Question

Bromination of isobutane is a two-step reaction, as shown below. Using the table of bond dissociation energies, calculate the enthalpy of each step and the enthalpy of the overall reaction.

I got the first enthalpy right but the second and the third one wrong. Help!

Answer 1

Concepts and reason

The concept used to solve this problem is based on enthalpy of reaction.

Enthalpy of a reaction is defined as the enthalpy change when stoichiometric quantities of reactants react to form the products in the reaction. The energy required to break a bond is known as bond dissociation energy and the same amount energy released while forming a bond. It may be noted that bond dissociation as well as bond formation take place in chemical reaction.

Fundamentals

Enthalpy of reaction can be calculated by using bond dissociation energy as follow.

$\Delta H = \sum \Delta H\left( {{\rm{reactants}}} \right) - \sum \Delta H\left( {{\rm{products}}} \right)$

Here, $\Delta H$ is heat of reaction.

Part 1

The reaction for combustion of acetylene can be written as follow.

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{\rm{C}} - {\rm{H}} + {\rm{Br}} \to {\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{{\rm{C}}^ \bullet }{\rm{ + H}} - {\rm{Br}}$

Enthalpy of reaction for this reaction can be written as follow.

$\Delta {H_1} = \Delta H\left( {{{\left( {{\rm{C}}{{\rm{H}}_3}} \right)}_3}{\rm{C}} - {\rm{H}}} \right) - \Delta H\left( {{\rm{H}} - {\rm{Br}}} \right)$

Substitute $400{\rm{ kJ mo}}{{\rm{l}}^{ - 1}}$ for $\Delta H\left( {{{\left( {{\rm{C}}{{\rm{H}}_3}} \right)}_3}{\rm{C}} - {\rm{H}}} \right)$ and ${\rm{366 kJ mo}}{{\rm{l}}^{ - 1}}$ for $\Delta H\left( {{\rm{H}} - {\rm{Br}}} \right)$ .

$\begin{array}{c}\\\Delta {H_1} = \left( {400 - 366} \right){\rm{ kJ mo}}{{\rm{l}}^{ - 1}}\\\\ = + 34{\rm{ kJ mo}}{{\rm{l}}^{ - 1}}\\\end{array}$

Part 2

The reaction for combustion of acetylene can be written as follows:

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{{\rm{C}}^ \bullet } + {\rm{Br}} - {\rm{Br}} \to {\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{\rm{C}} - {\rm{Br}} + {\rm{Br}}$

Enthalpy of reaction for this reaction can be written as follows:

$\Delta {H_2} = \Delta H\left( {{\rm{Br}} - {\rm{Br}}} \right) - \Delta H\left( {{{\left( {{\rm{C}}{{\rm{H}}_3}} \right)}_3}{\rm{C}} - {\rm{Br}}} \right)$

Substitute ${\rm{292 kJ mo}}{{\rm{l}}^{ - 1}}$ for $\Delta H\left( {{{\left( {{\rm{C}}{{\rm{H}}_3}} \right)}_3}{\rm{C}} - {\rm{Br}}} \right)$ and ${\rm{193 kJ mo}}{{\rm{l}}^{ - 1}}$ for $\Delta H\left( {{\rm{Br}} - {\rm{Br}}} \right)$ .

$\begin{array}{c}\\\Delta {H_2} = \left( {193 - 292} \right){\rm{ kJ mo}}{{\rm{l}}^{ - 1}}\\\\ = - 99{\rm{ kJ mo}}{{\rm{l}}^{ - 1}}\\\end{array}$

Part 3

The reaction for combustion of acetylene can be written as follows:

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{\rm{C}} - {\rm{H}} + {\rm{Br}} - {\rm{Br}} \to {\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{\rm{C}} - {\rm{Br}} + {\rm{H}} - {\rm{Br}}$

Enthalpy of overall reaction is calculated by adding enthalpy of step 1 and step 2.

$\Delta H = \Delta {H_1} + \Delta {H_2}$

Substitute ${\rm{34 kJ mo}}{{\rm{l}}^{ - 1}}$ for $\Delta {H_1}$ and $- 99{\rm{ kJ mo}}{{\rm{l}}^{ - 1}}$ for $\Delta {H_2}$ .

$\begin{array}{c}\\\Delta H = \left( {34 - 99} \right){\rm{ kJ mo}}{{\rm{l}}^{ - 1}}\\\\ = - 65{\rm{ kJ mo}}{{\rm{l}}^{ - 1}}\\\end{array}$

Ans: Part 1

Enthalpy of the reaction is $+ 34{\rm{ kJ mo}}{{\rm{l}}^{ - 1}}$ .