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Q.

If ΔfHC2H6=84kJmol1,Δsub HC, graphite )=720kJmol1,ΔatH2=435kJmol1 and the bond enthalpy εCH=414kJmol1 then the bond enthalpy εCC will be about

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a

345 kJ mol-1

b

315 kJ mol-1

c

295 kJ mol-1

d

375 kJ mol-1

answer is C.

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Detailed Solution

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According to Hess's lawΔfH=2Δsub H(C)+3ΔatHH2εCC6εCH εCC=2ΔsubH(C)+3ΔatHH26εCHΔfH=[2×720+3×4356×414+84]kJmol1=345kJmol1

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