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Calculating heat

  1. Nov 11, 2009 #1
    1. The problem statement, all variables and given/known data

    When 2.0g of NaOH were dissolved in 53.0g water in a caloremeter at 24.0 degrees Celsius, the temperature of the solution went up to 33.7 degrees Celsius.

    2. Relevant equations

    q = c * m * delta t

    -(c * m * delta t) = (c * m * delta t)

    c = specific heat, m = mass, t = temperature

    3. The attempt at a solution

    When 2.0g of NaOH were dissolved in 53.0g water in a calorimeter at 24.0 degrees Celsius, the temperature of the solution went up to 33.7 degrees Celsius.

    a) Calculate q H20.

    I used the equation q = cmt, so

    q = (4.18)(53.0)(9.70) = 2150J

    b) Find the change of enthalpy for the reaction as it occured in the calorimeter.

    = -2150J

    c) Find the change of enthalpy for the solution of 1.00g NaOH in water.

    For this I just halved -2150J and got -1080 (3 significant figures) but I'm not sure if this is correct.

    d) Find the change of enthalpy for the solution of one mole NaOH in water.

    For this I found the mass of one mole of NaOH, which is 40.00g. Then I multiplied this by my answer to part c and got -43200J/mol

    e) Using enthalpies of formation as given in thermodynamic tables, calculate the change of enthalpy for the reaction NaOH yields Na^+ + OH^-

    I looked at a thermodynamic data table and found the delta H value of Na^+ to be -240100J/mol, OH^- to be -230000J/mol, and NaOH to be -425600J/mol. This obviously isn't what I got for my answer in part d. What have I done wrong? Please Help!
     
    Last edited: Nov 12, 2009
  2. jcsd
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