[butters]

The question is:
"Confirm the statement in the text that oxidation of 1.0 L of methanol to form CO2 (g) and H2O (l) in a fuel cell will provide at least 5.0 kW-h of energy. (The density of methanol is 0.787 g/mL.)"

The correct answer is given as: "Methanol provides 23 kj per gram of 1.8 x 10⁴ kj/L. At 100% efficiency the cell produces 5.0 kW-h of energy."

This is what I have:

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There are a few things that  I do not understand; I don't know how liter per kj was determined, I don't know how the kj amount was converted into 5.0 kW-h andwhy is the density and volume of methanol given? There doesn't seem to be any use for that information, or is there? Please explain these things in detail..
Thank you.

[Borek]

Where is the 787g taken from?

How are Joule and Watt related?

[butters]

The volume and density are given, so I multiplied the two to get the mass: (1.0L) 1000 mL x 0.787 g/mL = 787g.

And I don't know how Joule and Watt are related. The chapter does not talk about watts, nor does it give any examples of such problems. This is not even a chapter, it's just titled "The Chemistry of Fuels and Energy Sources." It's is just a few pages and it talks about the current energy crisis, this is followed by a few questions. This will supposedly be on the exam. The Prof might have gone over it, but I don't remember and I cannot find any notes on this..

Edit: okay, this is what a quick search on Google reveals:
"1 Watt = 1 Joule / 1 sec

Essentially, watts are how many joules you consume in a second."

[Borek]

The volume and density are given, so I multiplied the two to get the mass: (1.0L) 1000 mL x 0.787 g/mL = 787g.

So why do you ask what are density and volume for if you have used them?

Edit: okay, this is what a quick search on Google reveals:
"1 Watt = 1 Joule / 1 sec

Essentially, watts are how many joules you consume in a second."

Check what is kWh and everything should be obvious.

[butters]

So why do you ask what are density and volume for if you have used them?

I did not use them.
I calculated the standard enthalpy (H values are given at the end of the book) for the reaction then I divided it by 32.042g/mol of CH3OH.
Here's how I calculated the H: H2O + Co2 - CH3OH = -726.769

Check what is kWh and everything should be obvious.

You mean: 1 kWh=3600 kJ?
That gives me the wrong answer...

What is the definition of kWh in relation to Kj?

[Borek]

The volume and density are given, so I multiplied the two to get the mass: (1.0L) 1000 mL x 0.787 g/mL = 787g.

So why do you ask what are density and volume for if you have used them?

I did not use them.

How can you multiply two numbers - volume and density - not using them?

You mean: 1 kWh=3600 kJ?
That gives me the wrong answer...

What is the definition of kWh in relation to Kj?

So there must be some other mistake, 1kWh means kW used for an hour and that means 3600 kJ.

Not 2H₂O?

[butters]

How can you multiply two numbers - volume and density - not using them?

Because initially, I did not know how to solve this problem. The book does not teach how to solve this type of problem. So at first I was just fiddling around with the numbers, trying to make some sense out of it. I thought, since this information is given, it must be of some use... apparently not.

So there must be some other mistake, 1kWh means kW used for an hour and that means 3600 kJ.

Well, even if my method is wrong, I am getting the right answer, and you can check for yourself. Here is the book, the answer for this question is on page 4 (appendix p) # 11:
http://books.google.com/books?id=jcn6sgt7RpoC&printsec=frontcover#v=onepage&q&f=false
Page 255 is the section where this question comes from, and I have read the whole thing, nowhere does it show you how to solve this thing.

Not 2H₂O?

No, I did use 2H2O, 1CO2 and 1CH3OH, but I just didn't write that to save time.

[Borek]

You calculated 23 kJ/g - taht's OK. But that's not the final answer that the question asks for.

Now you have to calculate how much energy will be released when burning 1L of methanol and convert this amount to kWh.

[butters]

You calculated 23 kJ/g - taht's OK. But that's not the final answer that the question asks for.

Now you have to calculate how much energy will be released when burning 1L of methanol and convert this amount to kWh.

Yeah, but what I was saying is that at-least we know 23 Kj is right.
What answer are you getting for the KW-h figure?

[rabolisk]

The answer given is correct. You know the energy in kJ when 1 gram of methanol is oxidized. You need to go from that to the energy in kWh when 1 liter of methanol is oxidized. Put it another way, you're going from kJ/g to kWh/L. Simple dimensional analysis should get you the correct answer.

BTW, you understood the chemistry correctly, so that's good. The kWh probably confused you, which is understandable.

[butters]

Simple dimensional analysis should get you the correct answer.

Okay, so you are saying that I should first convert it into Kj/L and then into kW-h/L?
Is this why the density has been given?
Can the density be used as a conversion factor (in this case)?
Is this next step correct: 23 Kj/g (0.787 g/0.001L) ?

Edit: bingo!
Yes, it worked!
So, the next step was: 18101 Kj/L (1 kW-h/3600 K/j)
Which gives: 5.028 = 5.0

It worked.
Thank you so much!
This annoying problem had plagued me for like two days..