Question

Heat evolved in the reaction ${ H }_{ 2 }+{ Cl }_{ 2 }\rightarrow 2HCl$ is 182 KJ. Bond energies of H-H and Cl-Cl are 430 and 242 KJ/mol respectively.
Thus H-Cl bond energy is:
(a) 245 KJ/mol
(b) 427 KJ/mol
(c) 336 KJ/mol
(d) 154 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:
In the reaction given in the question:
${ H }_{ 2 }+{ Cl }_{ 2 }\rightarrow 2HCl$

Two moles of HCl is produced from one mole of hydrogen and one mole of chlorine gas. The heat of this reaction is: -182 KJ. But this enthalpy change accounts for two moles of HCl. We need to find the heat of formation of HCl which is in terms of KJ/mol. Hence we will multiply the above reaction by $\cfrac { 1 }{ 2 } $ and calculate the heat of formation of HCl using that reaction. Hence the heat of formation of HCl will be: $\cfrac { 1 }{ 2 } { H }_{ 2 }+\dfrac { 1 }{ 2 } { Cl }_{ 2 }\rightarrow HCl$
$\Delta { H }_{ f }=\cfrac { -182\quad KJ }{ 2\quad mol } =-91KJ/mol$

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,
$-91=(\dfrac { 1 }{ 2 } \times 242\quad +\cfrac { 1 }{ 2 } \times 430)-\Delta { H }_{ B(H-Cl) }$
$\Rightarrow \Delta { H }_{ B(H-Cl) }=91+215+121=427KJ/mol$
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. It is not meant for the formation of 2 moles of a substance therefore we divided the enthalpy change -182 KJ by 2 in order to get the enthalpy of formation of HCl.