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Thermodynamic equilibrium qu

  1. Apr 30, 2004 #1
    Determine the thermodynamic equilibrium constant of the following reaction to explain why graphite is not an easily converted, inexpensive source of diamond at 90 F… C(graphite) = C(diamond). Given this evidence, need diamond owners worry about disintegration of their diamond jewelry to graphite on a hot summer day? How about the loss of diamond through oxidation by the air… C(diamond)+)2 = CO2? Verify with the thermodynamic equilibrium constant for this reaction.
     
  2. jcsd
  3. Jul 16, 2004 #2
    To find the eq constant you need to know some useful equations. Delta G=Delta G naught+RTLn(K) where DG (delta G) is the gibbs free energy at any moment in time and DGN (delta G naught) is the standard state energy of the reaction. At equilibrium DG always=0. Thus 0=DGN+RTln(k) or DGN=-RTln(K). THis is true for any reaction at equilibrium. You can find DGN by looking at tables of thermodynamic information, your book should have it in the back. Remember DGN= delta G formation of products-delta G formation of reactants. The delta G of formations of the graphite(reactant) and delta G formation of the product(diamond) should be in the back. Thus you know DGN, R is a constant in J/Kmol, and T is given in C which you should convert to Kelvin. Thus solve DGN=-RTlnK for K, the equilibrium constant. You will see that the equilibrium constant shows that the reactions asked do not happen under reasonsable conditions. Just do the same procedure for all the questions, just solving for the different DGN's. Ahhh graphite-->diamond, brings me back to my school days, its a classic problem.
     
  4. Jul 16, 2004 #3
    While you can proceed to find the values for the Gibbs energy--I believe the word "free" is used less often these days due to a change in the convention-- (and use the fact that a more negative value of delta G at some (normal) temperature implies greater spontaneity of reaction, ie. the reaction has a greater tendency to proceed in the direction involving a more negative delta G value) I think you should make a note of the structural changes involved in the (possible) conversion of graphite to diamond:

    If supposing such a change could take place easily, it would mean the following (in brief):

    1. sp2 hybridization state of C in graphite converts to sp3 hybridization in diamond
    2. the layered structure involving delocalized pi electrons is converted to a rigid tetrahedral structure (as a result of sp2 to sp3 change)

    What I really mean is that even though you get a correct mathematical answer which explains why the reaction doesn't take place, it is useful to understand what might happen in case the reaction proceeded under the conditions given and make a qualitative prediction (perhaps even entropy change...but I am not very firm about the idea applied here yet since I do not have values to support my arguments).
     
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