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D_2(g)+1/2 O_2(g) → D₂O_(l), Δ H= -294 KJ. If bond enthalpies of D₂ and O₂ respectively are 440 and 498 KJ/mole, then bond enthalpy of D-O is (KJ)

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491.5

495.1

419.5

415.9

answer is A.

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Detailed Solution

ΔH = [BE_D-D + 1/2 BE_O=O] - 2BE_D-O
-294=[440+1/2 (498)]-2BE_D-O
-294=(440+249)-2BE_D-O
-294=689-2BE_D-O
2BE_D-O = 689+294 = 983
BE_D-O=983/2=491.5

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