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 for the isomerization of glucose 1-phosphate to glucose 6-phosphate at 37°C, given ΔG° = -1.74 kcal/mole, is closest to 0.09.

Step-by-step explanation:

The question asks us to determine the equilibrium constant (K) for the isomerization of glucose 1-phosphate to glucose 6-phosphate at 37°C, given that the standard free energy change (ΔG°) is -1.74 kcal/mole. To find K, we can use the relationship ΔG° = -RTlnK, where R is the gas constant (1.987 cal/(mol·K)) and T is the temperature in Kelvin. First, we convert ΔG° to cal by multiplying -1.74 kcal/mol by 1000 cal/kcal, giving us -1740 cal/mol. Next, we convert 37°C to Kelvin by adding 273 to get 310 K. Plugging these values into the formula, we get:

ΔG° = - (1.987 cal/(mol·K)) × (310 K) × lnK

We isolate lnK and solve for K:

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

lnK ≈ -2.857

K = e-2.857

K ≈ 0.057, but since we are given choices, we should round K to two decimal places. The closest value to our calculated K is (B) 0.09.