Question

The amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by $s$-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below:

For the following reaction
$CH_4(g) + Cl_2(g) \xrightarrow{\text{light}}CH_3Cl(g) + HCl(g)$
The correct statement is

• A. Initiation step is exothermic with $\Delta H ^{\circ}=-58 \,kcal \,mol ^{-1}$.
• B. Propagation step involving ${ }^{\circ} CH _{3}$ formation is exothermic with $\Delta H ^{\circ}=-2\, kcal\, mol ^{-1}$.
• C. Propagation step involving $CH _{3} Cl$ formation is endothermic with $\Delta H ^{\circ}=+27\, kcal\, mol ^{-1}$.
• D. The reaction is exothermic with $\Delta H ^{\circ}=-25\, kcal\, mol ^{-1}$.

Answer 1

Answer: (D)

$CH _{4}+ Cl _{2} \xrightarrow{\text { light }} CH _{3} Cl + HCl$ this reaction is obtained from given reaction.
$CH _{3}- H \longrightarrow CH _{3}^{\odot}+ H ^{\circ} \ldots(1)$
$Cl - Cl \longrightarrow Cl ^{\circ}+ Cl ^{\circ} $...(2)
$CH _{3}- Cl \longrightarrow CH _{3}{ }^{\circ}+ Cl ^{\circ}$ ...(3)
$H - Cl \longrightarrow H ^{\circ}+ Cl ^{\circ}$ ...(4)
$(1)+(2)-(3)-(4)$
Hence $\Delta H =105+58-85-103=-25 \,KCal / mole$