Question

Look up the enthalpy of formation of atomic hydrogen in the back of this book. This is the enthalpy change when a mole of atomic hydrogen is formed by dissociating $1/2$mole of molecular hydrogen (the more stable state of the element). From this number, determine the energy needed to dissociate a single $H_2$molecule, in electron-volts.

Short answer

The enthalpy formation of atomic hydrogen molecule is$\mathrm{\Delta }{H}_{\text{molecule}}=4.525\mathrm{ev}.$

Step-by-step solution

Step: 1 Definition of Enthalpy formation:

The enthalpy change that occurs when one mol of a chemical is formed from its constituent elements, such as carbon dioxide from carbon and oxygen. Any molecule can be created as a result of a reaction between the ingredients involved:

Step: 2 Enthalpy change reaction:

The Enthalpy formation of atomic hydrogen gas from molecular hydrogen $H_2$is $\mathrm{\Delta }H=217.97\mathrm{kJ}\text{per mole}$of hydrogen atoms.

The reaction equation is

$\frac{1}{2}{\mathrm{H}}_2\to \mathrm{H}$

The molecular hydrogen of one mole is

${\mathrm{H}}_2\to 2\mathrm{H}$

The difference between the enthalpy of products and reactants is

$\begin{array}{l}\mathrm{\Delta }H=\mathrm{\Delta }{H}_{\text{prod}}-\mathrm{\Delta }{H}_{\text{react}}\\ \mathrm{\Delta }H=2\mathrm{\Delta }{H}_H-\mathrm{\Delta }{H}_{H_2}\end{array}$

Step: 3 Finding value of enthalpy hydrogen molecules:

From the above equations,$\mathrm{\Delta }{H}_H=217.97\mathrm{kJ};{\mathrm{\Delta H}}_{{\mathrm{H}}_2}=0$.

$$\begin{gathered} \mathrm{\Delta }H=2\times 217.97-0 \\ \mathrm{\Delta }H=435.94\mathrm{kJ}. \end{gathered}$$

where,$∆H$is enthalpy per mole.

$$\begin{gathered} \mathrm{\Delta }{H}_{\text{molecule}}=\frac{\mathrm{\Delta }H}{N_A} \\ \mathrm{\Delta }{H}_{\text{molecule}}=\frac{435.94\times {10}^3\mathrm{J}}{6.022\times {10}^{23}} \\ \mathrm{\Delta }{H}_{\text{molecule}}=7.24\times {10}^{-19}\mathrm{J}. \end{gathered}$$

If one electron volt is $1.6\times {10}^{-19}\mathrm{J}$,the enthalpy change in electron volt is

$\begin{array}{l}\mathrm{\Delta }{H}_{\text{molecule}}=7.24\times {10}^{-19}\times \mathrm{J}\frac{\mathrm{ev}}{1.6\times {10}^{-19}\mathrm{J}}\\ \mathrm{\Delta }{H}_{\text{molecule}}=4.525\mathrm{ev}.\end{array}$