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.H2C = CH2(g) + H2(g) ----- H3C-CH3(g) at 298 K will be

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.H2C = CH2(g) + H2(g) -----> H3C-CH3(g) at 298 K will be (A) +250 kJ (B) -250 kJ (C) +125 kJ (D) -125 kJ

Tardigrade Admin · Accepted Answer

Δ H = ∑ B.E. of reactants - ∑ B.E. of products = B.E.(C = C)+ 4 × B.E.(C-H) + B.E.(H-H) -B.E.(C-C) - 6 × B.E.(C-H) = B.E.(C = C) + B.E.(H-H) - B.E.(C-C) - 2 × B .E.(C -H ) = 615+435-347 - 2 × 414 = 1050 - 1175 = -125 kJ.

Q. 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 $m o l^{- 1}$ the value of enthalpy change for the reaction. $H_{2} C = C H_{2} (g) + H_{2} (g) - - - - - > H_{3} C - C H_{3} (g)$ at 298 K will be

+250 kJ

-250 kJ

+125 kJ

-125 kJ

$Δ H = \sum$ 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.

Q. 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.H2​C=CH2​(g)+H2​(g)−−−−−>H3​C−CH3​(g) at 298 K will be