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Question

The enthalpy change for the reaction H2(g)+C2H4(g)C2H6(g) is:

The bond energies are, HH=103, CH=99, CC=80 and C=C=145 K cal mol1

A
10 Kcalmol1
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B
+10 Kcalmol1
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C
30 Kcalmol1
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D
+30 Kcalmol1
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Solution

The correct option is D +30 Kcalmol1
H2 has a sigle HH bond. So ΔfHH2=103kcalmol1
C2H4 has structure HC|H=C|HH So, ΔfH=4×BECH+BEC=C
ΔfHC2H4=4×99+145=541kcalmol1
C2H6 has 6 CH bonds and one CC bond. So, ΔfHC2H6=6×99+80=674kcalmol1
Enthalpy change for the given reaction ΔH=674[541+103]
ΔH=+30kcalmol1

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