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Calorimetry help (chemistry)

  1. Apr 29, 2010 #1
    Hi, here is a basic summary of what we did in a lab; there were 3 reactions:

    The procedure:
    Reaction 1: Solid sodium hydroxide dissolves in water to form an aqueous solution of ions.
    NaOH(s)-> Na+(aq) + OH-(aq) ΔH1=-34.121kJ

    Reaction 2: Solid sodium hydroxide reacts with an aqueous solution of HCl to form water and an aqueous solution of sodium chloride.
    NaOH(s) + H+(aq) + Cl-(aq) -> H2O + Na+(aq) + Cl-(aq) ΔH2=-83.602kJ

    Reaction 3: An aqueous solution of sodium hydroxide reacts with an aqueous solution of HCl to form water an an aqueous solution of sodium chloride.
    H+(aq) + OH-(aq) + Na+(aq) + Cl-(aq) -> H2O + Na+(aq) + Cl-(aq) ΔH3= -50.2kJ

    The ΔH values were calculated by dividing the heat gained by the number of moles (each reaction had 0.05moles of NaOH)

    The problem:

    Net ionic equations for reaction 2 & 3:

    2: NaOH(s) + H+(aq) -> H2O + Na+(aq)
    3: H+(aq) + OH-(aq) -> H2O

    i) In reaction 1, ΔH1 represents the heat evolved as solid NaOH dissolves. Look at the net ionic equations for reactions 2 and 3 and make similar statements as to what ΔH2 and ΔH3 represent.

    ii) Compare ΔH2 with (ΔH1 + ΔH3). Explain in sentences the similarity between these two values by using your answer to #5 above.

    Attempt at answering:
    i) Firstly, ΔH2 represents the heat evolved as the hydrogen ion displaces the sodium ion, creating a single displacement reaction. ΔH3 represents the heat evolved as the hydrogen and hydroxide ion form water via a neutralization reaction.

    ii) ΔH2 is equal to (or supposed to be, this is a source of error while calculating) (ΔH1 + ΔH3). The similarity between these two values is that .. (this is where I get confused!)
     
  2. jcsd
  3. Apr 30, 2010 #2

    Borek

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    Staff: Mentor

    These were two different experiments, yet what you state is in both cases heat evolved is just that of the neutralization reaction.

    Think again about what must happen in the first case (that is, second reaction).
     
  4. May 2, 2010 #3
    I thought about it and I'm still hazy by what you mean. "both cases heat evolved is just that of the neutralization reaction."
     
  5. May 2, 2010 #4

    Borek

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    Staff: Mentor

    I agree that taken out of context it doesn't make sense, but in the context in which I wrote it I still think it is OK. But let's try to reword it, and remember that what I wrote refers to the quote from your post:

    The way you wrote it there is no difference between sources of heat evolved in both experiments - if you try to write reaction equations for each process as described, you will see that in both cases it is H+ + OH- -> H2O. This is obviously wrong - you are omitting something.

    --
     
  6. May 2, 2010 #5
    Hi, I'm still unsure what I am omitting.

    Part I: I assume reaction 3 is the energy/heat gained as the H and OH molecules react because they are the only two in the net ionic equation.

    Part II:
    Perhaps I'm not understanding the purpose of the net ionic equations other than that if there is a product created, it creates energy and the net ionic equation eliminates a substance that is on both sides.

    This is exactly what I wrote in a word file, for the lab:

     
  7. May 2, 2010 #6

    Borek

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    Staff: Mentor

    Google heat of dissolution or heat of solvation.

    --
     
  8. May 2, 2010 #7
    I googled it and even searched through threads on this forum, but I still can't get a solid answer.

    What I know is H1 + H3 = H2

    Why? I don't know. I don't understand why they add up. I also don't understand what I am omitting in the 3rd net ionic equation.

    In the lab it says to explain these using the net ionic equations only.
     
  9. May 3, 2010 #8

    Borek

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    Staff: Mentor

    What about

    NaOH(s) -> Na+(aq) + OH-(aq)

    --
    methods
     
  10. May 3, 2010 #9
  11. May 3, 2010 #10

    Borek

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    Staff: Mentor

    When solid NaOH dissolves, heat evolves. That's your 1st experiment - heat of dissolution. When dissolved NaOH reacts with acid (neutralization), heat evolves - that's your 3rd experiment, heat of neutralization.

    What are sources of heat in your 2nd experiment?

    --
     
  12. May 3, 2010 #11
    Alright, so would this be fair to say?

    Reaction 2 represents the heat evolved as the sodium ion(s) in sodium hydroxide(s) is displaced with the hydrogen ion(aq), producing water and a sodium ion(aq). It also represents the heat evolved from the dissolved NaOH and the neutralization of water.

    Reaction 3 represents the heat evolved as a hydrogen ion and hydroxide ion undergo a neutralization reaction, producing water.


    The solid NaOH dissolves into a liquid NaOH in reaction 1. This is the heat of dissolution. In reaction 3, the reaction undergoes a neutralization reaction, creating water. In reaction 2, the NaOH dissolves into a liquid and also forms water, thus, both reactions occur doing reaction 2, making H2 = (H1+H3)
     
  13. May 3, 2010 #12

    Borek

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    Staff: Mentor

    Right :smile:

    --
    methods
     
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