For the reaction A  B, ΔG'° = - 60 kJ/mol. The reaction is started with 10 mmol of A; no B is initially present. After 24 hours, analysis reveals the presence of 2 mmol of B, 8 mmol of A. Which is the most likely explanation? A) A and B have reached equilibrium concentrations. B) An enzyme has shifted the equilibrium toward A. C) B formation is kinetically slow; equilibrium has not been reached by 24 hours. D) Formation of B is thermodynamically unfavorable. E) The result described is impossible, given the fact that ΔG'° is - 60 kJ/mol.

Option A (A and B have reached equilibrium concentrations)

Explanation:

for the reaction

A→B, with ΔG'° = - 60 kJ/mol

Since the presence of enzyme does not affect the equilibrium concentrations (ΔG is not affected), only affect the kinetics → option B is not possible.

For the same reason C is not possible since ΔG is related with the thermodynamics of the reaction (equilibrium concentrations and feasibility) and not the kinetics.

Since the reaction always proceeds in direction of minimum ΔG, and the reaction lowers ΔG →the process is thermodynamically favorable → D and E are not possible

Since there is always at least a small conversion from pure reactants to products since the entropy increases when pure A is mixed with B (entropy production due to mixing), since also the process is thermodynamically favorable and besides the analysis reveals there was conversion from A to B → option A is the one that is more likely (A and B have reached equilibrium concentrations).