Question

Draw a curve that describes the energy of a system with H and CI atoms at varying distances. Then, find the minimum energy of this curve two ways.

(a) Use the bond energy found in Table 8.1 to calculate the energy for one single HCl bond (Hint: How many bonds are in a mole?)

(b) Use the enthalpy of reaction and the bond energies for ${\mathrm{H}}_2$and $\mathrm{C}{\mathrm{l}}_2$to solve for the energy of one mole of $\mathrm{H}\mathrm{C}\mathrm{l}$bonds.

${\text{H}}_{\text{2}}\text{(g) + C}{\text{l}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2HCl(g)}$ ΔH°_rxn = −184.7 kJ/mol

Short answer

(a.) The energy for one single HCl bond is $7.16\times 10^{-19}\mathrm{J}/\mathrm{b}\mathrm{o}\mathrm{n}\mathrm{d}$.

(b.) The energy of one mole of HCI bonds is $431.9\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$.

Step-by-step solution

Definition of Concept

Hybridization occurs when two or more different pure orbitals with comparable energy are mixed together to produce an equivalent amount of impure orbitals with equal energy and definite geometry, which are referred to as hybrid orbitals.

Find the energy for one single HCl bond

(a)

Energy of a system with H and CI atoms at varying distances

Energy required to break 1 mole of$\mathrm{H}-\mathrm{C}\mathrm{l}$bonds

$\begin{array}{rl}& =431kJ/mol\\ & =431\times 10^3\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}\end{array}$

1 mol HCl contains$6.02\times 10^{23}$molecules

Hence, energy required to break$1\mathrm{H}-\mathrm{C}\mathrm{l}$molecule is

$\begin{array}{rl}& =431\times 10^3\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}\times \frac{1\mathrm{m}\mathrm{o}\mathrm{l}}{6.022\times 10^{23}\mathrm{b}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{s}}\\ & =7.16\times 10^{-19}\mathrm{J}/\mathrm{b}\mathrm{o}\mathrm{n}\mathrm{d}\end{array}$

Therefore, the required value is $7.16\times 10^{-19}\mathrm{J}/\mathrm{b}\mathrm{o}\mathrm{n}\mathrm{d}$.

Find the energy of one mole of HCI bonds

(b)

$\mathrm{\Delta }{H}_{\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}^{\circ }=\sum \mathrm{\Delta }{H}_{\mathrm{f}}^{\circ }(\mathrm{b}\mathrm{r}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{n})-\sum \mathrm{\Delta }{H}_{\mathrm{f}}^{\circ }(\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{e}\mathrm{d})$b

$-184.7\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}=\mathrm{\Delta }{H}_{\mathrm{f}}^{\circ }(\mathrm{H}-\mathrm{H})+\mathrm{\Delta }{H}_{\mathrm{f}}^{\circ }(\mathrm{C}\mathrm{l}-\mathrm{C}\mathrm{l})-\mathrm{\Delta }{H}_{\mathrm{f}}^{\circ }(\mathrm{H}-\mathrm{C}\mathrm{l})$

$-184.7\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}=(436\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}+243\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l})-2\times \mathrm{\Delta }{H}_{\mathrm{f}}^{\circ }(\mathrm{H}-\mathrm{C}\mathrm{l})$

$2\mathrm{\Delta }{H}_{\mathrm{f}}^{\circ }(\mathrm{H}-\mathrm{C}\mathrm{l})=436\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}+243\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}+184.7\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$

$=863.7\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$

$\mathrm{\Delta }{H}_{\mathrm{f}}^{\circ }(\mathrm{H}-\mathrm{C}\mathrm{l})=\frac{863.7\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}}{2}$

$=431.9\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$

Therefore, the required value is$431.9\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$.