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.

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).