smaerd
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Bomb Calorimeter Question
This isn't homework it's just a question on a practice test, but I am a little stuck! Any help would be greatly appreciated.
6. When 0.7521 g of benzoic acid was burned in a calorimeter containing 1,000. g of water, a temperature rise of 3.60C was observed. What is the heat
capacity of the bomb calorimeter, excluding the water? The heat of combustion of benzoic acid is -26.42 kJ/g.
1.34 kJ/C <- this is the answer
So here's what I have worked out so far.
Q = Qcalorimeter + QH2O
QH2O = -1*(4.184 J/g C * 1000g * 3.6C) = -15062.4 J/g
-15052.4 J/g * (1kJ/1000J) = 15.0624 kJ/g
Here is where I am stuck -
So then? : QCalorimeter = -26.42 kJ/g - (-15.0624 kJ/g) = -11.3676 kJ/g
Qcalorimeter = -1 * (heat capacity of calorimeter * change in temperature)
-1* (x kJ/C * 3.6C) = -11.3676 kJ/G
3.157kJ/C - which is not the right answer
If I were given the molar mass of benzoic acid I could solve this another way, but that must not be the point of the exercise because it is not
included. Thanks again.
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psychokinetic
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Perhaps you are meant to figure out the molar mass of benzoic acid yourself. We had that often as part of exercises.
“If Edison had a needle to find in a haystack, he would proceed at once with the diligence of the bee to examine straw after straw until he found
the object of his search.
I was a sorry witness of such doings, knowing that a little theory and calculation would have saved him ninety per cent of his labor.”
-Tesla
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smaerd
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this is a chem1 class, so we haven't been taught much about organic molecules yet. I do know about some basic ones and obviously with the web I could
quickly figure this out. Though if a similar question shows up on a test, I won't have that luxury  .
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psychokinetic
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Fair enough. We had it somewhat drilled into us in chem 101. But then, chem 101 was mostly organic and nomenclature. Perhaps I only find it easy
because I'm a nerd.
Sorry I don't know how to figure out the problem at hand. Good luck with that.
“If Edison had a needle to find in a haystack, he would proceed at once with the diligence of the bee to examine straw after straw until he found
the object of his search.
I was a sorry witness of such doings, knowing that a little theory and calculation would have saved him ninety per cent of his labor.”
-Tesla
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Magpie
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Watch those units - they must be as consistent as the math. If they aren't consistent something is wrong.
Q(comb) = (0.7521g)((-26.42KJ/g) = -19.87KJ
Q(H20) = (3.6C)(1000g)[4.184J/(g-C)] = 15.062KJ
Q(cal) = 19.87KJ - 15.062KJ = 4.81KJ
SpH = 4.81KJ/3.6C = 1.34 KJ/C
The single most important condition for a successful synthesis is good mixing - Nicodem
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Nicodem
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And that is where you made the mistake. You neglected that you don't have 1 g of benzoic acid, but 0.7521 g instead. Thus the heat liberated was 26.42
kJ/g * 0.7521 g = 19,87 kJ. You would have spotted the error in the equation, if you would have done the units correctly (it is -15.06 kJ, and not
kJ/g). Thus the correct QCalorimeter = 4,81 kJ. The heat capacity of the calorimeter is thus 4.81 kJ / 3.6 °C = 1.34 kJ/K.
Molar mass is irrelevant as you have the enthalpy in kJ/g units and the mass in g units.
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smaerd
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Thanks magpie and nicodem!
Ahhh! Unit's are where I make the most mistakes. I can't believe I forgot it wasn't a full gram .
Thermochem is pretty tricky but I think I'm starting to get a better grasp on the basics. Thanks guys.
[Edited on 27-10-2010 by smaerd]
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Ozone
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I attached an early version of a lab module I assembled and taught back in 2007. I made them use the Paar bomb in a foam cooler with a lab stirrer,
thermistor and a Dionex ACI with Peaknet to catch the data. They got it in mV...
Then, I showed them the Paar 6200 LE automated isoperibol  calorimeter.
Anyway, I hope this is helpful,
O3
Attachment: Calorimetry Supplementary Material_01.pdf (154kB)
This file has been downloaded 675 times
-Anyone who never made a mistake never tried anything new.
--Albert Einstein
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smaerd
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Ahh yes, thank you  seeing the diagram and some extra info helped out a lot to
see what was going on better. Thank you.
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