Ncert Exercises & Solutions

Use the following data to calculate $∆_{lattice} H^{⊝}$ for NaBr. $∆_{sub} H^{⊝}$ for sodium metal $= 108.4 kJ {mol}^{- 1}$ , ionisation enthalpy of sodium $= 496 kJ {mol}^{- 1}$ , electron gain enthalpy of bromine $= - 325 kJ {mol}^{- 1}$ , bond dissociation enthalpy of bromine $= 192 kJ {mol}^{- 1}$ , $∆_{f} H^{⊝}$ for NaBr(s) $= - 360.1 kJ {mol}^{- 1}$

Given that,

∆_{sub} H^{⊝}

for Na metal

= 108.4 k J {mol}^{- 1}

IE of Na

= 496 k J {mol}^{- 1}

∆_{eg} H^{⊝}

of Br

= - 325 k J {mol}^{- 1}

∆_{diss} H^{⊝}

of Br

= 192 kJ {mol}^{- 1}

∆_{f} H^{⊝}

for NaBr

= - 360.1 k J {mol}^{- 1}

Born-Haber cycle for the formation of NaBr is as

By applying Hess's law,

∆_{f} H^{⊝} = ∆_{sub} H^{⊝} + IE + ∆_{diss} H^{⊝} + ∆_{eg} H^{⊝} + U

- 360.1 = 108.4 + 496 + 96 + (- 325) - U

U = + 735.5 kJ {mol}^{- 1}

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