The enthalpy change for the reaction H2(g)+C2H4(g)→C2H6(g) is:
The bond energies are, H−H=103,C−H=99,C−C=80 and C=C=145Kcalmol−1
A
−10Kcalmol−1
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B
+10Kcalmol−1
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C
−30Kcalmol−1
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D
+30Kcalmol−1
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Solution
The correct option is D+30Kcalmol−1 H2 has a sigle H−H bond. So ΔfHH2=103kcalmol−1 C2H4 has structure H−C|H=C|H−H So, ΔfH=4×BEC−H+BEC=C ΔfHC2H4=4×99+145=541kcalmol−1 C2H6 has 6 C−H bonds and one C−C bond. So, ΔfHC2H6=6×99+80=674kcalmol−1 Enthalpy change for the given reaction ΔH=674−[541+103] ΔH=+30kcalmol−1