PLEASE HELP! - Tilted Forum Project Discussion Community

Stiltzkin · 12-01-2004, 10:18 PM

I need an answer to this quick, quick, quick. Well moreso, an explanation of how to solve it, because I have the answer.

From the following heats of combustion,
CH3OH(l) + 3/2O2(g) --> CO2(g) + 2H2O(l) enthalpy of reaction=-726.4 KJ/mol
C(graphite) + O2(g) --> CO2(g) enthalpy of reaction=-393.5 KJ/mol
H2(g) + 1/2O2(g) --> H2O(l) enthalpy of reaction=-285.8 KJ/mol

calculate the enthalpy of formation of methanol (CH3OH) from it's elements:
C(graphite) + 2H2(g) + 1/2O2(g) --> CH3OH(l)

How the heck do I find enthalpy of formation?
I KNOW this is in the wrong board, I know where this type of question belongs, but let's face it, it could take a whole day before anyone sees it.

Please leave this here long enough for someone to help me.

I went to a review session but the TA didn't know how to explain it, she just solved it. She solved it by multiplying the first reaction's enthalpy by (-1) and the third reaction's enthalpy by (2). Afterwards she added up all the enthalpies. The problem is that I don't understand why she did that and she couldn't explain it either >.<

Please help!

Hard8s · 12-01-2004, 10:45 PM

Check this out.
http://wine1.sb.fsu.edu/chm1045/notes/Energy/HeatForm/Energy05.htm
Hope it helps.

Hard8s · 12-01-2004, 10:48 PM

here is another link
http://dbhs.wvusd.k12.ca.us/webdocs/...Formation.html

Stiltzkin · 12-01-2004, 10:59 PM

They didn't help, but I do appreciate the effort. I *think* I got it now, though.

Well, I looked at an example in the book and I worked it through, and it turns out that it's not that hard. The reason why the TA couldn't explain it is because it's kind of a process of trial and error, and she seemed like she was in a hurry.

I apologize for the out-of-context-ness of this post, though. Feel free to move it to the appropriate board now.

Hard8s · 12-01-2004, 11:01 PM

Moogle on then my brother!! Moogle on.

12-01-2004, 10:18 PM	#1 (permalink)
Stiltzkin Junkie	PLEASE HELP! I need an answer to this quick, quick, quick. Well moreso, an explanation of how to solve it, because I have the answer. From the following heats of combustion, CH<sub>3</sub>OH<sub>(l)</sub> + 3/2O<sub>2(g)</sub> --> CO<sub>2(g)</sub> + 2H<sub>2</sub>O<sub>(l)</sub> enthalpy of reaction=-726.4 KJ/mol C<sub>(graphite)</sub> + O<sub>2(g)</sub> --> CO<sub>2(g)</sub> enthalpy of reaction=-393.5 KJ/mol H<sub>2(g)</sub> + 1/2O<sub>2(g)</sub> --> H<sub>2</sub>O<sub>(l)</sub> enthalpy of reaction=-285.8 KJ/mol calculate the enthalpy of formation of methanol (CH3OH) from it's elements: C<sub>(graphite)</sub> + 2H<sub>2(g)</sub> + 1/2O<sub>2(g)</sub> --> CH<sub>3</sub>OH<sub>(l)</sub> How the heck do I find enthalpy of formation? I KNOW this is in the wrong board, I know where this type of question belongs, but let's face it, it could take a whole day before anyone sees it. Please leave this here long enough for someone to help me. I went to a review session but the TA didn't know how to explain it, she just solved it. She solved it by multiplying the first reaction's enthalpy by (-1) and the third reaction's enthalpy by (2). Afterwards she added up all the enthalpies. The problem is that I don't understand <i>why</i> she did that and she couldn't explain it either >.< Please help! __________________ The most important thing in this world is love. Last edited by Stiltzkin; 12-01-2004 at 10:57 PM..

12-01-2004, 10:45 PM	#2 (permalink)
Hard8s Addict Location: Wherever I am!	Check this out. http://wine1.sb.fsu.edu/chm1045/notes/Energy/HeatForm/Energy05.htm Hope it helps. __________________ If ignorance is bliss, then wipe this smile off my face!

12-01-2004, 10:48 PM	#3 (permalink)
Hard8s Addict Location: Wherever I am!	here is another link http://dbhs.wvusd.k12.ca.us/webdocs/...Formation.html __________________ If ignorance is bliss, then wipe this smile off my face!

12-01-2004, 10:59 PM	#4 (permalink)
Stiltzkin Junkie	They didn't help, but I do appreciate the effort. I think I got it now, though. Well, I looked at an example in the book and I worked it through, and it turns out that it's not that hard. The reason why the TA couldn't explain it is because it's kind of a process of trial and error, and she seemed like she was in a hurry. I apologize for the out-of-context-ness of this post, though. Feel free to move it to the appropriate board now. __________________ The most important thing in this world is love. Last edited by Stiltzkin; 12-01-2004 at 11:14 PM..

12-01-2004, 11:01 PM	#5 (permalink)
Hard8s Addict Location: Wherever I am!	Moogle on then my brother!! Moogle on. __________________ If ignorance is bliss, then wipe this smile off my face!