHº = –436.7 kJ mol–1 sub.Hº=89.2 K(g) ionHº=419 K+(g) bondHº=243/2 Cl(g) Cl–(g) ele. gainHº=–348.6 latticeHº so, ƒH = ƒH(KCl) = subH + 1st ionisationH(K) + ½bondH(Cl2(g)) + electron gainH + latticeH – 436.7 = 89.2 + 419 + ½(243) + (–348.6) + latticeH latticeH = – 717.8 kJ mole–1  – 718 kJ mole–1

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