91Ó°ÊÓ

The chemical reaction in hydrogen fuel cell is ${\text{H}}_2{\text{°¿â†’H}}_2\text{+}\frac{1}{2}{\text{O}}_2$.

The Gibbs free energy for ${\text{H}}_2$and ${\text{O}}_2$at $25°\text{C}$is zero each and for ${\text{H}}_2\text{O}$at $25°\text{C}$is -237 KJ.

Here, Gibbs free energy,

$$\begin{gathered} \text{G=U-TS+PV} \\ \text{where} \\ \text{G=Gibbsfreeenergy} \\ \text{T=absolutetemperature} \\ \text{S=entropy} \\ \text{P=Pressure} \\ \text{V=Volume} \end{gathered}$$

The infinitesimal change in U is

$\text{»å±«=°Õ»å³§-±Ê»å³Õ+μ»å±·Â·Â·Â·Â·Â·}\left(1\right)$

The infinitesimal change in $G$is

localid="1647330926956" $\text{»å³Ò=»å±«-°Õ»å³§-³§»å°Õ+±Ê»å³Õ+³Õ»å±Ê·····}\left(2\right)$

From (1) and (2)

$$\begin{gathered} \text{»å³Ò=°Õ»å³§-±Ê»å³Õ+μ»å±·-°Õ»å³§-³§»å°Õ+±Ê»å³Õ+³Õ»å±Ê} \\ =\text{μ»å±·-³§»å°Õ+³Õ»å±Ê·····}\left(3\right) \end{gathered}$$

AS, P and N are constant.

So, equation (3) reduces to

$\text{dG=-SdT}$.

And relation between Gibbs energy at two different temperatures $25°\text{C}$and $75°\text{C}$is:

${\text{dG}}_{75}{\text{=dG}}_{25}\text{-³§»å°Õ······}\left(4\right)$

Using

$$\begin{gathered} {\text{dG}}_{25}\left({\text{H}}_2\right)\text{=0} \\ {\text{³§=131°ì´³Â·°­}}^{-1} \\ \text{dT=50K} \end{gathered}$$

in equation (4)

$$\begin{gathered} {\text{dG}}_{75}\left({\text{H}}_2\right)\text{=0-}\left(131{\text{°ì´³Â·°­}}^{-1}\right)\left(50\text{K}\right) \\ \text{=-6550kJ} \end{gathered}$$

Similarly

$$\begin{gathered} {\text{dG}}_{25}\left({\text{O}}_2\right)\text{=0} \\ {\text{S=205°ì´³Â·°­}}^{-1} \\ \text{dT=50K} \end{gathered}$$

So, equation (4) becomes

$$\begin{gathered} {\text{dG}}_{75}\left({\text{O}}_2\right)\text{=0-}\left(205{\text{°ì´³Â·°­}}^{-1}\right)\left(50\text{K}\right) \\ \text{=-10250kJ} \end{gathered}$$

Now,

$$\begin{gathered} {\text{dG}}_{25}\left({\text{O}}_2\right){\text{=-237×10}}^3\text{J} \\ {\text{S=70°ì´³Â·°­}}^{-1} \\ \text{dT=50K} \end{gathered}$$

Hence equation (4) becomes

$$\begin{gathered} {\text{dG}}_{75}\left({\text{H}}_2\text{O}\right)\text{=}\left(-237×{10}^3\text{J}\right)\text{-}\left(70{\text{°ì´³Â·°­}}^{-1}\right)\left(50\text{K}\right) \\ \text{=-240500kJ} \end{gathered}$$

The Gibbs free energy for ${\text{H}}_2$is -6550 kJ, for ${\text{O}}_2$is -10250 kJ and for ${\text{H}}_2\text{O}$is -240500 kJ.

The chemical reaction in hydrogen fuel cell is ${\text{H}}_2{\text{°¿â†’H}}_2\text{+}\frac{1}{2}{\text{O}}_2$.

The Gibbs free energy for ${\text{H}}_2$and ${\text{O}}_2$at $25°\text{C}$is zero each and for ${\text{H}}_2\text{O}$at $25°\text{C}$is -237 KJ.

Gibbs energy at$75°\text{C}$is

As,

So,

$$\begin{gathered} {\text{dG}}_{75}\text{=-240500kJ+6550kJ+}\frac{1}{2}\left(10250\text{kJ}\right) \\ \text{=-228825kJ} \end{gathered}$$

Hence, the electrical work done for each mole of hydrogen fuel is 228825 kJ.