The K_eq is 0.503 at 25°C for the following reaction.What is the $\Delta$G°' for this reaction? D-Glucose-6-phosphate $\to$ D-Fructose-6-phosphate $\Delta$G°'= ? A)-2.87 × 10³ J/mol B)+1.70 kJ/mol C)+143 J/mol D)The $\Delta$G°' cannot be determined with the given information. E)The $\Delta$G°' can be determined with the given information but the correct answer is not shown.

Question

Quizplus · Accepted Answer

The standard free energy change (ΔG°') can be calculated using the equation:

ΔG°' = -RTln(Keq)

where R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin, and Keq is the equilibrium constant.

Given: Keq = 0.503, T = 25°C = 298 K

ΔG°' = - (8.314 J/mol K)(298 K) ln(0.503)

ΔG°' = +1.70 kJ/mol

Therefore, the answer is B.

The Keq Is 0 Δ\DeltaΔ G°' for This Reaction? D-Glucose-6-Phosphate →\to→ D-Fructose-6-Phosphate Δ\DeltaΔ

The K_eq Is 0 $\Delta$ G°' for This Reaction?
D-Glucose-6-Phosphate $\to$ D-Fructose-6-Phosphate $\Delta$