The average S–F bond energy in kJ mol of SF is _______. (Rounded off to the nearest integer) [Given: The values of standard enthalpy of formation of SF (g), S(g) and F(g) are - 1100, 275 and 80 kJ mol

Question

The average S–F bond energy in kJ mol of SF is _______. (Rounded off to the nearest integer) [Given: The values of standard enthalpy of formation of SF (g), S(g) and F(g) are - 1100, 275 and 80 kJ mol respectively.] –1 6 6 -1

Prashant Kotian · Accepted Answer

Correct answer: 309.00. **Step 1: Identify the reaction for SF₆ dissociation** SF₆(g) → S(g) + 6F(g) **Step 2: Apply Hess's Law** ΔH°rxn = Σ ΔH°f(products) - Σ ΔH°f(reactants) ΔH°rxn = [ΔH°f(S) + 6·ΔH°f(F)] - [ΔH°f(SF₆)] **Step 3: Substitute the given values** ΔH°rxn = [275 + 6(80)] - (-1100) ΔH°rxn = [275 + 480] + 1100 ΔH°rxn = 755 + 1100 = 1855 kJ/mol **Step 4: Relate reaction enthalpy to bond energy** The energy required to break all 6 S–F bonds equals ΔH°rxn: Total bond energy (6 bonds) = 1855 kJ/mol **Step 5: Calculate average S–F bond energy** Average S–F bond energy = 1855 ÷ 6 = 309.17 kJ/mol **Step 6: Round to the nearest integer** 309.17 ≈ 309 kJ/mol Therefore, the answer is 309.00.

💡 Solution & Explanation

Practice 22,000+ questions like this