Use the following data to calculate \Delta _{lattice}  H^{\ominus } for NaBr.
\Delta _{sub} H^{\ominus } 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
\Delta _{f}H^{\ominus } for NaBr (s) = – 360.1 kJ mol-1

Answers (1)

In order to calculate the lattice enthalpy of NaBr,

(i) Na(s) \rightarrow Na(g) ; \Delta _{sub}H^{\circ } =108.4\; kJ mol^{-1}

(ii) Na\rightarrow Na^{+} + e^{-} ; \Delta _{i}H^{\circ } = 496\; kJ mol^{-1}

(iii) \frac{1}{2}Br_{2}\rightarrow Br, \frac{1}{2}\Delta_{diss} H^{\circ } =96\; kJ\; mol^{-1}

(iv) Br+e^{-}\rightarrow Br^{-};\Delta egH^{\circ } = - 325 kJ\; mol^{-1}

\Delta _{f} H^{\circ } = \Delta _{sub} H^{\circ } + \frac{1}{2}\Delta _{diss}H^{\circ }+ \Delta _{i}H^{\circ } + \Delta _{eg}H^{\circ } + \Delta _{latttice}H^{\circ }

\Delta _{latttice}H^{\circ } = -360.1 -108.4-96-496+325 = -735.5KJ mol^{-1}

 

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