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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!
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Last edited by Stiltzkin; 12-01-2004 at 10:57 PM..
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