Chemical Forums
Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: ainoko_hikaru on September 18, 2008, 09:30:13 AM
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i'm a bit confused in this problem. i think i know how to solve it, but i don't know if what i am doing is correct. please help.
here's the problem:
A 1.50g sample of NH4NO3 (s) is added to 35.0g of h2O
in a coffee cup calorimeter and stirred until it dissolves. The
temperature of the solution drops from 22.7°C to 19.4°C. What
is the heat of solution of NH4NO3, that is, what is the ΔH for
the process?
I tried using the formula qrxn = m*s*ΔT.
my value for m is the sum of the grams of ammonium nitrate and water, which is 36.5g.
my value for s is the sum of the specific heats of ammonium nitrate (1.77J/g°C) and water (4.186 J/g°C) which would be 5.956J/g°C.
ΔT would be the difference of 19.4°C and 22.7°C.
therefore, the value for q would be 220.694J.
since ΔH = qrxn / nsolute ,
ΔH = 220.694J / (1.50g / 80.04g/mol)
= 11.8 kJ/mol
am i correct? thanks... :)
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my value for s is the sum of the specific heats of ammonium nitrate (1.77J/g°C) and water (4.186 J/g°C) which would be 5.956J/g°C.
Everything else looks OK to me, but that's not correct. Specific heat of solution depends on concentration and is NOT sum of specific heats of solvent and solute. I can be missing something, but I don't recall any systhematic method of calculating specific heat of solution, I believe these things have to be measured experimentally.
You can assume spefic heat of your solution is that of pure water.
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so instead of 5.956J/g°C, I will only use 4.186 J/g°C?
the answer would then be...
qrxn = 504.2037J
ΔH = 504.2037J / (1.50g / 80.04g/mol)
= 26.9 kJ/mol
is this correct? :)