A possible reaction of ethane with chlorine is shown below. $$\mathrm { CH } _ { 3 } \mathrm { CH } _ { 3 } + \mathrm { Cl } _ { 2 } \longrightarrow \mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } \mathrm { Cl } + \mathrm { HCl }$$ This reaction could conceivably occur by the following chain mechanisms [1], [2], and [3]. Determine $\Delta$H for step [2]. [1] $\quad \mathrm { Cl } _ { 2 } \longrightarrow 2 \mathrm { Cl } \cdot$
[2] $\mathrm { Cl } ^ { \cdot } + \mathrm { CH } _ { 3 } \mathrm { CH } _ { 3 } \stackrel { \text { slow } } { \longrightarrow } \mathrm { HCl } + \mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } ^ \cdot$
$[ 3 ] \mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } ^ \cdot + \mathrm { Cl } _ { 2 } \stackrel { \text { fast } } { \longrightarrow } \mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } \mathrm { Cl } + \mathrm { Cl } ^\cdot$ Bond dissociation energies (kcal/mol): $$\begin{array} { | l l | }
\hline \text { Bond A-B } & \Delta \mathrm { H } \
\hline \mathrm { Cl } - \mathrm { Cl } & 58 \
\mathrm { CH } _ { 3 } - \mathrm { H } & 104 \
\mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } - \mathrm { H } & 98 \
\mathrm { CH } _ { 3 } \mathrm { CH } _ { 3 } & 88 \
\mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } - \mathrm { Cl } & 81 \
\mathrm { H } - \mathrm { Cl } & 103 \
\hline
\end{array}$$

Question

A possible reaction of ethane with chlorine is shown below. $$\mathrm { CH } _ { 3 } \mathrm { CH } _ { 3 } + \mathrm { Cl } _ { 2 } \longrightarrow \mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } \mathrm { Cl } + \mathrm { HCl }$$ This reaction could conceivably occur by the following chain mechanisms [1], [2], and [3]. Determine $\Delta$H for step [2]. [1] $\quad \mathrm { Cl } _ { 2 } \longrightarrow 2 \mathrm { Cl } \cdot$
[2] $\mathrm { Cl } ^ { \cdot } + \mathrm { CH } _ { 3 } \mathrm { CH } _ { 3 } \stackrel { \text { slow } } { \longrightarrow } \mathrm { HCl } + \mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } ^ \cdot$
$[ 3 ] \mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } ^ \cdot + \mathrm { Cl } _ { 2 } \stackrel { \text { fast } } { \longrightarrow } \mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } \mathrm { Cl } + \mathrm { Cl } ^\cdot$ Bond dissociation energies (kcal/mol): $$\begin{array} { | l l | } 
\hline \text { Bond A-B } & \Delta \mathrm { H } \
\hline \mathrm { Cl } - \mathrm { Cl } & 58 \
\mathrm { CH } _ { 3 } - \mathrm { H } & 104 \
\mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } - \mathrm { H } & 98 \
\mathrm { CH } _ { 3 } \mathrm { CH } _ { 3 } & 88 \
\mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } - \mathrm { Cl } & 81 \
\mathrm { H } - \mathrm { Cl } & 103 \
\hline
\end{array}$$

Quizplus · Accepted Answer

The enthalpy change for step [2] is equal to the difference between the bond dissociation energies of the reactants and the products. The reactants are Cl-Cl and CH3-CH3, and the products are HCl and CH3-CH2-. The bond dissociation energies are:Cl-Cl: 58 kcal/molCH3-CH3: 88 kcal/molHCl: 103 kcal/molCH3-CH2-: 98 kcal/molThe enthalpy change for step [2] is therefore:ΔH = (103 kcal/mol + 98 kcal/mol) - (58 kcal/mol + 88 kcal/mol) = -5 kcal/mol

A Possible Reaction of Ethane with Chlorine Is Shown Below