1. Tardigrade
  2. Question
  3. Chemistry
  4. 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 Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

Thermodynamics

Solution:

$\Delta H = \sum$ 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.