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Heat of Reaction question

  1. Jan 25, 2009 #1
    1. The problem statement, all variables and given/known data
    When 50.0 mL of 1.0 M H2SO4 at 26.1 celcius was added to 50.0 mL of .976 M NaOH also at 26.1 celcius, the temperature rose to 32.6 celcius. Calculate the heat of reaction.

    2. Relevant equations

    H2SO4 (aq) + 2NaOH(aq) --> Na2SO4 (aq) + 2H2O(l)

    3. The attempt at a solution
    First, I converted both given amount of solutions into moles and found that NaOH was the limiting reactant (made .0488 mol h20 while the sulfuric acid made .1 mol). Then I converted the 50 mL of NaOH into grams (1 g/mL) and plugged that into the q=mcDeltaT equation. I did:
    50g*4.184J/gcelcius*6.5 celcius = 1359.8 J
    Then I divided that by the mols of NaOH
    1359.8 J/.0488 mol = 27864.7541 J/mol

    Can anyone confirm if I did this correctly?
    Last edited: Jan 25, 2009
  2. jcsd
  3. Jan 25, 2009 #2
    1 g/mL is the density of water, therefore I don't think that you can convert 50 mL of NaOH into 50g of NaOH using it. The density of NaOH should be something like 2.130 g/mL, but I'm not sure if that is the correct value or not.
  4. Jan 25, 2009 #3
    oh i assumed 4.184 because it is a dilute solution
  5. Jan 25, 2009 #4
    I think 4.184 is fine for the spec heat. I'm talking about the density of the material that reacted. But maybe I am misunderstanding.
  6. Jan 25, 2009 #5
    ohh good point.. hmm maybe I don't know... thanks for the heads up on that
  7. Jan 28, 2009 #6
    so im doing a lab on acid and bases and acid and metal reactions, and i have to calculate the energy change, heat

    -what would be a difference in the reactions with the heat??
    - which one would be faster??
  8. Jan 28, 2009 #7
    ditto_299, you should start a new thread and follow the template. We need more information about your problem (and you should show us what you've tried to do so far) before we can help you. Cheers.
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