Ochem energy calculation - correct answer?

[tandoorichicken]

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
$$\Delta G° = -RT \ln{K_{eq}}$$.

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.

 

[Bystander]

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.

 

[ravenprp]

It is possible that the solution manual is incorrect for this problem. However, it is always important to double check your calculations and make sure that all the values and units used are correct. It is also possible that the solution manual used different values or assumptions in their calculation, which could account for the discrepancy in the answer. It would be best to consult with your instructor or a tutor to verify the correct answer and understand any differences in calculations. In any case, it is always good practice to question and critically evaluate solutions, so keep up the good work!