Question

The bond dissociation energy of gaseous \[{H_2}\],\[C{l_2}\] and HCl are 104, 58 and 103 kcal/mol respectively. Calculate the enthalpy of formation of HCl gas.
A.-75 Kcal
B.-22 Kcal
C.-45 Kcal
D.None of these

Answer 1

Hint: The standard enthalpy of formation can be defined as the change in enthalpy when one mole of a substance in the standard state is formed from its pure elements in the similar conditions. The bond dissociation energy is the amount of energy required to break apart one mole of covalently bonded gases into a pair of ions.

Complete step by step answer:
Enthalpy of formation can be calculated from the bond dissociation energy of its gaseous constituents. Knowing the reactions involved in the chemical process and using addition subtraction to get the final equation results in the formation of enthalpy change.
We can arrange the given bond dissociation energies as per their dissociated chemical equations:
\[{H_2}(g) \to 2{H^ + }(g),\,\,\Delta H = \,104\,Kcal\]…………..(I)
\[C{l_2}(g) \to 2C{l^ + }(g),\,\,\Delta H = \,58\,Kcal\]……………..(II)
\[HCl(g) \to {H^ + }(g) + C{l^ - }(g),\,\,\Delta H = \,103\,Kcal\]………..(III)
The enthalpy of formation of HCl can be calculated as:
\[\dfrac{1}{2}{H_2}(g) + \dfrac{1}{2}C{l_2}(g) \to HCl(g)\]
On dividing equation (I) and (II) by 2 and adding them, we get
\[\dfrac{1}{2}{H_2}(g) + \dfrac{1}{2}C{l_2}(g) \to H(g) + Cl(g),\,\,\,\Delta H = \,81\,Kcal\]………..(IV)
On subtracting the enthalpy change of (III) equation from (IV), we obtain \[\Delta H\]= -22 Kcal

Hence, the correct option is (B).

Note:
Another method to solve it is without involving all these equations and just putting the given energy values in the formula of enthalpy of formation which is the subtraction of bond energy of products from sum of bond energies of reactants.