Solveeit Logo
Login

Question

Question: Bond dissociation enthalpy of\({{H}_{2}}\),\(C{{l}_{2}}\) and \(HCl\)are 434, 242, and 431 KJ/mol re...

Bond dissociation enthalpy ofH2{{H}_{2}},Cl2C{{l}_{2}} and HClHClare 434, 242, and 431 KJ/mol respectively. Enthalpy of formation of HCl is:
A. −93kJmol−1-93kJmo{{l}^{-1}}
B. 245kJmol−1245kJmo{{l}^{-1}}
C. 93kJmol−193kJmo{{l}^{-1}}
D. −245kJmol−1-245kJmo{{l}^{-1}}

Explanation

Solution

To solve this question we will first write the equation for 1 mole of HCl. As we know that enthalpy of formation ΔH\Delta His given by: ΔH=∑HR−∑HP\Delta H=\sum{{{H}_{R}}}-\sum{{{H}_{P}}}
where, HR{{H}_{R}}= Enthalpy of reactant
HP{{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 ΔH\Delta H.
- We will firstly write the equation for the formation of HCl as:
H2+Cl2→2HCl{{H}_{2}}+C{{l}_{2}}\to 2HCl
Now, we can write the equation of 1 mole of HCl as:
H22+Cl22→HCl2\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:

& \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{aligned}$$ **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.