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Topic: Heat of formation  (Read 8781 times)

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lbm_2005

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Heat of formation
« on: December 14, 2004, 09:01:45 AM »
Please help...how do you find the heat of formation with the reactants of potasium and oxygen. The quantity of potasium is 50g.

50g
K2+O2->K2O2

K2 -> 123.7 (Gas)

O2-> 0 (gas)
 

Offline Donaldson Tan

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Re:Heat of formation
« Reply #1 on: December 14, 2004, 11:38:26 AM »
experimentally or theoretically?
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lbm_2005

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Re:Heat of formation
« Reply #2 on: December 14, 2004, 01:14:20 PM »
I guess theoretically..i really dont know..but i think that i had balanced my equation wrong and it is supposed to be
2K2+o2->2K2O
i have to find the heat of formation using mole to mole setup!

Mr Amino

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Re:Heat of formation
« Reply #3 on: December 14, 2004, 01:26:45 PM »
Its just delta H (product) - delta H (reactants)

There is no mass involved.

So delta H (K2O2)-123.7= your answer
« Last Edit: December 14, 2004, 01:28:20 PM by Mr Amino »

Offline Mitch

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Re:Heat of formation
« Reply #4 on: December 14, 2004, 02:25:35 PM »
Should be   K  + (1/2)O2 -----> K2O
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Offline jdurg

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Re:Heat of formation
« Reply #5 on: December 14, 2004, 02:47:30 PM »
Should be   K  + (1/2)O2 -----> K2O

I think it would help if we knew the NAME of the product expected.  If it's potassium oxide, then the formula would be K2O.  If it's potassium peroxide, then it would be K2O2.  If it were potassium superoxide, then it would be KO2.   ;D  So then you would know how it should be properly balanced and what you need to find.  The way you could calculate the heat of formation is to use a bomb calorimeter and perform the experiment.  Then using the proper math you can calculate the heat generated per mass of product formed, and from that you can get the heat generated per mole of product which is the standard heat of formation.  

When attempting to calculate the delta H of formation, mass IS important.  If you react 50 grams of potassium completely, you need to calculate the mass of product that will be formed.  Then you take the mass of the product that is formed and figure out the moles of product.  You can then take the heat measured for that number of moles of product and convert to heat of formation for ONE mole of product.
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Mr Amino

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Re:Heat of formation
« Reply #6 on: December 14, 2004, 10:45:33 PM »
Well if im wrong, this wouldnt be the first time.  :P  However, I still wonder why you need the mass?  You say you react 50g (X moles) of K.  According to mitch's equation, wouldn't you just end up with X/2 moles of K2O if the K reacts completely like you said?  I assume this is all just theoretical and whatnot.
« Last Edit: December 14, 2004, 10:46:52 PM by Mr Amino »

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