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).

the correct choice is a .

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