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