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The hydration enthalpies of Li+(g) and Br-(g) are -500 kJ mol-1and -350 kJ mol-1 respectively and the lattice energy of LiBr(s) is 807 kJmol-1. Write the thermochemical equations for enthalpy of solution of LiBr(s) and calculate its value. |
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Answer» Given : Enthalpy of hydration of Li+(g) = ΔhydH1 = -500 kJmol-1 Enthalpy of hydration of Br-(g) = ΔhydH2 = -350 kJ mol-1 Lattice energy of LiBr(s) = ΔLH3 = 807 kJ mol-1 Enthalpy of solution of LiBr(s) = ΔsolnΔH = ? The thermochemical equation for the dissolution of LiBr(s) forming a solution is, LiBr(s) + aq → Li+(aq) + Br-(aq)(I) ΔsolH = ? This takes place in two steps as follows : (i). LiBr(s) → Li+(g) + Br-(g) ΔLH3 (ii).(a). Li+(g) + aq → Li+(aq) ΔhydH1 (b) Br-(g) + aq → Br-aq ΔhydH2 Hence by adding equations (i) and (ii) (a) and (b) we get equation . ∴ ΔsolH = ΔLH3 + ΔhydH1 + ΔhydH2 = 807 + (-500) + (-350) = -43 kJ mol-1 ∴ Heat of solution of LiBr(s) = ΔsolH = -43 kJ mol-1 |
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