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Enthalapy of Formation - HELP

  1. Oct 16, 2006 #1

    OOPS I didn't see this forum I posted this earlier in the "Other Sciences > Chem." section...Sorry about that!!!! My original post is below...

    Hey guys this is my 1st post here...Need help with a question thats been troubling me for a long time...

    A student determined experimentally that 2.06 kJ of heat energy were released when 0.365 g of copper reacted with excess chloride. From these data, the enthalapy of formation of copper (II) chloride would be...
    a. -206 kJ/mol
    b. -1.09 kJ/mol
    c. +2.71 kJ/mol
    d. +165 kJ/mol

    please help because its frustrating me - when i picked b. my teacher said its WRONG !!!! I don't get it at all. I think area it is focussing on is Hess's Law! Please help me!!! :confused:

    Thanks in advance,
  2. jcsd
  3. Oct 16, 2006 #2


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    Hi ashooazmi and welcome to PF!

    In your question, why did you pick option b?

    You don't need to apply Hess's law for this question. Just use the definition of the http://dbhs.wvusd.k12.ca.us/webdocs/Thermochem/StandardEnthalpyFormation.html" [Broken].

    Were you able to get the correct answer now?

    (That's a very colorful post!)
    Last edited by a moderator: May 2, 2017
  4. Oct 17, 2006 #3
    hey - thanks for the nice welcome - I am still very much confused I cant figure out what i am doing wrong!!! PLease help
  5. Oct 20, 2006 #4
    2.06 kJ of heat energy were released when 0.365 g of copper reacted with excess chloride.

    From here, you can calculate the number of moles of copper, let say x.
    Since x number of moles of copper will react with chlorine to give 2.6 KJ (negative sign should be included, i.e. -2.06KJ, since the energy is released).

    Then, you can find the energy that is released when 1 mole of copper reacts with chlorine. This would give the enthalapy of formation of copper (II) chloride.
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