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# Bond dissociation enthalpy of${{H}_{2}}$,$C{{l}_{2}}$ and $HCl$are 434, 242, and 431 KJ/mol respectively. Enthalpy of formation of HCl is:A. $-93kJmo{{l}^{-1}}$B. $245kJmo{{l}^{-1}}$C. $93kJmo{{l}^{-1}}$D. $-245kJmo{{l}^{-1}}$

Last updated date: 20th Jun 2024
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Hint: To solve this question we will first write the equation for 1 mole of HCl. As we know that enthalpy of formation $\Delta H$is given by: $\Delta H=\sum{{{H}_{R}}}-\sum{{{H}_{P}}}$
where, ${{H}_{R}}$= Enthalpy of reactant
${{H}_{P}}$= Enthalpy of product

Complete step by step solution:
- Enthalpy of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, in their most stable standard states. It is denoted by the symbol $\Delta H$.
- We will firstly write the equation for the formation of HCl as:
${{H}_{2}}+C{{l}_{2}}\to 2HCl$
Now, we can write the equation of 1 mole of HCl as:
$\dfrac{{{H}_{2}}}{2}+\dfrac{C{{l}_{2}}}{2}\to \dfrac{HCl}{2}$
We wrote this equation because enthalpy of formation is basically for one mole only.
- Now, as we are being provided with the bond dissociation enthalpy, so we will write the formula and solve the answer:
\begin{align} & \Delta H=\sum{{{H}_{R}}}-\sum{{{H}_{P}}} \\ & =\left[ \dfrac{{{H}_{2}}}{2}+\dfrac{C{{l}_{2}}}{2} \right]-\left[ 431 \right] \\ & =\left[ 218+121 \right]-\left[ 431 \right] \\ & =338-431 \\ & =-93KJ/mol \\ \end{align}

Hence, we can say that the correct option is (A), that is Enthalpy of formation of HCl is $-93kJmo{{l}^{-1}}$