Question

sed on the following energy values, determine the AH- for sodium chlo Na (s) + 12 Cl2 (g) → NaCl (s) AHsub for Na: Bond Energy of Cl2: Lattice energy of NaCl: Ionization energy for Na: Electron Affinity of Cl: Na (s) → Na (g) Cl2 (g) → 2C1 (g) NaCl (s) → Na" (g) + C (g) Na (g) → Na* (g) + le Cl (g) + le → Cl (g) u u edaeftheriodietwandeta determine whe nde to determine whether LiFor

Answer 1

Answer:

Step 1: Explanation

The analysis of the formation of an ionic compound from its elements is commonly discussed in terms of a Born-Haber cycle, which breaks the overall process into a series of steps of known energy

Step 2: Calculation

Step 1: Na(s) --> Na(g) ------------> (  ΔHsub = 108 kJ/mol )

Step 2: 1/2 Cl2(g) --> Cl(g) --------> ( BE = 244/2 kJ/mol ) = (122 kJ/mol )

[since given BE for Cl₂ = 242 kJ/mol coversion but we need Cl hence EA will multiply by 1/2]

Step 3: Na(g) --> Na⁺(g) + e- ----------> ( IE₁ = 496 kJ/mol )

Step 4: Cl(g) + e- --> Cl^-(g) ----------->    ( ΔH_EA = -349 kJ/mol )

Step 5: M⁺(g) + Cl^-(g) --> NaCl(s) ------------>     (  ΔU = -788 kJ/mol )

Step 6: Na⁺(g) +1/2 X^-(g) --> MX ------------>     (  ΔH⁰_f = we need to find )

Overall reaction: Na(s)+ 1/2 Cl₂(g) --> NaCl(s)

Overall energy

ΔH⁰_f    = U + ΔHsub + IE + BE + ΔH_EA  

ΔH⁰_f   = -788 kJ/mol + 108 kJ/mo + 496 kJ/mol + 122 kJ/mol + (-349 kJ/mol)

ΔH⁰_f   = - 411 kJ/mol

Note: -ve sign is thermodynamically correct but most report Enthalpy of formation without -ve sign it is understood that it is -ve. So change the sign according to your need