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Prelab Hess problem

  1. Apr 12, 2013 #1
    1. The problem statement, all variables and given/known data

    ΔH for the rxn 3MgO(s) + 6H+(aq) --> 3Mg2+(aq) + 3H20(l) = 409.5 kj/mol
    What is ΔH for Mg2+(aq) + H20(l) --> Mg0(s) + 2H+(aq)






    3. The attempt at a solution
    We just went over this in class I'm confused as ever. I don't know if it is a
    ΔH = Ʃreactants - Ʃproducts
    Or if I had to somehow manipulate the equations which is what we just went over . I don't really understand so if you point me in the right direction I would appreciate it. I'm confused because all the worksheets where we did the latter choice I mentioned we were given a target equation. Here I don't see one. Explain your thinking please.
    Thanks ,
    J
     
  2. jcsd
  3. Apr 13, 2013 #2

    Borek

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    Staff: Mentor

    Compare both equations reagent by reagent.
     
  4. Apr 13, 2013 #3
    Can you elaborate? I don't understand. Thank you.
     
  5. Apr 13, 2013 #4
    Hi
    I looked at it again. I thought how can I get from the first equation to the second
    ΔH for the rxn 3MgO(s) + 6H+(aq) --> 3Mg2+(aq) + 3H20(l) = 409.5 kj/mol
    What is ΔH for Mg2+(aq) + H20(l) --> Mg0(s) + 2H+(aq)
    If I divide the first equation by 3
    I get MgO(s) + 2H+(aq) --> Mg2+(aq) + H20(l) = 136.5 kj/mol
    Then I can flip this and I will get
    Mg2+(aq) + H20(l) --> Mg0(s) + 2H+(aq) = -136.5 kj/ mol
    Which is the first second equation. So I just transformed the first into the second. I'm not sure if this is even proper I just kind of went with it lol. What does someone think?
     
  6. Apr 13, 2013 #5

    Borek

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    Staff: Mentor

    That's exactly what you were expected to do. Note that it is a direct application of the Hess law.
     
  7. Apr 13, 2013 #6
    Thanks dude we just went over it and it didn't sink in yet. Appreciate your help.
     
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