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Hess' Law

Use the follo...

Question

Use the following data to calculate $Δ_{l a t t i c} H^{⊖}$ for NaBr. $Δ_{s u b} H^{⊖}$ for sodium metal = 108.4 kJ $m o l^{- 1}$ , ionisation enthalpy of sodium = 496 kJ $m o l^{- 1}$ , electron gain enthalpy of bromine = -325 kJ $m o l^{- 1}$ , bond dissociation enthalpy of bromine = 192 kJ $m o l^{- 1}$ , $Δ_{f} H^{⊖}$ for NaBr(s) = -360. 1 kJ $m o l^{- 1}$

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Solution

Given that, $Δ_{s u b} H^{⊖}$ for Na metal = 108.4 k J $m o l^{- 1}$
IE of Na= 496 kJ $m o l^{- 1}$ , $Δ_{e g} H^{⊖}$ of Br = - 325 kJ $m o l^{- 1}$ , $Δ_{d i s s} H^{⊖}$ of Br= 192 kJ $m o l^{- 1}$ , $Δ_{f} H^{⊖}$ for NaBr = - 360.1 kJ $m o l^{- 1}$
Born-Haber cycle for the formation of NaBr is as

By applying Hess's law,
$Δ_{f} H^{⊖} = Δ_{s u b} H^{⊖} + I E + Δ_{d i s s} H^{⊖} + Δ_{e g} H^{⊖} + U$
-360.1 = 108.4 + 496 + 96 + (-325)-U
U=+735.5kJ $m o l^{- 1}$

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