91Ó°ÊÓ

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:

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

The reaction equation is

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

The molecular hydrogen of one mole is

${\mathrm{H}}_2→2\mathrm{H}$

The difference between the enthalpy of products and reactants is

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

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

$$\begin{gathered} \mathrm{Δ}H=2×217.97−0 \\ \mathrm{Δ}H=435.94\mathrm{kJ}. \end{gathered}$$

where,$∆H$is enthalpy per mole.

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

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

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