Question

Given that, $\vartriangle {H_f}\left( H \right) = 218\;KJ/mol$, express the H - H bond energy in Kcal/mol.
A) 52.15
B) 911
C) 104
D) 52153

Answer 1

Hint: The $\vartriangle {H_f}$ is the enthalpy of formation of Hydrogen. It is the heat change (either evolved or absorbed) accompanying the formation of 1 mole of a substance from its elements under a given condition of temperature and pressure.

Complete step by step answer:
We have given that $\vartriangle {H_f}\left( H \right) = 218\;KJ/mol$,
i.e., The heat evolved while forming 1 mole of hydrogen from its elements under a given condition of temperature and pressure is $218\;KJ/mol$.
Now we have to find the bond energy of H - H.
This can also be written as ${H_2}$.
So the energy required for the formation of H - H will be twice the energy for the formation of H.
This energy is the bond energy between H - H.
i.e., Bond energy of $H - H\; = \;2 \times \vartriangle {H_f}\left( H \right)$
$ \Rightarrow \;Bond\;energy\;of\;H - H\;\; = 2 \times 218\;KJ/mol\;$
Therefore the energy required to break the $H - H$ bond will be $436\;KJ/mol$.
Now for the conversion of KJ/mol to Kcal/mol
We know that 1 cal = 4.184 J
$\Rightarrow$ 1KJ = 0.239 Kcal
Therefore, 436 KJ/mol = 0.239 $\times$ 436 Kcal/mol
= 104.2Kcal/mol

Therefore the option (c) is correct.

Note: It is to be noted that the higher the negative value of enthalpy of formation, greater is the stability of the compound.
Enthalpy of reaction can be calculated from standard heat of formation.
$\vartriangle H_{Reaction}^0 = \vartriangle H_{f\left( {products} \right)}^0 - \vartriangle H_{f\left( {Reac{\text{tan}}ts} \right)}^0$
So here, we can see that the enthalpy of formation of Hydrogen is highly positive and it is very unstable. That’s why it is always in the form ${H_2}$.
While converting the units, we have to keep in mind that only units which need to change only have to change. Here, from KJ/mol to Kcal/mol the mol is common in both that is why we only changed KJ to Kcal.