Question

Calculate the enthalpy change for the process: $CC{l_4}\left( g \right) \to C\left( g \right) + 4Cl\left( g \right),$ and calculate bond enthalpy of $C - Cl\;$in $\;CC{l_4}$.
Given:${\Delta _{vap}}{H^\Theta }\left( {CC{l_4}} \right) = 30.5{\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}}$, ${\Delta _f}{H^ \ominus }(CC{l_4}) = - 135.5{\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}}$, ${\Delta _a}{H^\Theta }(C) = 715.0{\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}}$ where ${\Delta _a}{H^\Theta }$is enthalpy of atomization, ${\Delta _a}{H^\Theta }(C{l_2}) = 242{\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}}$

Answer 1

To answer this question, you should recall the concept of bond dissociation enthalpy and heat of reaction. Bond dissociation enthalpy is a measure of the strength of the chemical bond between any two species.

Complete step by step answer:
It is important to note that the breaking of a chemical bond is always an endothermic process. Thus, the enthalpy change associated with the breaking of a chemical bond is always positive. On the other hand, the formation of a chemical bond is almost always an endothermic process. In such cases, the enthalpy change will have a negative value.

The given chemical reactions are:
$1.\,CC{l_4}\left( l \right) \to CC{l_4}\left( g \right){\text{ }}{\Delta _{vap}}{H^ \ominus } = 30.5\;{\text{kJ mo}}{{\text{l}}^{{\text{ - 1}}}}$
$2.\,C\left( g \right) + 2C{l_2}\left( g \right) \to CC{l_{4}}\left( g \right){\text{ }}{\Delta _f}{H_ \ominus } = - 135.5{\text{kJ mo}}{{\text{l}}^{{\text{ - 1}}}}$
$3.\,C\left( s \right) \to C\left( g \right){\text{ }}{\Delta _a}H \ominus = 715.0\;{\text{kJ mo}}{{\text{l}}^{{\text{ - 1}}}}$
$4.C{l_2}\left( g \right) \to 2Cl\left( g \right){\text{ }}{\Delta _a}H \ominus = 242\;{\text{kJ mo}}{{\text{l}}^{{\text{ - 1}}}}$
Now to calculate the required enthalpy for the process $CC{l_4}\left( g \right) \to C\left( g \right) + 4Cl\left( g \right),$ we have to follow following algebraic process:
${\text{Equation}}\left( {\text{2}} \right){{ + 2 \times Equation}}\left( {\text{3}} \right){\text{ - Equation}}\left( {\text{1}} \right){\text{ - Equation}}\left( {\text{4}} \right)$.
Now, $\Delta H = {\Delta _a}{H^ \ominus } + 2{\Delta _a}{H^ \ominus }C{l_2} - {\Delta _{vap}}{H^ \ominus } - \Delta l{H^ \ominus }$.
Putting the values, we have
$= 715 + 2 \times 242 - 30.5 - \left( { - 135.5} \right)$.
Solving this equation, we get:
$\Delta H = 1304{\text{kJ mo}}{{\text{l}}^{{\text{ - 1}}}}$.
The bond enthalpy of $C - Cl\;{\text{ in }}\;CC{l_4} = \dfrac{{1304}}{4} = 326{\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}}$

Additional information:
- In ideal cases, it can be said that any bond is hundred percent ionic or covalent. But in reality, any bond shows partial ionic and covalent character. The components of any molecular: the cations polarize the anions and try to pull the electronic charge towards themselves and this increases the electric charge between the two ions. This is the case with a covalent bond, which is an electron charge density built up around the nuclei. Fajan’s rule is based on the fact that:
- Smaller the size of cation, larger the size of the anion will result in a more covalent character of the ionic bond.
- More the charge density of cation, greater is the covalent character of the ionic bond.
- Electronic configuration: For cations with same charge and size, the one, with $\left( {n - 1} \right){d^{n}}n{s^o}$ which is found in transition elements have greater covalent character than the cation with $n{s^{2}}n{p^6}$ electronic configuration, generally present s- block elements.

Note:
The value of bond enthalpy assesses the energy required to break or form the bond. Combined bond enthalpy for all broken and newly formed bonds during the reaction describes a total change in the energy of the system which is called change in enthalpy. This value of bond enthalpy i.e positive or negative, can determine whether the reaction is endothermic or exothermic.