Question

Isomerization of glucose 1-phosphate to glucose 6-phosphate is energetically favorable. At 37°C, ΔG° = -1.42 log10K. What is the equilibrium constant for this reaction if ΔG° = -1.74 kcal/mole at 37°C?

(a) 16.98
(b) 0.09 (c) -0.09 (d) 0.39

Answer 1

The equilibrium constant (K) for the isomerization of glucose 1-phosphate to glucose 6-phosphate at 37°C, with ΔG° = -1.74 kcal/mol, is approximately 17.1, indicating an energetically favorable reaction.

Step-by-step explanation:

To find the equilibrium constant (K) for the isomerization of glucose 1-phosphate to glucose 6-phosphate, we can use the standard Gibbs free energy change (ΔG°) formula:

ΔG° = -RTlnK

Where:

• ΔG° = standard Gibbs free energy change
• R = universal gas constant (1.987 cal/mol·K or 8.314 J/mol·K)
• T = temperature in Kelvin
• K = equilibrium constant

First, convert ΔG° from kcal/mole to cal/mole:

ΔG° = -1.74 kcal/mole × 1000 cal/kcal = -1740 cal/mole

Then, plug in the values (T = 310K since 37°C = 310K):

-1740 cal/mole = -1.987 cal/mol·K × 310K × lnK

× lnK = -1740 cal/mole / (-1.987 cal/mol·K × 310K)

lnK = 2.84

K = e^2.84

K = 17.1 (rounded to two decimal places)

Therefore, the equilibrium constant for this reaction is approximately 17.1, indicating that the reaction is energetically favorable under standard conditions.