Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

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. $-245kJmo{{l}^{-1}}$

seo-qna
Last updated date: 20th Jun 2024
Total views: 54.6k
Views today: 0.54k
Answer
VerifiedVerified
54.6k+ views
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 $-93kJmo{{l}^{-1}}$

Additional information:
- We can say that Enthalpy of formation is a special case of standard enthalpy of reaction where two or more reactants combine to form one mole of product.
- It is found that the Bond dissociation enthalpy is the change in enthalpy when one mole of covalent bonds of a compound is broken to form products in the gaseous phase.

Note:- We have seen that here the enthalpy of formation is negative, which indicates that the formation of a compound is exothermic, that is it takes less amount of energy to break bonds than the amount of energy that is released while making the bonds.
- If there is a positive enthalpy of formation then it indicates that the formation of a compound is endothermic, that is it takes a greater amount of energy to break bonds than the amount of energy that is released while making the bonds.