Question
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 are 434,242 and 431 KJ mol−1 respectively.
Enthalpy of formation of HClis:
A: −93KJ mol−1
B: 245 KJ mol−1
C: 93 KJ mol−1
D: -245 KJ mol−1
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 . 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:-
21H2(g)+21Cl2(g)→HCl(g) ΔH=? ----(1)
Now bond dissociation enthalpy of H2 is 434 KJ mol−1. When H2 bond H-H is break, it releases two atoms of hydrogen
H2(g)→2H(g) ; ΔH= 434 KJ mol−1 ----(2)
The bond Cl2 dissociates/breaks to give 2 atoms of Cl and the bond dissociation enthalpy of Cl2 given is 242 KJ mol−1
Cl2(g)→2Cl(g) ; ΔH= 242 KJ mol−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 mol−1
HCl(g)→H(g)+Cl(g) ; ΔH= 431 KJ mol−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 21mole of H2 and 21 mole of Cl2 in order to form products of equation (1). So multiply Equation (2) and (3) by 21 and add both of them.
[H2(g)→2H(g) ΔH= 434 KJ mol−1 ] ×21
[Cl2(g)→2Cl(g) ΔH= 242 KJ mol−1] ×21
Overall reaction
21H2(g)+21Cl2→H(g)+Cl(g);ΔH=21×434+21×242
Therefore 21H2(g)+21Cl2→H(g)+Cl(g);ΔH=338 KJ mol−1 ----(5)
The enthalpy of this reaction is 338 KJ mol−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)
21H2(g)+21Cl2→HCl(g);ΔH=−93KJ/mol−1
Which is the required equation along with its enthalpy of formation, ΔH. Thus we found enthalpy of formation of HCl as -93 KJ mol−1.
Hence option (A) is the correct answer.
Note:
The sign for the ΔH has a great significance. The negative value of ΔHsignifies that the heat is released in the reaction and the positive value of ΔHsignifies that the heat is absorbed/required during the reaction.