Calculate the enthalpy change of the following reaction H2C=CH2(.g.)+H2(.g.) arrow H3C-CH3(.g.) The bond energy of C-H,C-C,C=C,H-H are 414, 347, 615 and 453 KJ mol- 1 respectively.

Question

Calculate the enthalpy change of the following reaction H2C=CH2(.g.)+H2(.g.) arrow H3C-CH3(.g.) The bond energy of C-H,C-C,C=C,H-H are 414, 347, 615 and 453 KJ mol- 1 respectively. (A) +125 kJ (B) -12.5 kJ (C) -125 kJ (D) +12.5 kJ

Tardigrade Admin · Accepted Answer

Δ H= Sigma bond energy of reactant - Sigma bond energy of product to write bond energy use proper symbols textE textC text= textC textor Δ textC text= textC textH =[.1(.C=C.)+4(.C-H.)+1(.H-H.)].-[.1(.C-C.)+6(.C-H.)]. =-1(.C-C.)-2(.C-H.)+1(.C=C.)+1(.H-H.) =-347-2(.414.)+1(.615.)+1(.435.) =-125 kJ.

Q. Calculate the enthalpy change of the following reaction H2​C=CH2​(g)+H2​(g)→H3​C−CH3​(g) The bond energy of C−H,C−C,C=C,H−H are 414, 347, 615 and 453 KJ mol−1 respectively.

+125 kJ

-12.5 kJ

-125 kJ

+12.5 kJ

ΔH=Σ bond energy of reactant −Σ bond energy of product to write bond energy use proper symbols EC=C​orΔC=C​H =[1(C=C)+4(C−H)+1(H−H)]−[1(C−C)+6(C−H)] =−1(C−C)−2(C−H)+1(C=C)+1(H−H) =−347−2(414)+1(615)+1(435) =−125 kJ.

Q. Calculate the enthalpy change of the following reaction

$H_{2} C = C H_{2} (g) + H_{2} (g) \to H_{3} C - C H_{3} (g)$

The bond energy of $C - H, C - C, C = C, H - H$ are 414, 347, 615 and 453 KJ $m o l^{- 1}$ respectively.