For the reaction C?H? → C?H? + H? the reaction enthalpy ?rH = ____ kJ mol?¹. (Round off to the nearest integer).
[ Given: Bond enthalpies in kJ mol?¹ : C-C: 347, C=C:611; C-H:414,H-H:436]

For the reaction C? H? → C? H? + H? , calculate the enthalpy change (ΔH).

ΔH = [Bond energy (C-C) + 6 × Bond energy (C-H)] - [Bond energy (C=C) + 4 × Bond energy (C-H) + Bond energy (H-H)]

ΔH = 347 + 2 (414) - 611 - 436 = 128 kJ/mol.

<p>For the reaction C? H? → C? H? + H? , calculate the enthalpy change (ΔH).</p><div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1756803129phpzu3A5T.jpeg" alt="" width="514" height="100" loading="lazy"></picture></div></div><p>ΔH = [Bond energy (C-C) + 6 × Bond energy (C-H)] - [Bond energy (C=C) + 4 × Bond energy (C-H) + Bond energy (H-H)]</p><p>ΔH = 347 + 2 (414) - 611 - 436 = 128 kJ/mol.</p>

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