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
$C{H}_4\left(g\right)+C{l}_2\left(g\right)\stackrel{\text{light}}{\to }C{H}_{3}Cl\left(g\right)+HCl\left(g\right)$

The correct statement is

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

Answer 1

The correct option is D The reaction is exothermic with $\Delta H^{\circ }=-25\text{kcal}mo{l}^{-1}.$

$C{H}_4+C{l}_2\stackrel{\text{light}}{\to }C{H}_{3}Cl+HCl$ this reaction is obtained from given reaction

$C{H}_3-H\to C{H}_3^{⨀}+H^{⨀}...\left(1\right)$
$Cl-Cl\to C{l}^{⨀}+C{l}^{⨀}...\left(2\right)$
$C{H}_3-Cl\to CH{{}_3}^{⨀}+C{l}^{⨀}...\left(3\right)$
$H-Cl\to H^{⨀}+C{l}^{⨀}...\left(4\right)$
Hence $\Delta H=105+58-85-103=-25KCal/mole$