Reaction enthalpy from bond enthalpies
Using bond enthalpies of H–H = 436, Cl–Cl = 243, and H–Cl = 431 kJ/mol, what is ΔH for H2 + Cl2 → 2HCl?
Answer B: explanation for every choice
A
This forms only one H–Cl bond and reverses the sign.
B
Correct. ΔH = bonds broken − bonds formed = (436 + 243) − 2(431) = 679 − 862 = −183 kJ.
C
This calculates bonds formed − bonds broken, reversing the sign.
D
This forms only one H–Cl bond: 679 − 431 = +248 kJ.
TakeawayΔH = Σ bonds broken − Σ bonds formed.
Equation usedEnthalpy from bond energies
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