Question

Consider the reaction X!Y in a cell at 37°C. At equilibrium, the concentrations of X and Y are 50 µM and 5 µM, respectively. Use this information and the equations below to answer questions A-E.

ΔG° = -0.616 ln Keq ΔG = ΔG° + 0.616 ln [Y]/[X]

Recall that the natural log of a number z will have a negative value when z < 1, positive when z > 1, and 0 when z = 1.

A. What is the value of Keq for this reaction?
B. Is the standard free-energy change of this reaction positive or negative? Is the reaction X-->Y an energetically favorable or unfavorable reaction under standard conditions?
C. What is the value of the standard free energy? Refer to Table 3-1 in the textbook or use a calculator.
D. Imagine circumstances in which the concentration of X is 1000 µM and that of Y is 1 µM. Is conversion of X to Y favorable? Will it happen quickly? E. Imagine starting conditions in which the reaction X!Y is unfavorable, yet the cell needs to produce more Y. Describe two ways in which this may be accomplished.

Answer 1

A. Keq for the reaction X!Y at 37°C is 0.1.

B. The standard free-energy change (∆G°) of this reaction is negative, indicating it's energetically favorable under standard conditions.

C. The value of the standard free energy (∆G°) is -2.447 kcal/mol.

D. With a concentration of X at 1000 µM and Y at 1 µM, the conversion of X to Y is favorable, but the reaction might occur slowly due to the significant difference in concentrations.

E. To drive the production of Y in an initially unfavorable X!Y reaction, increasing the concentration of Y's reactants or decreasing the concentration of X can promote the production of Y.

Step-by-step explanation:

A. Keq is calculated using the equation ΔG° = -0.616 ln Keq. Rearranging the equation to solve for Keq, we get Keq = e^(-ΔG°/0.616). Substituting the given value of ΔG° = -2.447 kcal/mol, Keq = e^(-(-2.447)/0.616) = e^(3.974) ≈ 0.1.

B. The negative value of ΔG° (-2.447 kcal/mol) implies an energetically favorable reaction under standard conditions. This suggests that at equilibrium, there is a tendency for the reaction to proceed towards the formation of product Y.

C. Using the equation ΔG = ΔG° + 0.616 ln [Y]/[X], and the given concentrations at equilibrium ([Y] = 5 µM, [X] = 50 µM), ΔG = -2.447 + 0.616 ln(5/50) = -2.447 + 0.616 ln(0.1) = -2.447 + (-0.616) = -3.063 kcal/mol, indicating the standard free energy.

D. With [X] = 1000 µM and [Y] = 1 µM, the reaction is thermodynamically favorable (as Keq = [Y] / [X] = 1/1000 < Keq = 0.1), but due to the large concentration gradient, the reaction might occur slowly.

E. To promote Y production in an initially unfavorable reaction, either increasing the concentration of Y's reactants (decreasing Y and increasing X) or reducing the concentration of X would shift the equilibrium toward Y production, helping the cell produce more Y.