Question

Given that, bond energies of $H-H$ and $Cl-Cl$ are $430Kj/mol$ and $240Kj/mol$ respectively. $\Delta {{H}_{f}}$ for HCl is $-90Kj/mol$. Bond enthalpy of HCl is:
A. $380KJmo{{l}^{-1}}$
B. $425KJmo{{l}^{-1}}$
C. $245KJmo{{l}^{-1}}$
D. $290KJmo{{l}^{-1}}$

Answer 1

Hint: Bond Enthalpy can also be known by the name bond energy which is a quantity which generally tells about the strength of a chemical bond or we can say that it tells us about the stability of any compound.

Complete Step by step solution: The bond enthalpy of a chemical bond can be defined as the total amount of energy required to break 1 mole of that chemical bond. Breaking of a chemical bond is always an endothermic process in which heat is given to the system as energy required to break the bond is very high so these types of reactions are kept in the category of endothermic process. The enthalpy change associated with the breaking of a chemical bond is always positive while in case of the formation of a chemical bond is almost always an endothermic process and in such cases the enthalpy change will have a negative value. Now we know that
\[H-H+Cl-Cl\to 2HCl\]
From this equation $\Delta {{H}_{f}}$for HCl can be calculated by:
$\Delta {{H}_{f}}(HCl)=($Bond energy of $H)+($Bond Energy Of $Cl-Cl)-2($Bond energy of $HCl)$
$-90=(430)+(240)-2($Bond energy of $HCl)$
therefore Bond enthalpy of $HCl=\dfrac{430+240+90}{2}=380KJmo{{l}^{-1}}$

Hence option A is the correct answer.

Note: The higher the value of value of bond enthalpy shows stronger the bond is and high energy is required to break the bond. Bond enthalpy can be calculated directly if everything in the reaction is in a gaseous state. If it is in the liquid state, it needs extra energy to convert it from liquid state to gaseous state.