Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: nikita on October 25, 2008, 08:26:26 PM
-
Calculate the enthalpy of the reaction
4B(s) + 3O2(g) :rarrow: 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
Express your answer numerically in kilojoules per mole.
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 :rarrow: 2B2O3+ 6H2O :delta: H -4070kj
B. multiply all by 2:
4B+6H2 :rarrow: 2B2H6 :delta: H +72kj
C. multiply all by -6:
6H20 :rarrow: 6H2+3O2 :delta: H +1710
D. leave as is
H2O(l) :rarrow: H2O(g) :delta: H +44
IMO everything cancels except the original equation, 4B + 3O2 :rarrow: 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?
-
You have to reverse D and multiply times 6 as well, other wise your waters do not cancel.
-
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.