September 07, 2024, 11:21:44 PM
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Topic: Finding the heat of formation when given multiple equations, this is hard!  (Read 3818 times)

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Offline petstar21

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I usually am very good at these problems even though its math which is difficult for me. Ive been working at this problem so many times youd think the answer would puke out of me but i cant get it!

Calculate the heat of formation of NO form following equations:
4NH3 + 5O2 --> 4NO + 6 H2O  (delta H/ heat of formation= -1170 KJ)
4NH3 + 3O2---> 2N2 +6 H2O (delta H/ heat of formation = -1530 KJ)

I know the eqaution for formation of No is N2 + O2---> 2NO and then youd divide by two to get the heat fo formation for NO ( 1 mol)

therefore, you clearly reverse the last equation and everything cancells out except the O2's bc ones 5 and ones 3 so how do u do it??

Offline Borek

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everything cancells out except the O2's

And N2 and NO, so hardly everything...
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Offline english

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For me, everything is canceling to the heat of formation equation you gave.  You're over-thinking it.

Offline petstar21

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o wow im dumb!
but then when you have 2N2 + 2 O2---> 4 NO you have to divide by two, but do you therefore divide the heat of formnations by two as well and then your final answer by two???

Offline english

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This is actually a simplification of Hess's Law!  If you divide one reaction step by an integer, then yes the heat of reaction must be divided by that same integer.

If you half the molar amounts of the species present, it makes sense to half the heat given off, no?

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