Question

Given the following data: l r Reaction & Energy change (kJ)
Li(s)→ Li(g) & 161
Li(g)→ Li⁺(g) & 520
(1)/(2)F₂(g)→ F(g) & 77
F(g)+e⁻→ F⁻(g) & x
Li⁺(g)+F⁻(g)→ LiF(s) & -1047
Li(s)+(1)/(2)F₂(g)→ LiF(s) & -617 Based on the data provided, the value of electron gain enthalpy of fluorine would be:

• A. \(-300\,\text{kJ mol}^{-1}\)
• B. \(-350\,\text{kJ mol}^{-1}\)
• C. \(-328\,\text{kJ mol}^{-1}\)
• D. -228kJ mol⁻1

Hint:

Electron gain enthalpy is obtained using a Born–Haber cycle by balancing Δ Hf = sum (steps)

Answer 1

The Correct Option is C

Step 1: Apply Hess’s law to the Born–Haber cycle: -617 = 161 + 520 + 77 + x - 1047
Step 2: Simplify: -617 = -289 + x
Step 3: x = -328kJ mol⁻1