Answer:
ΔG = -1.53 kJ/mol
Explanation:
The given reaction is:
3-Phosphoglycerate → 2-Phosphoglycerate
The standard Gibbs free energy, ΔG°=+4.40 kJ
[2-Phosphoglycerate] = 0.290 mM
[3-Phosphoglycerate] = 2.90 mM
Temperature T = 37 C = 310 K
The standard Gibbs free energy, ΔG° is related to the free energy change ΔG at a given temperature by the following equation:

In this reaction:
![\Delta G =\Delta G^{0}+RTln\frac{[2-Phosphoglycerate]}{[3-Phosphoglycerate]}](https://tex.z-dn.net/?f=%5CDelta%20G%20%3D%5CDelta%20G%5E%7B0%7D%2BRTln%5Cfrac%7B%5B2-Phosphoglycerate%5D%7D%7B%5B3-Phosphoglycerate%5D%7D)
![\Delta G = 4.40kJ/mol +0.008314 kJ/mol-K*310Kln\frac{[0.290]}{[2.90]}=-1.53 kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20G%20%3D%204.40kJ%2Fmol%20%2B0.008314%20kJ%2Fmol-K%2A310Kln%5Cfrac%7B%5B0.290%5D%7D%7B%5B2.90%5D%7D%3D-1.53%20kJ%2Fmol)
I believe you mean 1.02*10^24 so I did
1.02*10^24/(6.02*10^23) * 54.938. I got d) 9.31*10^1 g as my answer.