Q2E
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 \({{\rm{H}}_{\rm{2}}}\)and \({\rm{C}}{{\rm{l}}_{\rm{2}}}\) to solve for the energy of one mole of \({\rm{HCl}}\)bonds.
\({{\text{H}}_{\text{2}}}{\text{(g) + C}}{{\text{l}}_{\text{2}}}{\text{(g)}} \rightleftharpoons {\text{2HCl(g)}}\) ΔH°rxn = −184.7 kJ/mol
Step-by-Step Solution
Verified(a.) The energy for one single HCl bond is \({\rm{7}}{\rm{.16 \times 1}}{{\rm{0}}^{{\rm{ - 19}}}}{\rm{\;J/ bond }}\).
(b.) The energy of one mole of HCI bonds is \(431.9\;{\rm{kJ}}/{\rm{mol}}\).
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.
(a)
Energy of a system with H and CI atoms at varying distances
Energy required to break 1 mole of \({\rm{H - Cl}}\) bonds
\(\begin{aligned}{\underline{\phantom{xx}}}{\rm{ }}& = 431\;kJ/mol \\{\rm{ }} &= 431 \times 1 {{\rm{0}}^{\rm{3}}}{\rm{\;J/mol}}\end{aligned}\)
1 mol HCl contains \({\rm{6}}{\rm{.02 \times 1}}{{\rm{0}}^{{\rm{23}}}}\) molecules
Hence, energy required to break \({\rm{1H - Cl}}\)molecule is
\(\begin{aligned}{\underline{\phantom{xx}}}{\rm{ }}& = 431 \times 1 {{\rm{0}}^{\rm{3}}}{\rm{\;J/mol \times }}\frac{{{\rm{1\;mol}}}}{{{\rm{6}}{\rm{.022 \times 1}}{{\rm{0}}^{{\rm{23}}}}{\rm{ bonds }}}}\\{\rm{ }}&= 7{\rm{.16 \times 1}}{{\rm{0}}^{{\rm{ - 19}}}}{\rm{\;J/ bond }}\end{aligned}\)
Therefore, the required value is \({\rm{7}}{\rm{.16 \times 1}}{{\rm{0}}^{{\rm{ - 19}}}}{\rm{\;J/ bond }}\).
(b)
\(\Delta H_{{\rm{reaction}}}^\circ {\rm{ }} = \sum \Delta H_{\rm{f}}^\circ {\rm{ (broken) }} - \sum \Delta H_{\rm{f}}^\circ {\rm{ (formed) }}\)b
\( - 184.7\;{\rm{kJ}}/{\rm{mol }} = \Delta H_{\rm{f}}^\circ ({\rm{H}} - {\rm{H}}) + \Delta H_{\rm{f}}^\circ ({\rm{Cl}} - {\rm{Cl}}) - \Delta H_{\rm{f}}^\circ ({\rm{H}} - {\rm{Cl}})\)
\( - 184.7\;{\rm{kJ}}/{\rm{mol }} = (436\;{\rm{kJ}}/{\rm{mol}} + 243\;{\rm{kJ}}/{\rm{mol}}) - 2 \times \Delta H_{\rm{f}}^\circ ({\rm{H}} - {\rm{Cl}})\)
\(2\Delta H_{\rm{f}}^\circ ({\rm{H}} - {\rm{Cl}}){\rm{ }} = 436\;{\rm{kJ}}/{\rm{mol}} + 243\;{\rm{kJ}}/{\rm{mol}} + 184.7\;{\rm{kJ}}/{\rm{mol}}\)
\( = 863.7\;{\rm{kJ}}/{\rm{mol}}\)
\(\Delta H_{\rm{f}}^\circ ({\rm{H}} - {\rm{Cl}}){\rm{ }} = \frac{{863.7\;{\rm{kJ}}/{\rm{mol}}}}{2}\)
\( = 431.9\;{\rm{kJ}}/{\rm{mol}} \)
Therefore, the required value is \(431.9\;{\rm{kJ}}/{\rm{mol}}\).