What is the value of $\Delta H ^ { \circ } \text { in } \mathrm { kcal } / \mathrm { mol }$ for the reaction shown below?
$\left( \mathrm { CH } _ { 3 } \right) _ { 3 } \mathrm { C } - \mathrm { H } + \mathrm { Cl } - \mathrm { Cl } \rightarrow \left( \mathrm { CH } _ { 3 } \right) _ { 3 } \mathrm { C } - \mathrm { Cl } + \mathrm { H } - \mathrm { Cl }$
Bond energies are:
$\begin{array} { l }
\quad \left( \mathrm { CH } _ { 3 } \right) 3 \mathrm { C } - \mathrm { H } = 91 \mathrm { kcal } / \mathrm { mol } \
\left( \mathrm { CH } _ { 3 } \right) 3 \mathrm { C } - \mathrm { Cl } = 78.5 \mathrm { kcal } / \mathrm { mol } \
\mathrm { Cl } - \mathrm { Cl } = 58 \mathrm { kcal } / \mathrm { mol } \
\mathrm { H } - \mathrm { Cl } = 103 \mathrm { kcal } / \mathrm { mol }
\end{array}$

Question

What is the value of $\Delta H ^ { \circ } \text { in } \mathrm { kcal } / \mathrm { mol }$ for the reaction shown below? 
$\left( \mathrm { CH } _ { 3 } \right) _ { 3 } \mathrm { C } - \mathrm { H } + \mathrm { Cl } - \mathrm { Cl } \rightarrow \left( \mathrm { CH } _ { 3 } \right) _ { 3 } \mathrm { C } - \mathrm { Cl } + \mathrm { H } - \mathrm { Cl }$
Bond energies are:
$\begin{array} { l } 
\quad \left( \mathrm { CH } _ { 3 } \right) 3 \mathrm { C } - \mathrm { H } = 91 \mathrm { kcal } / \mathrm { mol } \
\left( \mathrm { CH } _ { 3 } \right) 3 \mathrm { C } - \mathrm { Cl } = 78.5 \mathrm { kcal } / \mathrm { mol } \
\mathrm { Cl } - \mathrm { Cl } = 58 \mathrm { kcal } / \mathrm { mol } \
\mathrm { H } - \mathrm { Cl } = 103 \mathrm { kcal } / \mathrm { mol }
\end{array}$

Quizplus · Accepted Answer

The value of $\Delta H ^ { \circ }$ is calculated by subtracting the sum of the bond energies of the products from the sum of the bond energies of the reactants. Here, $\Delta H ^ { \circ } = (91 + 58) - (78.5 + 103) = 149 - 181.5 = -32.5 \, \mathrm{kcal/mol}$.

What Is the Value Of ΔH∘ in kcal/mol\Delta H ^ { \circ } \text { in } \mathrm { kcal } / \mathrm { mol }ΔH∘ in kcal/mol

What Is the Value Of $\Delta H ^ { \circ } \text { in } \mathrm { kcal } / \mathrm { mol }$