Question

Use the molar bond enthalpy data in the table to estimate the value of Delta H degree rxn for the equation CCI4(g) + 2F2(g) rightarrow CF4(g) + 2XCI2(g) Average molar bond enthalpies. (Hbond) The bonding in the molecules is shown here.

Answer 1

Concepts and reason

This problem can be solved by using the general equation for the heat of a reaction.

Fundamentals

The heat of a reaction or change in enthalpy of a reaction can be estimated from bond energy. Heat energy is estimated after balancing the given equation.

Bond energy is defined as the amount of energy required to break the bond of 1 mole of a gaseous compound to produce its constituents.

The change in enthalpy of a reaction is the difference of summation of bond energies of bond broken and the summation of the bond energies of the bonds formed. It is written as follow.

${\rm{\Delta }}{H^o} = \sum {{\rm{\Delta }}{H_{\left( {{\rm{reactant bonds broken}}} \right)}}} - \sum {{\rm{\Delta }}{H_{\left( {{\rm{product bonds formed}}} \right)}}}$ …… (1)

The reaction is as follows:

${\rm{CC}}{{\rm{l}}_{\rm{4}}}\left( g \right) + 2{{\rm{F}}_2}\left( g \right) \to {\rm{C}}{{\rm{F}}_4}\left( g \right) + 2{\rm{C}}{{\rm{l}}_{\rm{2}}}\left( g \right)$

First calculate the bond energy of the reactants or the bonds broken as follow.

$\sum {{\rm{\Delta }}{H_{\left( {{\rm{reactant bonds broken}}} \right)}}} = 4{\rm{\Delta }}H\left( {{\rm{C}} - {\rm{Cl}}} \right) + 2{\rm{\Delta }}H\left( {{\rm{F}} - {\rm{F}}} \right)$

Substitute ${\rm{331 kJ mo}}{{\rm{l}}^{ - 1}}$ for ${\rm{\Delta }}H\left( {{\rm{C}} - {\rm{Cl}}} \right)$ and ${\rm{155 kJ mo}}{{\rm{l}}^{ - 1}}$ for ${\rm{\Delta }}H\left( {{\rm{F}} - {\rm{F}}} \right)$ .

$\begin{array}{c}\\\sum {{\rm{\Delta }}{H_{\left( {{\rm{reactant bonds broken}}} \right)}}} = 4\left( {{\rm{331 kJ mo}}{{\rm{l}}^{ - 1}}} \right) + 2\left( {{\rm{155 kJ mo}}{{\rm{l}}^{ - 1}}} \right)\\\\ = 1634{\rm{ kJ mo}}{{\rm{l}}^{ - 1}}\\\end{array}$

Now, calculate bond energy of product bonds formed.

$\sum {{\rm{\Delta }}{H_{\left( {{\rm{product bonds formed}}} \right)}}} = {\rm{4\Delta }}H\left( {{\rm{C}} - {\rm{F}}} \right) + 2{\rm{\Delta }}H\left( {{\rm{Cl}} - {\rm{Cl}}} \right)$

Substitute ${\rm{439 kJ mo}}{{\rm{l}}^{ - 1}}$ for ${\rm{\Delta }}H\left( {{\rm{C}} - {\rm{F}}} \right)$ and ${\rm{243 kJ mo}}{{\rm{l}}^{ - 1}}$ for ${\rm{\Delta }}H\left( {{\rm{Cl}} - {\rm{Cl}}} \right)$ .

$\begin{array}{c}\\\sum {{\rm{\Delta }}{H_{\left( {{\rm{product bonds formed}}} \right)}}} = 4\left( {{\rm{439 kJ mo}}{{\rm{l}}^{ - 1}}} \right) + 2\left( {{\rm{243 kJ mo}}{{\rm{l}}^{ - 1}}} \right)\\\\ = 2242{\rm{ kJ mo}}{{\rm{l}}^{ - 1}}\\\end{array}$

Now, substitute 1634 kJ for $\sum {{\rm{\Delta }}{H_{\left( {{\rm{reactant bonds broken}}} \right)}}}$ and 2242 kJ for $\sum {{\rm{\Delta }}{H_{\left( {{\rm{product bonds formed}}} \right)}}}$ in equation (1).

$\begin{array}{c}\\{\rm{\Delta }}{H^o} = \sum {{\rm{\Delta }}{H_{\left( {{\rm{reactant bonds broken}}} \right)}}} - \sum {{\rm{\Delta }}{H_{\left( {{\rm{product bonds formed}}} \right)}}} \\\\ = \left( {1634 - 2242} \right){\rm{ kJ mo}}{{\rm{l}}^{ - 1}}\\\\ = - 608{\rm{ kJ mo}}{{\rm{l}}^{ - 1}}\\\end{array}$

Ans:

The heat of the reaction is $- 608{\rm{ kJ mo}}{{\rm{l}}^{ - 1}}$ .