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
$C{H}_4(g)+C{l}_2(g)\stackrel{\text{light}}{\to }C{H}_{3}Cl(g)+HCl(g)$
The correct statement is

• A. Initiation step is exothermic with $\Delta H^{\circ }=-58kcalmo{l}^{-1}$.
• B. Propagation step involving ${}^{\circ }C{H}_3$ formation is exothermic with $\Delta H^{\circ }=-2kcalmo{l}^{-1}$.
• C. Propagation step involving $C{H}_{3}Cl$ formation is endothermic with $\Delta H^{\circ }=+27kcalmo{l}^{-1}$.
• D. The reaction is exothermic with $\Delta H^{\circ }=-25kcalmo{l}^{-1}$.

Answer 1

Answer: (D)

$C{H}_4+C{l}_2\stackrel{\text{ light }}{\to }C{H}_{3}Cl+HCl$ this reaction is obtained from given reaction.
$C{H}_3-H⟶C{H}_3^{\odot }+H^{\circ }\dots (1)$
$Cl-Cl⟶C{l}^{\circ }+C{l}^{\circ }$...(2)
$C{H}_3-Cl⟶C{H}_3{}^{\circ }+C{l}^{\circ }$ ...(3)
$H-Cl⟶H^{\circ }+C{l}^{\circ }$ ...(4)
$(1)+(2)-(3)-(4)$
Hence $\Delta H=105+58-85-103=-25KCal/mole$