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Question: Bond dissociation enthalpy of \({{H}_{2}}\) , \(C{{l}_{2}}\) and HCl are 434, 242 and 432 kJ/mol, re...

Bond dissociation enthalpy of H2{{H}_{2}} , Cl2C{{l}_{2}} and HCl are 434, 242 and 432 kJ/mol, respectively. Enthalpy of formation of HCl is:
(A) -93 kJ/mol
(B) 245 kJ/mol
(C) 93 kJ/mol
(D) -245 kJ/mol

Explanation

Solution

The general knowledge about bond dissociation energy, enthalpy of formation and Hess’s law, we can solve the given illustration easily.
- There are two possibilities according to the given options i.e. 245 kJ/mol and 93 kJ/mol; only the signs would be determined when we actually solve the problem.

Complete Solution :
Let us firstly understand the given terms;
- Bond dissociation enthalpy-
It is a measure of strength of a chemical bond. It can also be defined as the standard enthalpy change when the bond (A-B) is cleaved by homolysis to give fragments of A and B.

- Enthalpy of formation-
It is the change in the enthalpy during the formation of 1 mole of the substance from its constituent elements, with all the contributing substances in their standard states.

- Hess’s law-
The law states that the change in enthalpy for a reaction is the same whether the reaction takes place in one step or a series of steps regardless of path by which reaction occurs.
Now, we can describe the required equations in an illustration and solve it as follows,
Given that,
Bond dissociation enthalpy of H2{{H}_{2}} = 434 kJ/mol
Bond dissociation enthalpy of Cl2C{{l}_{2}} = 242 kJ/mol
Bond dissociation enthalpy of HCl = 431 kJ/mol

The reaction between hydrogen and chlorine molecules to form hydrogen chloride can be described as follows:
H2+Cl2→2HCl{{H}_{2}}+C{{l}_{2}}\to 2HCl
This can be divided into simpler steps as:

Step (I)- Dissociation of hydrogen and chlorine molecules into their respective atoms.
H2→2H Cl2→2Cl \begin{aligned} & {{H}_{2}}\to 2H \\\ & C{{l}_{2}}\to 2Cl \\\ \end{aligned}

Step (II)- Combination of these atoms to form hydrogen chloride.
2H+2Cl→2HCl2H+2Cl\to 2HCl
Since, the first step is an endothermic process thus, ΔH1>0\Delta {{H}_{1}}>0 . Whereas, the next step releases the energy thus, ΔH2<0\Delta {{H}_{2}}<0 .

So,
ΔH1=ΔHCl−Cl+ΔHH−H ΔH1=242+434=676kJ/mol \begin{aligned} & \Delta {{H}_{1}}=\Delta {{H}_{Cl-Cl}}+\Delta {{H}_{H-H}} \\\ & \Delta {{H}_{1}}=242+434=676kJ/mol \\\ \end{aligned}
And,
ΔH2=−2ΔHH−Cl ΔH2=−862kJ/mol \begin{aligned} & \Delta {{H}_{2}}=-2\Delta {{H}_{H-Cl}} \\\ & \Delta {{H}_{2}}=-862kJ/mol \\\ \end{aligned}

Thus, applying Hess’s law we get,
ΔH′=ΔH1+ΔH2 ΔH′=676−862=−186kJ/mol \begin{aligned} & \Delta H'=\Delta {{H}_{1}}+\Delta {{H}_{2}} \\\ & \Delta H'=676-862=-186kJ/mol \\\ \end{aligned}
This is the enthalpy of formation of two molecules of HCl. Enthalpy of formation of a molecule of HCl is -93 kJ/mol.
So, the correct answer is “Option A”.

Note: An alternative method to solve the same illustration can be the direct relationship of bond dissociation enthalpy and enthalpy of reaction as stated,
ΔHreaction=∑(B.E)reactant−∑(B.E)product\Delta {{H}_{reaction}}=\sum{{{\left( B.E \right)}_{reactant}}-\sum{{{\left( B.E \right)}_{product}}}} .
Solving by this method will give the same answer.