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Homework Help: Hess's Law - Thermochemical Equations

  1. Sep 20, 2010 #1
    Hello! I am a high school senior taking an advanced chemistry course. I have been working on this problem for two days straight, and cannot figure out our error. Here is the problem we are given:

    Calculate the standard enthalpy of formation for the reaction HCL (g) + NH3 (g) ---> NH4CL (s), given the following thermochemical equations:

    H2 (g) + Cl2 (g) ---> 2HCL (g), standard enthalpy of formation: -184kJ (-92kJ/mol)
    N2 (g) + 3H2 (g) ---> 2NH3 (g), standard enthalpy of formation: -92kJ (-46kJ/mol)
    N2 (g) + 4H2 (g) + Cl2 (g) ---> 2NH4CL (s), standard enthalpy of formation: -628kJ (-314kJ/mol)

    It wants us to combine the equations using Hess's Law to arrive at the desired equation, and thus, the desired enthalpy of formation. I did so, and I arrive at the desired equation given above ( HCL (g) + NH3 (g) ---> NH4CL (s) ). However, the number I have arrived at time and time again (-176kJ/mol) is not the accepted standard enthalpy of formation for NH4CL, -314.43kJ/mol.

    Someone please show me the error before I bash my head against the wall any further! Thanks!

    Note, the book that gives the problem lists the answer I get (-176kJ/mol) as correct. I am wondering why this is not the accepted value. I'm sure it is something simple that I am missing, but I have been staring at it for much too long now and must admit I need assistance.
  2. jcsd
  3. Sep 21, 2010 #2


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    You are mixing up enthapy of formation with enthapy of reaction.

    The chemical equation that you have does not define the enthapy of formation; what you have found is the enthalpy change associated with that particular reaction.

    The "enthapy of formation" of a particular compound is defined as the energy change associated with the production of that compound from its constituent elements in their naturally occuring form.

    For example, for ammonium chloride, the enthapy of formation is the energy change associated with:

    [tex]\frac{1}{2}{N_{2(g)}} + 2{H_{2(g)}} + \frac{1}{2}C{l_{2(g)}}\rightarrow N{H_4}C{l_{(s)}}[/tex]
    Last edited: Sep 21, 2010
  4. Sep 21, 2010 #3
    Hmm, then my textbook is simply incorrect in it's wording of the problem? I copied the problem verbatim, and it definitely says find the enthalpy of formation for that reaction.
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