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Free Energy of Dilution

  1. Nov 14, 2007 #1
    Calculate ΔG for the dilution of aqueous HCl from 0.89 M to 0.253 M at 25°C.

    THis was a question I had to do some time earlier for one of my assignments. I got some help from a classmate and he said THE FOLLOWING:

    "H+ and Cl- are equimolar so you have to put them to the
    power of two....

    so the equation reads
    Deltag = RTln Q
    delta g = 8.314472J/K/mol*298.15K * 1kJ/1000J *ln (second
    molarity^2/first molarity^2)"





    WHEN he says EQUIMOLAR how do i find this out like i have no idea how he figured that out. If i did this question i would just do products over reactions to find the value of Q.

    But he says put them to power of 2. WHY?? Can someone explain?
     
  2. jcsd
  3. Nov 15, 2007 #2
    HCl is a strong acid and therefore completely disassociates in water. Therefore, whatever the molarity of HCl = molarity of H+ = molarity of Cl-

    Hence equimolar
     
    Last edited: Nov 15, 2007
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