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Ochem energy calculation - correct answer?

  1. May 21, 2006 #1
    I think the solution manual is wrong for this problem.

    The problem:
    The naturally occurring sugar glucose exists in two isomeric cyclic forms. These are called [alpha] and [beta], and at equilibrium they are present in a ratio of approximately 64:36. Calculate the free energy difference that corresponds to this equilibrium ratio.

    I basically used the free energy equation
    [tex]\Delta G° = -RT \ln{K_{eq}}[/tex].

    I used a value of 0.001986 for R and 298K for T (I assumed STP) and got a value of -0.34 kcal/mol for DG. The solution manual lists a value of -0.81 in kcal/mol.
  2. jcsd
  3. May 21, 2006 #2


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    Looks about right --- the idiot doing the solution may have taken a natural log and multiplied it by 2.303 rather than realizing that's a hangover from the old days of decimal logs from slide rules.
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