energy of hydrogen gas combustion

[Antielectron]

I have a science project and I have decided to find out if the amount of energy (joules) produced when combustion of hydrogen gas in air takes place. I need someone to check if there are any problems, and here is what I worked out:

 

Volume of Hydrogen gas (H₂)collected: 51.2cm³

 

= 0.0000512dm³

 

No. of moles of H₂= Volume/Molar Volume

 

= 0.0000512/24 (at rtp)

 

= 0.00000213mol

 

Enthalpy of Combustion forHydrogen: -286 kJ/mol

 

Energy produced: 286 kJ/mol x 0.00000213mol

 

= 0.00060918kJ (H₂)

 

I'm also comparing it with 10 x 9volts battery, which also requires me to calculate the energy (joules) produced...

 

Battery (Eveready 9 Volts): 325mAh

 

= 0.325 Ah

 

Energy produced: IV = Js^-1

 

90(0.325) = (200 x 60)^-1J (in 200minutes)

 

29.25 = 12000^-1J

 

J = 351000

 

351000J = 351kJ

 

∴Energy (Battery) > Energy (combustion of Hydrogengas)

 

Looks flawed to me, and assuming that the battery have a constant mAh throughout. Would be glad for any help given, thanks!

Edited July 6, 2011 by Antielectron

[CaptainPanic]

There are 1000 cm3 in a dm3. Your conversion is wrong there (you have an order of 1,000,000). Didn't (yet) check the rest. May do that during another coffee break today. Maybe someone else beats me to it.

[Antielectron]

Thanks for that =D So it becomes 0.0512dm³ , 0.00213mol and 0.60918kJ for the first half....

[CaptainPanic]

Thanks for that =D So it becomes 0.0512dm³ , 0.00213mol and 0.60918kJ for the first half....

Yep. That seems right.

 

Energy (Battery) > Energy (combustion of Hydrogengas)

Looks flawed to me, and assuming that the battery have a constant mAh throughout. Would be glad for any help given, thanks!

 

I'm a little confused about your battery calculation (in other words: I don't know what you're doing, so I cannot say where you go wrong).

Regarding batteries, I am stepping outside my own field, but I think the calculation should be like this:

 

Assuming a constant voltage of the battery:

Ah*V=Joules

So: the energy of 1 battery is (325/1000*3600)*9 =10530 J.

(The mAh value of the battery is 325 mAh. This is divided by 1000 to arrive at Ah, and multiplied by 3600 because there are 3600 seconds in 1 hour, and we like to get seconds to make everything in SI units. And finally, it's multiplied by 9V.

 

The total energy of all 10 batteries is then 10*10530 = 105.3 kJ

 

Of course, this is still a LOT more than the hydrogen. But you should realize that the 9V battery has a weight of perhaps 40 grams, and all 10 batteries are perhaps 400 grams.

The hydrogen gas has a weight of only 0.004 grams...

[Antielectron]

Thanks a lot for ur help, it was useful