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Question: Bond dissociation enthalpy of \({H_2},C{l_2},HCl\) are 434,242 and 431 KJ \(mo{l^{ - 1}}\) respectiv...

Bond dissociation enthalpy of H2,Cl2,HCl{H_2},C{l_2},HCl are 434,242 and 431 KJ mol1mo{l^{ - 1}} respectively.
Enthalpy of formation of HClHClis:
A: 93 - 93KJ mol1mo{l^{ - 1}}
B: 245 KJ mol1mo{l^{ - 1}}
C: 93 KJ mol1mo{l^{ - 1}}
D: -245 KJ mol1mo{l^{ - 1}}

Explanation

Solution

Enthalpy of formation refers to the amount of heat released or absorbed when a compound is formed. Bond-dissociation energy is a measure of heat required to break a molecule into atoms.

Complete step by step answer:
We know that bond dissociation enthalpy/energy is the amount of energy required to break a chemical bond A-B. It defines the measure of strength of a chemical bond. Now let us understand the enthalpy range of a reaction. The amount of heat evolved or absorbed in a chemical reaction when the number of moles of the reactants as represented by the chemical equation have completely reacted, is called the heat of reaction or enthalpy range of reaction. It is represented by ΔH\Delta H . There are different types of heat/enthalpies of reaction like enthalpy of combustion of a substance enthalpy of formation, bond dissociation enthalpy etc.
Now according to the question, we have to find the enthalpy of formation of 1 mole of HCl which can be done as:-
12H2(g)+12Cl2(g)HCl(g)\dfrac{1}{2}{H_2}(g) + \dfrac{1}{2}C{l_2}(g) \to HCl(g) ΔH\Delta H=? ----(1)
Now bond dissociation enthalpy of H2{H_2} is 434 KJ mol1mo{l^{ - 1}}. When H2{H_2} bond H-H is break, it releases two atoms of hydrogen
H2(g)2H(g){H_2}(g) \to 2H(g) ; ΔH\Delta H= 434 KJ mol1mo{l^{ - 1}} ----(2)
The bond Cl2C{l_2} dissociates/breaks to give 2 atoms of Cl and the bond dissociation enthalpy of Cl2C{l_2} given is 242 KJ mol1mo{l^{ - 1}}
Cl2(g)2Cl(g)C{l_2}(g) \to 2Cl(g) ; ΔH\Delta H= 242 KJ mol1mo{l^{ - 1}} -----(3)
When bond dissociation of HCl is done, it releases 1 atom of H and 1 atom of Cl and its bond dissociation enthalpy is 431 KJ mol1mo{l^{ - 1}}
HCl(g)H(g)+Cl(g)HCl(g) \to H(g) + Cl(g) ; ΔH\Delta H= 431 KJ mol1mo{l^{ - 1}} ----(4)
Now we know that according to Hess’s law the total amount of heat evolved or absorbed in a reaction is the same whether the reaction takes place in one step or in a number of steps.
Thus we have to obtain the equation(1) by using equation (2), (3) and (4).
Now we need a 12\dfrac{1}{2}mole of H2{H_2} and 12\dfrac{1}{2} mole of Cl2C{l_2} in order to form products of equation (1). So multiply Equation (2) and (3) by 12\dfrac{1}{2} and add both of them.
[H2(g)2H(g){H_2}(g) \to 2H(g) ΔH\Delta H= 434 KJ mol1mo{l^{ - 1}} ] ×12 \times \dfrac{1}{2}
[Cl2(g)2Cl(g)C{l_2}(g) \to 2Cl(g) ΔH\Delta H= 242 KJ mol1mo{l^{ - 1}}] ×12 \times \dfrac{1}{2}
Overall reaction
12H2(g)+12Cl2H(g)+Cl(g);ΔH=12×434+12×242\dfrac{1}{2}{H_2}(g) + \dfrac{1}{2}C{l_2} \to H(g) + Cl(g);\Delta H = \dfrac{1}{2} \times 434 + \dfrac{1}{2} \times 242
Therefore 12H2(g)+12Cl2H(g)+Cl(g);ΔH=338\dfrac{1}{2}{H_2}(g) + \dfrac{1}{2}C{l_2} \to H(g) + Cl(g);\Delta H = 338 KJ mol1mo{l^{ - 1}} ----(5)
The enthalpy of this reaction is 338 KJ mol1mo{l^{ - 1}} as calculated. We have added equations because according to Hess’s law, Equations can be subtracted, added as per requirement.
Now we need HCl (g) on the product side but in equation (4), it is given on the reactant side. So in order to obtain HCl (g) as a product we have to subtract equation (4) from equation (5)
12H2(g)+12Cl2HCl(g);ΔH=93KJ/mol1\dfrac{1}{2}{H_2}(g) + \dfrac{1}{2}C{l_2} \to HCl(g);\Delta H = - 93KJ/mo{l^{ - 1}}
Which is the required equation along with its enthalpy of formation, ΔH\Delta H. Thus we found enthalpy of formation of HCl as -93 KJ mol1mo{l^{ - 1}}.
Hence option (A) is the correct answer.

Note:
The sign for the ΔH\Delta H has a great significance. The negative value of ΔH\Delta Hsignifies that the heat is released in the reaction and the positive value of ΔH\Delta Hsignifies that the heat is absorbed/required during the reaction.