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.$H_2C = CH_2(g) + H_2(g) -----> H_3C-CH_3(g)$ at 298 K will be

• A. +250 kJ
• B. -250 kJ
• C. +125 kJ
• D. -125 kJ

Answer 1

Answer: (D)

-125 kJ

Answer 2

$\Delta H = \sum$ B.E. of reactants - $\sum$ B.E. of products = B.E.(C = C)+ 4 $\times$ B.E.(C-H) + B.E.(H-H) -B.E.(C-C) - 6 $\times$ B.E.(C-H) = B.E.(C = C) + B.E.(H-H) - B.E.(C-C) - 2 $\times$ B .E.(C -H ) = 615+435-347 - 2 $\times$ 414 = 1050 - 1175 = -125 kJ.