I think the solution manual is wrong for this problem.(adsbygoogle = window.adsbygoogle || []).push({});

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 GÂ° = -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.

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# Ochem energy calculation - correct answer?

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