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NEETChemistryThermodynamics (C)

Use the following data to calculate ∆ lattice H ⊖ for NaBr. ∆ sub H ⊖ 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 ∆ f H ⊖ for NaBr (s) = – 360.1 kJ mol –1

Options

  1. A∆ lattice H ⊖ (NaBr) = +700.5 kJmol -1
  2. B∆ lattice H ⊖ (NaBr) = +786.5 kJmol -1
  3. C∆ lattice H ⊖ (NaBr) = +735.5 kJmol -1
  4. D∆ lattice H ⊖ (NaBr) = +816.5 kJmol -1

Correct answer

C. ∆ lattice H ⊖ (NaBr) = +735.5 kJmol -1

Step-by-step solution

Correct Option is : (C) ∆ lattice H ⊖ (NaBr) = +735.5 kJmol -1

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