Posted by HANNAH on Monday, September 22, 2008 at 8:13pm.
This is an exercise in the Born-Haber cycle. Look that up in your text or use the Internet (Google). You add the separate equations you have (you can multiply them or reverse them if needed) to get the final equation for the formation from the elements. If you multiply them, multiply the heats of formation. If you reverse an equation, change the sign of the heat of formation but keep the number the same. Just taking a quick look, it appears equation 2 stays as is, equation 1 is reversed, and equation 3 stays as is but you will need to check me out on that. I tried to do it in my head and sometimes these things don't work very well with that many numbers and symbols flying around.
REWRITE the three reactions and their ∆H's
(a) Reverse the 1st reaction and change the algebraic sign of its ∆H
(b) Leave the 2nd reaction and its ∆H unchanged.
(c) Double the 3rd reaction (multiply all its coefficients by 2). Double its ∆H, also.
(d) ADD the three rewritten reactions and do the cancellations of like terms. If you did it correctly, the simplified reaction should be:
C(graphite) + 2 H2(g) + 1/2 O2(g) -> CH3OH(l)
(e) COMBINE the ∆H's of the three rewritten reactions algebraically to get the ∆H of the formation reaction of CH3OH.
Your answer should be between -200kJ and -300kJ
Addendum:
This is an example of a Hess's Law problem. Your textbook should have some solved examples. Here is one solved example online:
http://answers.yahoo.com/question/index?qid=20080514171625AAsqOoC
And I need to correct my first response. After reading GK's response, I realize I called this the Born-Haber cycle. That is similar but this is an example of Hess's Law.