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Caculating the enthalpy change of reaction from delta T values(experimentally)

  1. Feb 6, 2012 #1
    Hi! Thanks for your interest in this this post :D!

    So two reactions I performed were: A) Adding 50 cm3 water to 0.025 moles CuS04 and measuring ΔT.
    B) Adding 50 cm3 water to 0.025 moles CuS04.5H2O and measuring ΔT.

    Well turns out the calculations were wrong, since I apparently didn't get kJ/mol but rather mol/40 or so since I didn't take 1 mole of each element.

    I want to know whether this was right since my calculations don't yield answers close to the real value otherwise.

    ∆H reaction = (49.736*4.186*-0.228) /0.025moles= -2 kJ/mol

    ∆H2 = (51.85*4.186*(-0.929)) /0.025moles= -8 kJ/mol

    -2kJ and -8kJ are totally wrong! So I was told about the whole multiply the answer by 40 business.

    That gives us: -80-(-320)=-240 kJ which is still not close to the real answer of -11.7 kJ(rough).

    I used a temperature sensor, a calorimeter too! What went wrong and how can I correct this? Thanks for your patience this was a lengthy post I hope you can help me solve this problem! :)
  2. jcsd
  3. Feb 8, 2012 #2
    Help please?
  4. Feb 9, 2012 #3
    Well any idea>
  5. Feb 9, 2012 #4


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    Staff: Mentor

    You listed some numbers but you never explained what is what of what. I can guess, but if you want to get help, you need to be precise describing what you are doing.
  6. Feb 9, 2012 #5
    Okay my bad, first off can you suggest how you would carry out this experiment?
    Here are the reactions for the hydration of CuSO4:


    1)CuSO4(s) + Aq -----> CuSO4(aq) ∆H reaction What I got for ∆H reaction= -2kJ/mol Theoretical: -66.5 kJ/mol

    2)CuSO4(s) -----> CuSO4.5H20(s) ∆H reaction What I got for ∆H reaction= TO BE DERIVED USING HESS'S LAW kJ/mol Theoretical: -11.7kJ/mol

    3)CuSO4.5H20(s) ----->CuSO4(aq) ∆H reaction What I got for ∆H reaction= -8kJ/mol Theoretical: - kJ/mol

    So I dissolved 0.025 moles of CuSO4.5H20, CuSO4 in 50 cm3 water. (I don't know the significance of these number! These might be the source of error.) And I measured the change in temperature as Delta T

    Reaction 1) delta T=-0.228
    Reaction 3) Delta T=-0.929

    Now I calculated the Enthalpy changes from: Q/no of moles= mcΔT( to give kJ/mol)

    and I got freak values.
  7. Feb 10, 2012 #6


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    Staff: Mentor

    Without looking at the rest - these numbers don't make sense. First of all, in both cases delta T should be positive, as solutions should warm up. Second (and more important), temperature change should be larger for the first reaction (of the anhydrous copper sulfate). It is not. That means there was something wrong with the experiment, or with your notes, whatever.
  8. Feb 10, 2012 #7
    Hmm, I think I went wrong when I was told to take 0.025 moles and 50 cm3 without any damn reason. Okay, I measured the temperatures only after I added water, that's the reason the values are negative. but in that case the enthalpy won't be negative when it IS an exothermic reaction.
  9. Feb 11, 2012 #8


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    Staff: Mentor

    I don't see anything wrong with these numbers.

    So you don't know what was the initial temperature of the system? If so, how do you know ΔT? No idea what you are trying to tell here.
  10. Feb 11, 2012 #9
    I hope I can drop the formality. But Dude! measuring the initial temperature would mean measuring the temperature of water at room temperature?

    Oh and I was told I had the wrong answer because 0.025 mols does NOT represent 1 mol obviously and so my units in actuality weren't kJ/mol but kJ/mol/0.025. Odd logic or sensible? I disagree with this since I accounted for this in the formula:

    (mass of the solution * specific heat * temperature change)/(no.of moles of salt reacting) = enthalpy change ( kJ/mol)

    And you didn't answer me on how else my answer will be negative, I mean, none of the variables in the above formula would seem to be negative, like:
    Mass-postive (duh!)
    c- 4200 per kg, still positive
    ΔT- positive, I see you point. Well this is odd ain't it?

    And thanks for taking the time Borek!
  11. Feb 11, 2012 #10


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    Staff: Mentor

    Yes. But without doing it you can't know what the temperature change was.

    If you divided by number of moles your result is already per mole. But it is not clear to me what you did, when and how, as your posts are slightly chaotic.

    ΔT should be positive, which doesn't mean enthalpy of hydration is positive as well. Hydration is an exothermic process, by convention enthalpy of the exothermic process is negative, as energy is given away (is lost by the system). So you should have positive ΔT as water gets warm, but you should flip the sign when calculating ΔH.
  12. Feb 11, 2012 #11
    Well, turns out I won't need to flip the sign. However I have hardly varied my independant variable hence I must perform the experiment with 0.025 M, 0.05M, 0.1M, 0.2M, 0.4M masses of the chemicals. That will indicate efficiency in a better manner, you know may be I could calculate the average enthalpy change from the individual answers I get.
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