Question

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

Answer 1

Hint: 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 $\Delta H$ . 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:-
$\dfrac{1}{2}{H_2}(g) + \dfrac{1}{2}C{l_2}(g) \to HCl(g)$ $\Delta H$=? ----(1)
Now bond dissociation enthalpy of ${H_2}$ is 434 KJ $mo{l^{ - 1}}$. When ${H_2}$ bond H-H is break, it releases two atoms of hydrogen
${H_2}(g) \to 2H(g)$ ; $\Delta H$= 434 KJ $mo{l^{ - 1}}$ ----(2)
The bond $C{l_2}$ dissociates/breaks to give 2 atoms of Cl and the bond dissociation enthalpy of $C{l_2}$ given is 242 KJ $mo{l^{ - 1}}$
$C{l_2}(g) \to 2Cl(g)$ ; $\Delta H$= 242 KJ $mo{l^{ - 1}}$ -----(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 $mo{l^{ - 1}}$
$HCl(g) \to H(g) + Cl(g)$ ; $\Delta H$= 431 KJ $mo{l^{ - 1}}$ ----(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 $\dfrac{1}{2}$mole of ${H_2}$ and $\dfrac{1}{2}$ mole of $C{l_2}$ in order to form products of equation (1). So multiply Equation (2) and (3) by $\dfrac{1}{2}$ and add both of them.
[${H_2}(g) \to 2H(g)$ $\Delta H$= 434 KJ $mo{l^{ - 1}}$ ] $ \times \dfrac{1}{2}$
[$C{l_2}(g) \to 2Cl(g)$ $\Delta H$= 242 KJ $mo{l^{ - 1}}$] $ \times \dfrac{1}{2}$
Overall reaction
   $\dfrac{1}{2}{H_2}(g) + \dfrac{1}{2}C{l_2} \to H(g) + Cl(g);\Delta H = \dfrac{1}{2} \times 434 + \dfrac{1}{2} \times 242$
Therefore $\dfrac{1}{2}{H_2}(g) + \dfrac{1}{2}C{l_2} \to H(g) + Cl(g);\Delta H = 338$ KJ $mo{l^{ - 1}}$ ----(5)
The enthalpy of this reaction is 338 KJ $mo{l^{ - 1}}$ 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)
$\dfrac{1}{2}{H_2}(g) + \dfrac{1}{2}C{l_2} \to HCl(g);\Delta H = - 93KJ/mo{l^{ - 1}}$
Which is the required equation along with its enthalpy of formation, $\Delta H$. Thus we found enthalpy of formation of HCl as -93 KJ $mo{l^{ - 1}}$.
Hence option (A) is the correct answer.


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