Thermodynamics Chemistry challenge problem?

The half-cell reactions in a fuel cell are:

H2(g) ==> 2H+(aq) + 2e- . . .Eo = 0.00 V

and

O2(g) + 4H+(aq) + 4e- ==> 2H2O(l) . . .Eo = +1.23 V

If you multiply the first reaction by 2 and add it to the second reaction, we get simply

2H2(g) + O2(g) ==> 2H2O(l) . . .Eo cell = +1.23 V (although a real fuel cell probably won't deliver that much)

In the above reaction, n = 4.

1 kWh x (3600 s / h) = 3600 kWs = 3,600,000 Ws x (1 J/s / 1 W) = 3.6 x 10^6 J

In a typical month, there are 31 days x 24 hrs/day = 744 hours.

1500 kWh x (3.6 x 10^6 J / 1 kWh) = 5.4 x 10^9 J

The maximum amount of work that we can derive from an electrochemical cell is approximated by delta G.

delta G = -nFE cell = -(4 moles e-)(96,500 C / mole e-)(+1.23 V) = 475,000 J per 2 moles of H2 reacted

Since we need 5.4 x 10^9 J of energy,

5.4 x 10^9 J x (2 moles H2 / 475,000 J) x (22.4 L H2 / 1 mole H2 at STP) = 5.1 x 10^6 L H2

CHECK MY MATH!!!

i do no longer think of you go with me to plug the numbers in, so i'm going to tell you what you're able to desire to think of roughly: F=d/dt P the place P is momentum P= 2np the place n is the type of molecules and p is the guy momentum- the two is there because of the fact the exchange in velocity is two*velocity of the molecules (they bounce top lower back, with the comparable velocity as they went in!) Molecular oxygen is O2 (regularly!) and so has a mass of roughly 32 AMU. Oh yeah- F=stress*section So, placed the stress in pascals, the section in sq. meters, the mass in kg, and do the equations and you gets an expression for dn/dt... I made it much less complicated for you than i actually meant! :)

Correct answer is actually 5.1 x 10^5 L