Bond enthalpies of P-P, O=O and P-O are p, q and s respectively. Enthalpy change for the process, P4+3O2→P4O6 is

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4p + 3q - 6s

6p + 3q - 16s

4p + 3q - 12s

6p + 3q - 12s

answer is D.

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Detailed Solution

6P-P+3O=O-12P-O=∆H+6P + 3q - 12s = ∆H

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