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Author Topic: heat of solution of ammonium nitrate added to water  (Read 9052 times)

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ainoko_hikaru

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heat of solution of ammonium nitrate added to water
« on: September 18, 2008, 03:30:13 AM »

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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Borek

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Re: heat of solution of ammonium nitrate added to water
« Reply #1 on: September 18, 2008, 04:45:24 AM »

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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ainoko_hikaru

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Re: heat of solution of ammonium nitrate added to water
« Reply #2 on: September 18, 2008, 07:11:13 PM »

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?  :)
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