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#### nikita

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##### Calculate enthalpy of reaction
« on: October 25, 2008, 02:26:26 PM »

Calculate the enthalpy of the reaction
4B(s) + 3O2(g)  2B2O3

given the following pertinent information:
A. B2O3(s) + 3H2O(g) -> 3O2(g) + B2H6(g)  :delta: H = 2035kJ
B. 2B(s) + 3H2(g) -> B2H6(g)  :delta: H = 36kJ
C. H2(g) + 1/2O2(g) -> H2O(l)  :delta: H = -285kJ
D. H2O(l) -> H2O(g)  :delta: H = 44kJ

i just want to make sure I am doing this correctly.  I am going to write what the equations become and the new  :delta:H.

A.  multiply all by -2:
6O2+2B2H6 2B2O3+ 6H2O   :delta: H  -4070kj
B.  multiply all by 2:
4B+6H2 2B2H6  :delta: H +72kj
C.  multiply all by -6:
6H20 6H2+3O2   :delta: H +1710
D.  leave as is
H2O(l)  H2O(g)  :delta: H +44

IMO everything cancels except the original equation, 4B + 3O2  2B2O3 . I am not getting the correct answer.  Have I been looking at this so long that I am missing something obvious?  I am starting to not be able to do it anymore, but Ive done it several times and always come out with -2244kj.  Any insight as to what I am doing wrong?
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#### enahs

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##### Re: Calculate enthalpy of reaction
« Reply #1 on: October 25, 2008, 05:43:45 PM »

You have to reverse D and multiply times 6 as well, other wise your waters do not cancel.
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#### nikita

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##### Re: Calculate enthalpy of reaction
« Reply #2 on: October 26, 2008, 08:09:47 AM »

I cannot believe that I can look at a problem that many times and not see the liquid and the gas.  thank you, i needed some fresh eyes.
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