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Homework Help: Constant-Pressure Calorimetry question

  1. Aug 21, 2017 #1
    1. The problem statement, all variables and given/known data
    A quantity of 4.00 3 10 2 mL of 0.600 M HNO 3 is mixed with 4.00 3 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°C. What is the final temperature of the solution? (Use the result in Example 6.8 for your calculation.)

    The result in example 6.8 is the heat of neutralization = -56.2 kJ/mol, I believe.

    2. Relevant equations

    q = msDelta-T

    3. The attempt at a solution
    I've asked almost everybody I know who knows anything about chemistry to give this a shot, but no luck. One got x = 120 (no idea where that came from). I've spent hours working on this problem, and have tried various methods, anything I can think of. Two of my closer shots were:

    Taking 6.8's 2810J, I tried writing (with C meaning "degrees Celsius"):

    2810J = (400mL + 400mL)(4.184 J/g * C)(x - 18.46C)
    2810 = 3347.2x - 61789.312
    x = 19.2995 (degrees Celsius).

    This is incorrect.

    I tried taking the number of moles, determining the masses, then combining those, to write; I continue to assume q = 2810J, as have other people in past forum posts I've found via Google. Not entirely sure what else it could be.):

    q = (35.684g)(4.184 J/g * C)(x - 18.46C)
    2810J = 149.3x - 2711.288
    x = 36.98
     
  2. jcsd
  3. Aug 21, 2017 #2
    I think I got it, but I'm off by .10 from the answer in the book. Gonna do one more iteration, but with more attention paid to significant figures; if I'm closer, I'll post what I got.
     
  4. Aug 21, 2017 #3
    The balanced equation:

    2HNO3 (aq) + Ba(OH)2 (aq) ---> 2H2O (l) + Ba(NO3)2 (aq)

    The molar masses:
    HNO3 = 63.018g
    Ba(NO3)2 = 261.32g
    Ba(OH)2 = 171.316g

    Since
    400mL = 400g
    we can write

    400g Ba(OH)2 * (1 mole Ba(OH)2 / 171.316g Ba(OH)2) = 2.34 mol Ba(OH)2

    Using the heat of neutralization found in the previous problem (-56.2 * 10^3 J), we can determine Qrxn as follows:

    2.34 mol Ba(OH)2 * ((5.62 * 10^3 J) / 1 mol Ba(OH)2) = -1.32 * 10^4 J

    Thus

    Qrxn = -1.32 * 10^4 J

    Qrxn = -Qsoln

    -1.32 * 10^4 J = - Qsoln

    Qsoln = msDelta-T = (400g + 400g)(4.184 J/g * C)(x - 18.46C) = (3347.2x - 6.18*10^4 J)

    -1.32 * 10^4 J = - (3347.2x - 6.18 * 10^4 J)

    -7.5 * 10^4 J = -3347.2x

    x = 22.41

    The back of the book gives 22.49.
     
  5. Aug 22, 2017 #4

    Borek

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    Staff: Mentor

    Nothing seems correct here (which is not necessarily your fault).

    First of all, without even trying to understand what you did: one can't use an approximate value of 56.2 kJ (three significant digits) and expect an exact answer accurate to 4 significant digits. SD are blatantly wrong in many cases, but sometimes the logic they provide helps see why something is conceptually wrong.

    Accuracy of the 1g=1mL assumption is also questionable. As a rule of thumb for diluted solutions it works, but not when you expect 4 SD accuracy. 0.6 M nitric acid has the density of 1.0185 g/mL, which is more than enough to make the difference between 22.41 and 22.49 ridiculous (if you assumed 1g=1mL, you have 2% error in one of the input data, don't expect better accuracy in the final answer).

    You are doing many things that I have problems to follow, one I got is quite wrong:

    No. You don't have 400 g of Ba(OH)2. You have 400 mL of 0.3 M solution. All you need is n=CV, no need for molar masses here.
     
  6. Aug 22, 2017 #5
    This is from the textbook. The associated example from which this problem is a practice exercise said to assume x mL = x g.

    I'm completely flummoxed by this problem, as is everybody else I've asked to help. Most of what I did here is derived from the examples immediately preceding this practice exercise (it is from example 6.8 from Chang's Chemistry), as well as from the Instructor's Solution manual's solution for the linked-to problem 6.38 (i.e. after doing practice exercise 6.8, it directs you to problem 6.38 at the end of the chapter). Unfortunately, this solution manual does not give any help for practice exercises, which is why I really have no idea what's right or wrong.
     
  7. Aug 22, 2017 #6

    Borek

    User Avatar

    Staff: Mentor

    x mL = x g is for mass of the SOLUTION. You do understand the difference between SOLUTION and SOLUTE?

    If you have 1000 mL of 1 M NaCl solution - how many moles of NaCl is there? Just one (which just follows the definition of molarity), or 1000/58?

    Nothing really difficult here, on the contrary - the problem seems quite trivial. Calculate number of moles from n=CV (in general you should do it for both substances and find the limiting reagent, but even without calculations it is obvious they are mixed in the stoichiometric ratio), knowing number of moles of water produced calculate amount of heat produced from the neutralization, then just plug and chug into q=mcΔT (assume density of 1g/mL - that's where the x mL = x g comes into play).

    You won't get high accuracy with this approach. For an exact number you would need tabulated data on the solution density and specific heat of the Ba(ON3)2 - good luck finding them (hint: don't even try to waste your time doing it).
     
  8. Aug 22, 2017 #7
    I can't tell you how satisfying it is to have somebody go on about how wrong you are, only to have it confirmed that you are right. Such an arrogant tone for a homework-help forum, especially from somebody who was dead wrong.
     
  9. Aug 22, 2017 #8
    My answer was wrong, but I had the right idea. I finally got the proper solution, and it shows that I did make a mistake - should have taken the 0.24 moles, as per the equation n=MV - and which I had written on one of my five pages of notes, but I overlooked in the frustration of spending five hours and two nights working on the problem - rather than taking the 400g and dividing it by the molar mass to get the number of moles as 2.34. Other than that, however, my entire procedure was correct. INCLUDING converting 400mL to 400g.
     
  10. Aug 23, 2017 #9

    Borek

    User Avatar

    Staff: Mentor

    I told you in my very first post - fact that things are wrong is not necessarily your fault. To me it rather looks like the question is poorly formulated, as the accuracy given in the book's answer is impossible to achieve with the data given.

    To repeat what I wrote above - converting 400 mL to 400 g is only an approximation. Mass of the 400 mL of 0.6 M nitric acid solution is not 400 g, it is 407.4 g. Finding it requires knowing the density, which in turn requires consulting the density tables, if you have not used them your answer is a bit off.

    Besides, please reread your original post - you refer to question 6.8, you quote it as giving

    and then in your calculations you use a number

    that seems unrelated (and you never explained where it came from). This is inconsistent and makes it impossible to follow your work.
     
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