Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Enthalpy Lab Explanation!

  1. Jun 14, 2016 #1
    I just conducted a lab in which I am testing varying amounts of magnesium to react with a consistent amount of hydrochloric acid, determining which amount of magnesium produces a reaction whose enthalpy is the closest to the theoretical/ideal enthalpy of this equation. I created 3 conditions using 0.025 mols Mg, 0.020 mols Mg, and 0.017 mols Mg. 0.017 mols produced the closest to the theoretical enthalpy, then 0.020, then 0.025 being the least accurate. I am completely lost as to why this would be! Could anyone provide some sort of scientific explanation that is responsible for my data? Why does the least amount of Mg yield the most accurate enthalpy?
    Additional info just in case you need it to answer:
    Balanced equation: Mg + 2HCl ------> MgCl2 + H2 (can't do subscript)
    I used 50 mL HCl with a concentration of 1 mol Hcl per 1 L HCl(aq)
    Thanks all! Also this is urgent because my lab is due tomorrow :)
     
  2. jcsd
  3. Jun 15, 2016 #2
    What would be the stoichiometric amount of Mg required?
     
  4. Jun 15, 2016 #3
    I can't think of any systematic error that would produce that data. I think you just got a strange set of random error in your data.
     
  5. Jun 15, 2016 #4

    Charles Link

    User Avatar
    Homework Helper

    I think @Chestermiller is on the right track. Unless the components are mixed very thoroughly, if you just have enough acid to react with the Mg, I don't think the reaction(of converting the Mg) will be complete. Thereby for .025 moles Mg, the reaction will be incomplete, etc. For .017 moles Mg, the reaction may have been complete. (Compute the number of moles of Cl available. HCl needs to be in excess.)
     
    Last edited: Jun 15, 2016
  6. Jun 15, 2016 #5

    Charles Link

    User Avatar
    Homework Helper

    The OP doesn't appear to have returned yet, but there are only .050 moles of Cl available in 50 ml of 1M HCl solution. The result is that for the .025 moles of Mg, it would use up every single Cl atom available for the reaction of the .025 moles of Mg to be complete. It is an interesting experimental result that appears to have a simple answer.
     
    Last edited: Jun 15, 2016
  7. Jun 16, 2016 #6
    Yes you guys are right. Only the lesser two trials have excess Cl.
     
  8. Jun 16, 2016 #7

    Borek

    User Avatar

    Staff: Mentor

    Technically - if the experiment is done right - fact that there is just a stoichiometric amount of the HCl should not change the result. I suspect half of the problem lies in the experimental procedure (slow reaction can mean relatively large heat loses during the experiment), but as we don't know the setup, we can't be sure.
     
  9. Jun 16, 2016 #8

    James Pelezo

    User Avatar
    Gold Member

    I've run this type lab many times and find it's more in the 'kind' of Mg used... Fine powder form Mg gives the most consistent data close to theoretical where ribbon or 'chunk' types or ribbon strips give random, inconsistent results. The stoichiometry works more effectively with lower wt amounts of powder form Mg to give smaller Standard Deviations than higher weight amounts of power form Mg. The common explanation is lower mass amounts of powder react faster than the higher mass amounts as a higher surface area (assuming constant acid concentration) is exposed to the reactive interface. Theoretically, all should produce the same heat of reaction but particle size (surface area effects) and concentration effects (both [Mg] or [H+] variations) will affect the precision-type data trends. Keep after it, you are getting a great education. Good luck.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Enthalpy Lab Explanation!
  1. Work in Enthalpy (Replies: 10)

  2. Enthalpy and work (Replies: 3)

Loading...