If ΔfH∘C2H6=−84kJmol−1,Δsub H∘C, graphite )=720kJmol−1,ΔatH2=435kJmol−1 and the bond enthalpy εC−H=414kJmol−1 then the bond enthalpy εC−C will be about

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

  1. A

    295 kJ mol-1

  2. B

    315 kJ mol-1

  3. C

    345 kJ mol-1

  4. D

    375 kJ mol-1

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    Solution:

    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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