For the aqueous reaction dihydroxyacetone phosphate↽−−⇀glyceraldehyde−3−phosp hate dihydroxyacetone phosphate↽−−⇀glyceraldehyde−3−phosp hate the standard change in Gibbs free energy is ΔG°′=7.53 kJ/molΔG°′=7.53 kJ/mol . Calculate ΔGΔG for this reaction at 298 K298 K when [dihydroxyacetone phosphate]=0.100 M[dihydroxyacetone phosphate]=0.100 M and [glyceraldehyde-3-phosphate]=0.0020 0 M[glyceraldehyde-3-phosphate]=0.002 00 M .

ΔG = -2.17 kJ/mol

Explanation:

ΔG of a reaction at any moment could be obtained thus:

ΔG = ΔG° + RT ln Q

Where ΔG° is standard change in free energy of a particular reaction (7.53kJ/mol for the reaction of the problem, R is gas constant (8.314×10⁻³kJ/molK), T is absolute temperature (298K) and Q is reaction quotient of the reaction.

For the reaction:

dihydroxyacetone phosphate ⇄ glyceraldehyde−3−phosphate

Q is defined as:

Q = [glyceraldehyde−3−phosphate] / [dihydroxyacetone phosphate]

Replacing values in ΔG formula:

ΔG = 7.53kJ/mol + 8.314×10⁻³kJ/molK × 298.15K ln [0.00200M] / [0.100M]

ΔG = -2.17 kJ/mol

a. hope i'm right

explanation: