Question

Use the following data to calculate \[\ce{∆_{lattice}H^Θ}\] for \[\ce{NaBr}\].

\[\ce{∆_{sub}H^Θ}\] for sodium metal = 108.4 kJ mol^–1

Ionization enthalpy of sodium = 496 kJ mol^–1

Electron gain enthalpy of bromine = – 325 kJ mol^–1

Bond dissociation enthalpy of bromine = 192 kJ mol^–1

\[\ce{∆_fH^Θ}\] for \[\ce{NaBr (s)}\] = – 325 kJ mol^–1 

Answer 1

In order to calculate the lattice enthalpy of \[\ce{NaBr}\],

(i) \[\ce{Na(s) -> Na(g); ∆_{sub} H^Θ = 108.4 kJ mol^{-1}}\]

(ii) \[\ce{Na -> Na^{+} + e^{-}; ∆_iH^Θ = 496 kJ mol^{-1}}\] 

(iii) \[\ce{1/2 Br2 -> Br, 1/2 ∆_{diss}H^Θ = 96 kJ mol^{-1}}\]

(iv) \[\ce{Br + e- -> ; ∆_{eg}H^Θ = - 325 kJ mol^{-1}}\]

\[\ce{∆_fH^Θ = ∆_{sub}H^Θ + 1/2 ∆_{diss}H^Θ + ∆_iH^Θ + ∆_iH^Θ + ∆_{eg}H^Θ + ∆_{lattice}H^Θ}\]

= – 360.1 – 108.4 – 96 – 496 + 325

= – 735.5KJ mol^–1.