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Q.

D2(g)+1/2 O2(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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a

491.5

b

495.1

c

415.9

d

419.5

answer is A.

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Detailed Solution

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

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