Question

The bond dissociation energies of gases ${{H}_{2}},C{{l}_{2}}$ and HCl are 104,58 and 103 $kcal\text{ }mo{{l}^{-1}}$respectively. Calculate the enthalpy of formation of HCl (g).

Answer 1

Enthalpy of formation $\Delta {{H}_{f}}$,is the basically the change in enthalpy that takes place when one mole of a compound is formed in their most stable state of aggregation (that is stable state of aggregation at pressure of 1atm and temperature of 298.15K) from its elements. We will use the formula for calculating the enthalpy of formation of HCl (g):
$\Delta {{H}_{f}}=-\left[ \Delta {{H}_{{{B}_{\left( HCl \right)}}}}-\dfrac{1}{2}\left( \Delta {{H}_{{{B}_{\left( {{H}_{2}} \right)}}}}+\Delta {{H}_{{{B}_{\left( C{{l}_{2}} \right)}}}} \right) \right]$

Complete Step by step solution:
The enthalpy of formation is denoted by $\Delta {{H}_{f}}$. We will write the reaction of formation of HCl as:
$\dfrac{1}{2}{{H}_{2}}+\dfrac{1}{2}C{{l}_{2}}\to HCl$
The enthalpy of formation of HCl is denoted by $\Delta {{H}_{f}}$, that will be equal to negative times bond dissociation energy of product side that is HCl minus bond dissociation energy of ${{H}_{2}}\text{ }and\text{ }C{{l}_{2}}$.
Bond dissociation energy of HCl, ${{H}_{2}}\text{ }and\text{ }C{{l}_{2}}$is denoted by $\Delta {{H}_{{{B}_{\left( HCl \right)}}}}$,$\Delta {{H}_{{{B}_{\left( {{H}_{2}} \right)}}}}+\Delta {{H}_{{{B}_{\left( C{{l}_{2}} \right)}}}}$
We will write the equation as:
$\Delta {{H}_{f}}=-\left[ \Delta {{H}_{{{B}_{\left( HCl \right)}}}}-\dfrac{1}{2}\left( \Delta {{H}_{{{B}_{\left( {{H}_{2}} \right)}}}}+\Delta {{H}_{{{B}_{\left( C{{l}_{2}} \right)}}}} \right) \right]$
We are being provided with the bond dissociation energies of gases ${{H}_{2}},C{{l}_{2}}$as 104 , 58, and of HCl as 103.
Now, we will put all the values given in the equation,

\implies & -\left[ 103-\dfrac{1}{2}\left( 104+58 \right) \right] \\\ \implies & -22\text{ }Kcal \\\ \end{aligned}$$ **Hence, we can conclude that the enthalpy of formation of HCl (g) is -22 Kcal.** **Note:** \- The main difference between Enthalpy of formation and bond dissociation enthalpy is that Enthalpy of formation is a property that shows how stable a substance is. It is extremely useful in calculating reaction enthalpies. \- Whereas bond dissociation enthalpy is a measure of the strength of an individual bond and can be used to estimate $\Delta H$in the calculations.