Question

The bond enthalpies of H-H, Cl-Cl and H-Cl are 435,243 and 431 kJ/mol respectively. The enthalpy of formation of HCl(g) will be:
(A) 92 kJ/mol
(B) -92 kJ/mol
(C) 247 kJ/mol
(D) 770 kJ/mol

Answer 1

We calculate the calculate the enthalpy of the HCl by the formula as $\Delta {{\text{H}}^{\circ }}=\text{(}\Sigma {{\Delta }_{f}}{{\text{H}}^{\circ }}\text{(reactants)-}\Sigma {{\Delta }_{f}}{{\text{H}}^{\circ }}\text{(reactants)}$, here $\Delta {{\text{H}}^{\circ }}~$is the total enthalpy of the reaction of formation of HCl and ${{\Delta }_{f}}{{\text{H}}^{\circ }}{{\text{H}}_{2(g)}}$= 435 KJ/mol , ${{\Delta }_{f}}{{\text{H}}^{\circ }}\text{C}{{\text{l}}_{2}}$= 243 KJ/mol and ${{\Delta }_{f}}{{\text{H}}^{\circ }}\text{HC}{{\text{l}}_{(g)}}$=431 kJ/mol. Now calculate its enthalpy.

Complete step by step solution:
First of all, what is the enthalpy of formation? From the enthalpy of formation, we simplify the total change in the enthalpy of the reaction when 1mole of the compound is formed from its constituents’ elements.
We can easily calculate the enthalpy of formation of HCl as;
The reaction of formation of HCl is
$\dfrac{1}{2}{{\text{H}}_{2}}\text{+}\dfrac{1}{2}{{\text{O}}_{2}}\to \text{HCl}$
As we know that,
$\Delta {{\text{H}}^{\circ }}=\text{(}\Sigma {{\Delta }_{f}}{{\text{H}}^{\circ }}\text{(reactants)-}\Sigma {{\Delta }_{f}}{{\text{H}}^{\circ }}\text{(reactants)}$
=$\dfrac{1}{2}$(${{\Delta }_{f}}{{\text{H}}^{\circ }}{{\text{H}}_{2(g)}}$+ ${{\Delta }_{f}}{{\text{H}}^{\circ }}\text{C}{{\text{l}}_{2}}$ ) -${{\Delta }_{f}}{{\text{H}}^{\circ }}\text{HC}{{\text{l}}_{(g)}}$ ---------(1)
Enthalpy of formation of H-H is${{\Delta }_{f}}{{\text{H}}^{\circ }}{{\text{H}}_{2(g)}}$= 435 KJ/mol
Enthalpy of formation of Cl-Cl is ${{\Delta }_{f}}{{\text{H}}^{\circ }}\text{C}{{\text{l}}_{2}}$= 243 KJ/mol
Enthalpy of formation of HCl is${{\Delta }_{f}}{{\text{H}}^{\circ }}\text{HC}{{\text{l}}_{(g)}}$=431 kJ/mol
Put these values in equation(1), we get:
$\Delta {{\text{H}}^{\circ }}$= $\dfrac{1}{2}$ $(435+ 243)$ $-431$
$=$ $\dfrac{1}{2}$ $(678)$ $-431$
$= 339 –$ $431$
$= -92$ $kJ/mol$
Thus, the standard enthalpy of formation of HCl of the reaction;
$\dfrac{1}{2}{{\text{H}}_{2}}\text{+}\dfrac{1}{2}{{\text{O}}_{2}}\to \text{HCl}$
is: -92 KJ/mol.

Hence, option (B) is correct.

Note: The standard enthalpy of formation of any substance in its standard state is taken as zero. Example: standard enthalpy of formation of oxygen at the standard conditions of temperature and pressure i.e. at 1 atm pressure and 25 $^{\circ }C$ is always taken as zero as they undergo no change in their formation.