Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: MitchTwitchita on January 26, 2008, 09:27:00 PM
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can anybody here help me out with this problem?
A quantity of 4.00 x 10^2 mL of 0.600 M HNO3 is mixed with 4.00 x 10^2 mL of 0.300 M Ba(OH)2 in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of both solutions is the same at 18.46 degrees C. What is the final temperature of the solution? (Use the result in Example 6.8 for your calculation).
Results for previous example: qrxn = -2.81 KJ.
qrxn + qsoln = 0
qrxn = -qsoln
therefore, qsoln = 2.81
qsoln = ms(deltaT)
2.81 = (400 g + 400 g)(4.184 KJ/g * degrees C)(x - 18.46 degrees C)
x = [2.81/(800 g)(4.184 KJ/g * degrees C)] + 18.46 degrees C
=18.46 degrees C?
Can anybody please show me where I'm going wrong?
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Results for previous example: qrxn = -2.81 KJ.
qrxn + qsoln = 0
qrxn = -qsoln
therefore, qsoln = 2.81
qsoln = ms(deltaT)
2.81 = (400 g + 400 g)(4.184 KJ/g * degrees C)(x - 18.46 degrees C)
x = [2.81/(800 g)(4.184 KJ/g * degrees C)] + 18.46 degrees C
=18.46 degrees C?
Can anybody please show me where I'm going wrong?
should be 4.184 J / g deg C
change 2.81 kJ to Joules first then do the math
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2810 J/(800 g)(4.184 J/g * degrees C)(x - 18.46 degrees C)
=19.30 degrees C
This still seems to be the wrong answer. I think I'm totally screwing up somewhere in the molarities of the the acid and base.
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molarities don't enter into the calculation unless you are calculating enthalpy per mole.
the acid and base concentrations are fairly dilute and there is a large volume of solution. water has a high specific heat so can absorb alot of energy before it increases temp.
the only point that is missing is that the question did not give a density for the solution, which you assumed as 1.00g/mL and is likely correct . otherwise the mass of the solution would be the only thing that would seem unaccounted for.
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O.K., so I figured the masses of each species.
HNO3 = 0.400 L x 0.600 mol/1L = 0.24 mol
0.24 mol x 63.02 g/1 mol = 15.1 g HNO3
Ba(OH)3 = 0.400 L x 0.300 mol/1 L = 0.12 mol
0.12 mol x 188.3g/1 mol = 22.60 g Ba(OH)3
But now I'm stuck again.
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O.K., so I figured the masses of each species.
HNO3 = 0.400 L x 0.600 mol/1L = 0.24 mol
0.24 mol x 63.02 g/1 mol = 15.1 g HNO3
Ba(OH)3 = 0.400 L x 0.300 mol/1 L = 0.12 mol
0.12 mol x 188.3g/1 mol = 22.60 g Ba(OH)3
But now I'm stuck again.
your previous calculation was correct. the mass of the solution was needed and you had that right. I just meant that to be complete, the question should have given the density of the solution. mass of each compound is absorbed into the volume of the solutions. as I see it the final temp of 19.30 deg is correct
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The back of the text has an answer of 22.49 degrees C, which I can't get for the life of me.
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Check the results from the previous question you had to use.
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The result for qsoln is 2.81 KJ
and the heat of neutralization is -56.2 KJ/mol
However, I don't know if heat of neutralization applies or how to apply it if it does.
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water has a high specific heat so can absorb alot of energy before it increases temp.
I would rather put it as "water has a high specific heat so its temp. increases slowly when absorbing energy".
"Before" is not the best word in this context, as it may suggest energy absorption without temperature change. No such animal.
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phase transitions?