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JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry

The ionization enthalpy of \(Na ^{+}\) formation from \(Na _{( g )}\) is \(495.8\, kJ\, mol ^{-1},\) while the electron gain enthalpy of \(Br\) is \(-325.0\, kJ\, mol ^{-1}\). Given the lattice enthalpy of \(NaBr\) is \(-728.4\, kJ\, mol ^{-1}\). The energy for the formation of NaBr ionic solid is \((-)\) ........... \(\times 10^{-1} kJ \,mol ^{-1}\)

  1. A \(5581\)
  2. B \(4856\)
  3. C \(5596\)
  4. D \(5576\)
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Answer & Solution

Correct Answer

(D) \(5576\)

Step-by-step Solution

Detailed explanation

\(\Delta H _{\text {formation }}= IE _{1}+\Delta Heg _{1}+ LE\) \(=495.8+(-325)+(-728.4)\) \(=-557.6\) \(=-5576 \times 10^{-1} KJ / mol .\) Note: The above calculation is not for \(\Delta H _{\text {formation }}\) but for \(\Delta H _{ Reaction}\) But on the basis of given data…
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