91Ó°ÊÓ

Given:

The transition is

${\mathrm{CH}}_{4}12{\mathrm{O}}_2,2{\mathrm{H}}_2\mathrm{O}âˆ\pounds {\mathrm{CO}}_2$

The temperature is 208k and the pressure is 1 bar.

Formula used:

Write the expression for Gibbs energy-

$G=H-TS$

Here, G is Gibbs energy, I is the enthalpy, T is the absolute Write the expression for the infinitesimal change in G.

$\mathrm{Δ}G=\mathrm{Δ}H-T\mathrm{Δ}{S}_{m…}\left(1\right)$

Write the expression for the change in enthalpy for the reaction

$\mathrm{Δ}G=2\mathrm{Δ}{G}_{{\mathrm{H}}_2\mathrm{O}}+\mathrm{Δ}{G}_{{\mathrm{CO}}_2}-\mathrm{Δ}{G}_{{\mathrm{CH}}_4}-2\mathrm{Δ}{G}_{{\mathrm{O}}_2}……..\text{(3)}$

Refer table at the back of the book.

Substitute $-393.51\mathrm{kJ}$for$\mathrm{Δ}{H}_{{\mathrm{CO}}_2},-285.83\mathrm{kJ}$for $\mathrm{Δ}{H}_{{\mathrm{H}}_2\mathrm{O}},0$for $\mathrm{Δ}{H}_{{\mathrm{O}}_2}$and $-74.81\mathrm{kJ}$for $\mathrm{Δ}{H}_{{\mathrm{CH}}_4}$from the table in expression (2).

Thus, the value in the change of enthalpy is $-890.36\mathrm{kJ}$ and the value in the change of Gibbs free energy is $-817.9\mathrm{kJ}$.

The reaction is ${\mathrm{CH}}_4+2{\mathrm{O}}_2→2{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$.

Temperature is 298 K.

Pressure is 1 bar.

Formula used:

Work done, $\mathrm{W}=∆\mathrm{G}$

where

G=Gibbs free energy

W=work done

Here, $∆\mathrm{G}=-817.9\mathrm{kJ}$.

So,$\mathrm{W}=-817.9\mathrm{kJ}$.

Hence, the electrical work done for each mole of methane fuel is -817.9 kJ.

The reaction is ${\mathrm{CH}}_4+2{\mathrm{O}}_2→2{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$.

Temperature is 298 K.

Pressure is 1 bar.

As the reaction is occuring at constant pressure.

So,

$$\begin{gathered} ∆\mathrm{Q}=∆{\mathrm{H}}_{\mathrm{r}}-∆{\mathrm{H}}_{\mathrm{p}} \\ \mathrm{where} \\ ∆\mathrm{Q}=\mathrm{Energy}\mathrm{difference} \\ ∆{\mathrm{H}}_{\mathrm{r}}=\mathrm{Enthalpy}\mathrm{change}\mathrm{for}\mathrm{reactants} \\ ∆{\mathrm{H}}_{\mathrm{p}}=\mathrm{Enthalpy}\mathrm{change}\mathrm{for}\mathrm{the}\mathrm{products} \end{gathered}$$

As, $∆{\mathrm{H}}_{\mathrm{r}}=890.36\mathrm{kJ}$and $∆{\mathrm{H}}_{\mathrm{p}}=817.9\mathrm{kJ}$.

So,$$\begin{gathered} ∆\mathrm{Q}=890.36\mathrm{kJ}-817.9\mathrm{kJ} \\ =72.46\mathrm{kJ} \end{gathered}$$

Hence the amount of waste heat produced for each mole of methane fuel is 72.46 kJ.

The reaction is ${\mathrm{CH}}_4+2{\mathrm{O}}_2→2{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$.

Temperature is 298 K.

Pressure is 1 bar.

Formula for the work done per each electron is

$$\begin{gathered} {\mathrm{W}}_{\mathrm{e}}=\frac{\mathrm{W}}{8{\mathrm{N}}_{\mathrm{A}}} \\ \mathrm{where} \\ \mathrm{W}=\mathrm{work}\mathrm{done}\mathrm{per}\mathrm{mole} \\ {\mathrm{N}}_{\mathrm{A}}=\mathrm{Avogadro}\text{'}\mathrm{s}\mathrm{number} \end{gathered}$$

Here

$$\begin{gathered} \mathrm{W}=817.9\mathrm{kJ} \\ {\mathrm{N}}_{\mathrm{A}}=6.623×{10}^{23} \end{gathered}$$

So,

$$\begin{gathered} {\mathrm{W}}_{\mathrm{e}}=\frac{817.9\mathrm{kJ}}{8\left(6.623×{10}^{23}\right)} \\ =1.061\mathrm{eV} \end{gathered}$$

Hence, the voltage of the cell is 1.061 V.