Question

The bond dissociation energies of gaseous $H_{2}$, $Cl_{2}$, and HCl are 104, 58, and 103 kcal/mol respectively. Calculate the energy of formation of HCl gas.

Answer 1

"Enthalpy:

It is defined as the "sum" of "internal energy" and "product" of "pressure and volume".

$H\quad =\quad U\quad +\quad PV$

Relation between internal energy and enthalpy is $\Delta H\quad =\quad \Delta U\quad +\quad \Delta nRT$.

"Standard enthalpy of formation":

The "enthalpy change" when "one mole" of a "substance" in "standard state" is formed from its "constituent element" in their standard state

From the given,

The "bond dissociation energy" of ${ H }_{ 2 }$= 104 kcal

The "bond dissociation energy" of${ Cl }_{ 2 }$ = 58 kcal

The "bond dissociation energy" of HCl = 103 kcal

The given chemical reaction is as follows.

$\frac { 1 }{ 2 } { H }_{ 2 }\quad +\quad \frac { 1 }{ 2 } { Cl }_{ 2 }\quad \rightarrow \quad HCl$

${ \Delta}_{f}H\quad (HCl)\quad =\quad { \Delta H}_{ Reactants }\quad -\quad { \Delta H}_{ products }$

$=\quad \frac { 1 }{ 2 } (104)\quad +\quad \frac { 1 }{ 2 } (58)\quad -\quad 103\quad{ \Delta}_{f}H\quad (HCl)\quad =\quad -22\quad kcal$"