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shortythree
04-21-2014, 10:11 PM
I have been stuck on this question for a long time and I can't get the right answer.
If you could help me out with this that would be awesome.


http://i.imgur.com/XjLmoHt.jpg






For your time:

http://i.imgur.com/4IshfIF.gif
http://i.imgur.com/HHfH0BY.gif
http://i.imgur.com/OMGe43S.gif
http://i.imgur.com/4VWJoka.gif
http://i.imgur.com/eyA3UZf.gif

cheesecake93
04-21-2014, 10:14 PM
so glad I'll never have to take this class again.. I honestly tried reading that problem to help you and whathephuckisgoingonhere

Pexterra
04-21-2014, 10:17 PM
so glad I'll never have to take this class again.. I honestly tried reading that problem to help you and whathephuckisgoingonhere

lol, we just had lecture on this today and will have to look at it online before i understand it. sorry man

shortythree
04-21-2014, 10:26 PM
so glad I'll never have to take this class again.. I honestly tried reading that problem to help you and whathephuckisgoingonhere

lol, we just had lecture on this today and will have to look at it online before i understand it. sorry man

thanks anyways

Domina8r
04-21-2014, 10:27 PM
Doing chemistry atm.

If I get 50% and pass I'll have a party......painful class....all the best OP.

GetBigBrah
04-21-2014, 10:33 PM
Follow this and you'll get your answer:

http://www.chm.davidson.edu/vce/calorimetry/heatofcombustionofmethane.html

shortythree
04-21-2014, 10:57 PM
Update: So apparently i'm suppose to get the bond energies from this chart:

http://i.imgur.com/fNZjzmC.jpg


and use this format .... I just don't know which bonds break and which ones don't

http://i.imgur.com/pyiL1Zm.jpg

niktak11
04-21-2014, 11:13 PM
I haven't done this is years but I'm guessing you calculate the total bond energy of the reactants and the products and take the difference

shortythree
04-21-2014, 11:18 PM
I haven't done this is years but I'm guessing you calculate the total bond energy of the reactants and the products and take the difference yeah i've been trying that.... lol i'm already on my last attempt.

http://i.imgur.com/8jSZDBz.jpg
http://i.imgur.com/mAUoMHL.jpg

niktak11
04-21-2014, 11:21 PM
Just make a balanced equation, find the difference in bond energy, and then divide the answer you got by the number of moles of the reactant in the balanced equation

If you post all of your calculations here someone will see what you messed up on

jgpkpbrah
04-21-2014, 11:26 PM
They did a chit job explaining that problem in the book. Essentially they assumed you would realize that net change in bonds broken and formed is what they show. However, that is the (net) result of the problem. Now lets consider your balanced equation...
2 CH3OH + 3 O2 --> 2 CO2 + 4H2O
It is far easier to think of the problem as ALL of your reactants breaking down and ALL of your products forming. Therefore we have...

2[ 3(C-H) + (C-O) + (O-H)] + 3(O=O) - 2[ 2(C=O)] - 4[ 2(H-O)]

Now substitute the bonds, i.e. C-H, with their appropriate bond energies. I'm not going to bother plugging in the numbers but you should find it pretty easy from here srs.

shortythree
04-21-2014, 11:30 PM
Just make a balanced equation, find the difference in bond energy, and then divide the answer you got by the number of moles of the reactant in the balanced equation

If you post all of your calculations here someone will see what you messed up on

This is one way I attempted it:

Balanced Equation:

2 CH3OH + 3 O2 ---> 2CO2 + 4 H2O

6 (C-H) +2 (O-H) +3 (O=O) ---> 4(C=O) + 8(H-O)

6(+414)+2(+360)+2(+464) + 3(+498) = +5626 kj/mol

-4(799) - 8(464) = -6908 kj/mol

+5626 - 6908 = -1282 kj/mol

~ -1280 kj/mol (3 sig figs)

jgpkpbrah
04-21-2014, 11:36 PM
This is one way I attempted it:

Balanced Equation:

2 CH3OH + 3 O2 ---> 2CO2 + 4 H2O

6 (C-H) +2 (O-H) +3 (O=O) ---> 4(C=O) + 8(H-O)

6(+414)+2(+360)+2(+464) + 3(+498) = +5626 kj/mol

-4(799) - 8(464) = -6908 kj/mol

+5626 - 6908 = -1282 kj/mol

~ -1280 kj/mol (3 sig figs)
ya forgot the 2 C-O bond energy. it's like this H3C-O-H

shortythree
04-21-2014, 11:39 PM
ya forgot the 2 C-O bond energy. it's like this H3C-O-H

Forgot to write it in the 2nd equation but I added it to the overall total.
The bond energy is the 2(+360)

shortythree
04-21-2014, 11:49 PM
ya forgot the 2 C-O bond energy. it's like this H3C-O-H also gave you my measly reps

niktak11
04-21-2014, 11:55 PM
Divide that by two to get it in kj/mol. So it'd be 6.41 x 10^2 kj/mol (assuming your other calculations were correct)

shortythree
04-21-2014, 11:57 PM
Divide that by two to get it in kj/mol divide what? the -1282 kj/mol?

niktak11
04-22-2014, 12:00 AM
divide what? the -1282 kj/mol?

Isn't that the total energy difference between the products and reactants of the balanced equation? If so then you need to divide by two because the balanced equation has two moles of methanol

shortythree
04-22-2014, 12:03 AM
Isn't that the total energy difference between the products and reactants of the balanced equation? If so then you need to divide by two because the balanced equation has two moles of methanol I'd still divide it even though I used the 2 moles to multiply the bond energies?

niktak11
04-22-2014, 12:05 AM
I'd still divide it even though I used the 2 moles to multiply the bond energies?

Yes, you calculated the energy of combustion of two moles of methanol

shortythree
04-22-2014, 12:20 AM
Yes, you calculated the energy of combustion of two moles of methanol alright thanks, I'll take this into consideration since I'm on my last attempt lol

shortythree
04-22-2014, 02:19 PM
Yes, you calculated the energy of combustion of two moles of methanol Update: You were right brah thanks

http://i.imgur.com/LcfNDBg.jpg

http://i.imgur.com/3h2T0Vt.jpg

niktak11
04-22-2014, 02:28 PM
Update: You were right brah thanks

http://i.imgur.com/LcfNDBg.jpg

http://i.imgur.com/3h2T0Vt.jpg

Aww yeah

NotoriousSRJ
04-22-2014, 02:33 PM
Is this Mastering Chemistry? I remember doing it back in first year. Pretty sure you could find answers to all the questions on yahoo answers.

shortythree
04-22-2014, 02:35 PM
Is this Mastering Chemistry? I remember doing it back in first year. Pretty sure you could find answers to all the questions on yahoo answers. Yes and not this question , I checked lol