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
- A∆ lattice H ⊖ (NaBr) = +700.5 kJmol -1
- B∆ lattice H ⊖ (NaBr) = +786.5 kJmol -1
- C∆ lattice H ⊖ (NaBr) = +735.5 kJmol -1
- 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