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If at 298 ~K the bond energies of C - H , C - C , C = C and H - H bonds are respectively 414,347,615 and 435 ~kJ mol ⁻¹ , the value of enthalpy change for the reaction H ₂ C = CH ₂( ~g )+ H ₂( ~g ) H ₃ C - CH ₃( ~g ) at 298 ~K will be

Options

  1. A-250 ~kJ
  2. B+125 ~kJ
  3. C-125 ~kJ
  4. D+250 ~kJ

Correct answer

C. -125 ~kJ

Step-by-step solution

CH ₂= CH ₂( ~g )+ H ₂( ~g ) CH ₃- CH ₃ H=1( C = C )+4( C - H )+1( H - H )-1( C - C )-6( C - H )=1( C = C )+1( H - H )-1( C - C )-2( C - H )=615+435-347-2 414=1050-1175=-125 ~kJ

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