Question

What is the bond dissociation energy of HCl, $\Delta { H }_{ f }(HCl)=-93$, B.E. (${ H }_{ 2 }$)=434, B.E.($ { Cl }_{ 2 }$)=242 (in KJ/mol).
(A) 532 KJ/mol
(B) 431 KJ/ mol
(C) 332 KJ/mol
(D) 232 KJ/mol

Answer 1

Hint: The enthalpy of formation of a substance is equal to the difference between the sum of the bond enthalpies of the reactants multiplied with their respective stoichiometric coefficient and the sum of the bond enthalpies of the products multiplied by their respective stoichiometric coefficients.

Complete step by step answer:
To solve this question we first need to understand bond energy. It refers to the energy which is required to break one mole of bonds present between the atoms in a gaseous molecule. For diatomic molecules, the bond dissociation energy is equal to their enthalpies of atomization. But for polyatomic molecules, the bond energy for a particular bond is not the same when it is present in different types of molecules or even in the same molecule. Hence in a molecule of methane, the bond dissociation energy for the first C-H bond is 427 KJ/mol, for the second C-H bond it is 439 KJ/mol and so on. Therefore for polyatomic molecules the bond dissociation energy is actually an average value. Now let us solve the question, We will first write the equation for the formation of 1 mole of HCl:
$ \cfrac { 1 }{ 2 } { H }_{ 2 }(g)+\cfrac { 1 }{ 2 } { Cl }_{ 2 }(g)\rightarrow HCl(g)$

One mole of HCl is produced from half mole of hydrogen gas and half mole of chlorine gas. The enthalpy of formation of HCl is -93 KJ/mol which is also equal to the heat of the reaction in this case.
Now the enthalpy of formation of HCl will be equal to the difference between the sum of the bond enthalpies of the reactants multiplied with their respective stoichiometric coefficient and the sum of the bond enthalpies of the products multiplied by their respective stoichiometric coefficients i.e.

$\begin{matrix} \Delta { H }_{ f } \\ Enthalpy\quad of \\ formation\quad of\quad HCl \end{matrix}=\begin{matrix} (\cfrac { 1 }{ 2 } \Delta { H }_{ B(Cl-Cl) }+\cfrac { 1 }{ 2 } \Delta { H }_{ B(H-H) }) \\ sum\quad of\quad bond\quad enthalpies \\ of\quad the\quad reactants \end{matrix}-\begin{matrix} \Delta { H }_{ B(H-Cl) } \\ Bond\quad enthalpy \\ of\quad HCl \end{matrix}$
Hence,
$-93=(\dfrac { 1 }{ 2 } \times 242\quad +\cfrac { 1 }{ 2 } \times 434)-\Delta { H }_{ B(H-Cl) }$
$\Rightarrow \Delta { H }_{ B(H-Cl) }=93+217+121=431KJ/mol$
So, the correct answer is “Option B”.

Note: The enthalpy of formation of a substance involves the formation of 1 mole of the substance from its elements under given conditions of temperature and pressure. Bond dissociation energy and bond dissociation enthalpy are actually different quantities. The bond dissociation energy values are measured at 0K using spectroscopic methods while the bond dissociation enthalpy values are calculated using heat capacities the pressure-volume work. But their values are close and hence are used interchangeably.