Question

For the reaction Y->X at standard conditions with [Y] = 1 M and [X] = 1 M, ΔG is initially a large negative number. As the reaction proceeds, [Y] decreases and [X] increases until the system reaches equilibrium. How do the values of ΔG and ΔG° change as the reaction equilibrates?

(a) ΔG becomes less negative and ΔG° stays the same.
(b) ΔG becomes positive and ΔG° becomes positive.
(c) ΔG stays the same and ΔG° becomes less negative.
(d) ΔG reaches zero and ΔG° becomes more negative.

Answer 1

As the reaction proceeds towards equilibrium, ΔG becomes less negative and ΔG° stays the same.

Step-by-step explanation:

As the reaction proceeds towards equilibrium, the concentrations of reactants and products change. In this case, [Y] decreases while [X] increases. Initially, ΔG is a large negative number, indicating that the reaction is spontaneous and favors the formation of products. However, as the reaction progresses towards equilibrium, ΔG becomes less negative. The value of ΔG°, which represents the standard free energy change, remains the same throughout the reaction. Therefore, the correct choice is (a) ΔG becomes less negative and ΔG° stays the same.