6. Using the bond energy data given in the parenthesis, what would be the expected
enthalpy for the following reaction
H2(g) + F2 (g) à 2HF (g)
(H-H: 432 kJ/mol; F-F: 154 kJ/mol; H-F: 565 kJ/mol)?
a. -544 kJ/mol
b. +544 kJ/mol
c. -21 kJ/mol
d. +21 kJ/mol
H2 + F2 → 2HF Bond Energy Enthalpy Calculation: CSIR NET Solved
The reaction H₂(g) + F₂(g) → 2HF(g) uses given bond energies (H-H: 432 kJ/mol, F-F: 154 kJ/mol, H-F: 565 kJ/mol) to find the enthalpy change.
Calculation Method
Enthalpy change (ΔH) equals energy to break bonds minus energy released forming bonds. Bonds broken are one H-H (432 kJ/mol) and one F-F (154 kJ/mol), totaling 586 kJ/mol. Bonds formed are two H-F bonds (2 × 565 = 1130 kJ/mol). Thus, ΔH = 586 – 1130 = -544 kJ/mol.
Option Analysis
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a. -544 kJ/mol: Correct, matches exact calculation showing exothermic reaction due to stronger H-F bonds.
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b. +544 kJ/mol: Wrong; this reverses the sign, treating bond formation as endothermic instead of exothermic.
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c. -21 kJ/mol: Incorrect; possibly from misadding bonds broken (e.g., 432 – 154 = 278, then flawed subtraction), ignores proper method.
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d. +21 kJ/mol: Wrong; combines sign error with small number miscalculation, not matching bond data.
