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## Homework Statement

1. [6] Methane (Natural Gas) Fuel Cell.

Consider a fuel cell that uses methane as fuel. The chemical reaction is

CH

_{4}+ 2O

_{2}= 2H

_{2}O + CO

_{2}+ energy

(a) [0.5] Measurements reveal that deltaH for this reaction is -890 kJ for each mole of

methane processed, assuming standard temperature and pressure (T =298 K, P =100 kPa).

Ignoring the volume of the liquid H20, use the ideal gas law to estimate the change in

volume during this reaction, and thus determine the amount of energy we get for “free”

from the collapsing environment.

(b) [2.5] Using the appropriate table at the back of the textbook (or elsewhere), look up

the entropies of the various chemicals involved in the above reaction (at standard

temperature and pressure) and thus compute deltaS, the change in entropy for each mole of methane processed. Thus evaluate deltaG, the change in the Gibbs free energy (at

standard temperature and pressure) for each mole of methane processed.

(c) [0.5] Assuming ideal performance, how much electrical work can you get out of the

cell for each mole of methane fuel?

(d) [1] How much waste heat must be dumped into the environment for each mole of

methane fuel?

(e) [0.5] What is the theoretical maximum “efficiency" of this fuel cell?

(f) [1] In more detail, the above chemical reaction proceeds in two steps that are

happening simultaneously at the positive and negative electrodes of the fuel cell:

at - electrode CH

_{4}+ 2H

_{2}O = CO

_{2}+ 8H

^{+}+ 8e

^{-}

at + electrode 20

_{2}+ 8H

^{+}+ 8e

^{-}= 4H

_{2}O

What this means is that for each methane molecule that reacts, eight electrons are pushed

around the circuit. It takes electrical work to do this. Considering the electrical work

computed in part (c), for one mole of methane, what is the electrical work per electron?

Express your answer in units of electron volts. Considering that 1 volt is the voltage

needed to give an electron 1 eV of energy, what is the voltage of the fuel cell?

## Homework Equations

Lots

## The Attempt at a Solution

a)

[tex]P \Delta V= \Delta n R T[/tex]

[tex]\Delta V = \Delta n R T / P[/tex]

We are going from 3 moles of gas to 1 mole:

[tex]\Delta V = -2 R T / P = -0.0496 L[/tex]

b)

[tex]\Delta S = \Delta S_{products} - \Delta S_{reactants} = [213.7+2(69.91)] - [2(205.1)+186.3] = -242.98 J/K[/tex]

I got the entropy values from my textbook

Then

[tex]\Delta G = \Delta H - T\Delta S = -890000 - 298(-242.98) = 817592 J[/tex]

c)

[tex]W = \Delta H = 890 kJ[/tex] ?

I'm just using the enthalpy of the reaction, which is the amount of heat released, so assuming it is all converted to useful work? Is this correct?

d)

Here I'm tempted to just use [tex]W = P \Delta V[/tex], since that is the amount of work the system does on the environment in expanding... ie the heat released to the environment?? Nevermind, I just saw that the volume is decreasing!

e)

If I can get the one before then this is easy:

[tex]e = workdone / heatsupplied = 1 - Qh / Qc[/tex]

right?

f)

Assuming c) is correct, then I just divide by 8 to get the work per electron, then convert to eV.

Thanks so much for any help you can offer!