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

Question

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

TestHub · Accepted Answer

CH 2 ​ = CH 2 ​ ( g ) + H 2 ​ ( g ) → CH 3 ​ − CH 3 ​ Δ 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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