Question

Calculate the ΔG°' for the reaction with 3 significant figures with no label for the dimension (just the number). fructose-6-phosphate → glucose-6-phosphate. Given the equilibrium constant is 1.97 and the physiological relevant temperature is 37° C.

Answer 1

Answer:

ΔG° = 1747.523

Explanation:

The parameters mentioned are;

Gibbs Free energy ΔG°

Equilibrium constant Kc

Temperature T = 37 + 273 = 310 (upon conversion to kelvin temperature)

The formular relating all three parameters is given as;

ΔG° =  -RTlnKc

Where; R = rate constant = 8.314 J⋅K−1⋅mol−1

Upon solving;

ΔG° = - 8.314 * 310 * ln(1.97)

ΔG° = 1747.523