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 \((-)\) ........... \(\times 10^{-1} kJ \,mol ^{-1}\)
- A \(5581\)
- B \(4856\)
- C \(5596\)
- D \(5576\)
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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