1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calorimetry problem

  1. Dec 5, 2015 #1
    1. The problem statement, all variables and given/known data
    When 1.50g of Ba(s) is added to 100.0g of water in a container open to the atmosphere, the reaction shown below occurs and the temperature of the resulting solution rises from 22.00 celsius to 33.10 celsius. If the specific heat of the solution is 4.18J/g*celsius, calculate the change of enthalpy for the reaction, as written below.

    Ba(s) + 2H2O(l) -> Ba(OH)2(aq) + H2(g)

    2. Relevant equations
    q = ms(delta)T
    q(surrounding) = -q(system)
    (delta)H = q (constant pressure)

    3. The attempt at a solution
    First, I calculated the limiting reagent of the solution (Ba(OH)2) using both the masses of Ba and H2O. The limiting reagent happened to be .0109 moles of Barium Hydroxide (also the same number of moles for solid Barium). After that, I multiplied the moles by the molar mass of Barium Hydroxide, which gave me the mass of 1.87g. After I found delta T, I used the following information to find the heat of the surrounding by performing the following equation:

    q(surrounding) = (1.87g)(4.18J/g*celsius)(11.10 celsius)
    q = 86.8 J

    Since the reaction is an exothermic reaction, the q will become negative, becoming -86.8J.

    After that, I calculate the entire reaction by using the following equation:

    [(.0109 mol Ba(OH)2) / -86.8J)] = [1 mol Ba(OH)2/x]
    .0109x mol Ba(OH)2 = -86.8J * mol Ba(OH)2
    x = -7963.30 J or -7.96 kJ

    The correct answer happens to be -431 KJ/mol. I have absolutely no clue how it got there. I even tried to work backwards from the answer, but it still makes no sense. I have my final exam next week and I really need to understand this 100%.
     
  2. jcsd
  3. Dec 5, 2015 #2

    DrClaude

    User Avatar

    Staff: Mentor

    Think carefully here.
     
  4. Dec 5, 2015 #3
    I triple checked the calculations and the wording, and I don't see what the problem is. The specific heat is the for aqueous solution. It is pretty obvious that the limiting reagent would involve the dimensional analysis from the Barium just by looking at the molar mass of that and water, and looking at the given masses. So after finding the limiting reagent, I multiplied it by the molar mass of Barium Hydroxide (since that is the solution we were given the specific heat for) and got the required mass to find the heat of the reaction for Barium Hydroxide. That equation is for the surrounding heat, which is positive, before I convert it to system heat, which will then become negative 86.8 J. Yes, it is joules because the given specific heat was in joules. I am not soo sure what you are implying since that response is pretty vague. I was hoping for maybe a more specific explanation for why that is wrong?
     
  5. Dec 5, 2015 #4

    DrClaude

    User Avatar

    Staff: Mentor

    One of the numbers in that equation was wrong. It should be obvious which one. I'll give you another hint anyway: the calorimeter is measuring the change in temperature of what?
     
  6. Dec 5, 2015 #5
    Change of heat: 33.10 celsius - 22.00 celsius = 11.10 celsius

    That is the change of temperature of the solution.

    The problem specifically said the specific heat of the solution is 4.18.

    Also, .0109 moles * [((137.327+2(16+1.008) g / 1 mol Ba(OH)2] = 1.87 g Ba(OH)2

    I don't see any need to convert any of the units in my calculation since everything will cancel out to give me just joules. Is there anything you are seeing that I am not?
     
  7. Dec 5, 2015 #6

    Borek

    User Avatar

    Staff: Mentor

    How much of the solution is there?
     
  8. Dec 5, 2015 #7
    1.87g of solution.
     
  9. Dec 5, 2015 #8

    Borek

    User Avatar

    Staff: Mentor

    And that's where you are wrong.
     
  10. Dec 5, 2015 #9
  11. Dec 5, 2015 #10
    Look, I understand you guys want people to figure it out themselves and not have others do their own HW, but you see, if I actually had a clue on what to do, I wouldn't be posting this problem on here. Hell this is the first time I have ever even used the HW help on this website because most of the time, I figure it out myself. Borek, could you please explain to me why my mass for the solution is wrong? I have already shown the dimensional analysis in my previous posts on how I got that number. The barium is the limiting reagent, so all I did was some stoichiometry to find out that the solution has the same number of moles as the solid barium. So unless you do something completely different, then I am lost.
     
  12. Dec 5, 2015 #11
    Wait...is the mass I'm multiplying supposed to be the mass of the water?
     
  13. Dec 6, 2015 #12

    Borek

    User Avatar

    Staff: Mentor

    Better, but still not correct. Solution is not just water - it is water and whatever is dissolved in it.
     
  14. Dec 6, 2015 #13
    Well that explains why my answer beforehand was slightly below the correct answer. Now, I just solved for the right answer. Thanks a lot for the help. In the future, I need to calculate the overall mass of the solution by taking the sum of all the masses given in the solution, and then follow the same procedures there after?
     
  15. Dec 6, 2015 #14

    Borek

    User Avatar

    Staff: Mentor

    Yes, mass of the mixture is sum of masses of all its components. It is an obvious conclusion of the mass conservation principle.
     
  16. Dec 6, 2015 #15
    Well, I feel pretty stupid right now. Nonetheless, I feel a little more comfortable with thermodynamics, so that is good I guess. I apologize about all the posts beforehand. I was kind of under pressure I have my exam coming up next week and I kind of panicked a little bit.
     
  17. Dec 6, 2015 #16

    Borek

    User Avatar

    Staff: Mentor

    No problem, we all start to crack now and then :wink:
     
  18. Dec 6, 2015 #17

    DrClaude

    User Avatar

    Staff: Mentor

    Let me add an epilogue. You can avoid this kind of mistake by thinking about the physical context, what is actually going on. As I asked above: the calorimeter is measuring the change in temperature of what? The answer is: of everything that is inside it. Therefore, the change in temperature is for the entire solution, and the heat released will increase the temperature of that solution, so it is that mass that must be included in the formula.
     
    Last edited: Dec 6, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Calorimetry problem
  1. Calorimetry Problem. (Replies: 2)

  2. Calorimetry Problem (Replies: 1)

Loading...