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Enthalpy, heat of neutralization, calorimetry

  1. Jun 29, 2010 #1
    1. The problem statement, all variables and given/known data
    A quantity of 2.00x10^2 mL of .862 M HCl is mixed with 2.00x10^2 mL of .431 M Ba(OH)2 in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the HCl and Ba(OH)2 solutions is the same at 20.48 C.

    For the process
    H+(aq) + OH-(aq) --> H2O (l)
    the heat of neutralization is -56.2 kJ/mol. What is the final temperature of the mixed solution?



    2. Relevant equations
    I'm thinking all I need is q=ms delta t


    3. The attempt at a solution
    q=(heat of neutralization per mol)(number of moles H2O)
    q= -56.2 kJ/mol x (200 mL*.862mol/1000mL) = -9.69 kJ

    If -9.69 = q = ms(T(f)-T(i))
    Then T(f) = q/ms + T(i)
    = -9690 J/(400g)(4.184g/J*C) + 20.48*C
    = 121.8 degrees C

    The answer should be 26.3 degrees C. Any help?
     
  2. jcsd
  3. Jun 29, 2010 #2

    Borek

    User Avatar

    Staff: Mentor

    Do you know how to use calculator?

    [tex]\frac{9690}{400\times4.184} + 20.48[/tex]

    and not

    [tex]\frac{9690}{400} 4.184 + 20.48[/tex]

    (not to mention fact that you have lost minus sign, in a way luckily for you).
     
    Last edited by a moderator: Aug 13, 2013
  4. Jun 29, 2010 #3
    well that's embarrassing. thanks.
     
  5. Jun 30, 2010 #4

    Borek

    User Avatar

    Staff: Mentor

    Happens :devil:
     
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