The first reaction in Glycolysis is the phosphorylation of Glucose:\[ \text{Pi} + \text{Glucose} \rightarrow \text{Glucose 6-Phosphate} + \text{water} \]This is a thermodynamically unfavorable process with a \(\Delta G_0' = +13.8 \, \text{kJ/mol}\). In a liver cell at 37°C, the concentrations of both phosphate and glucose are normally maintained at about 0.005 M each. What would be the equilibrium concentration of Glucose 6-Phosphate according to the above data? Note: In the biochemical standard state \([\text{H}_2\text{O}] = 1.0\).Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer.Options:- a: 0.00000012- b: 0.0000025- c: 0.0000025- d: 0.00474- e: 0.005- f: 1.0- g: 0.1- h: 0.25- i: 2.4

Glycolysis is a metabolic process that occurs in the cytoplasm irrespective of the presence or absence of oxygen. Glycolysis catabolizes the glucose into two molecules of pyruvate with the generation of NADH and ATP molecules. Pyruvate molecules are further oxidized through the PDH reaction and the TCA cycle into carbon dioxide, reducing equivalents, and GTP. The formula to calculate the standard free energy change is as follows: ∆G°=-2.303 RT log [Glucose-6-phosphate][Pi] [Glucose] Given Values are as follows: ∆G° = +13.8 kJ/molR = 8.314T = 37°C = 37+273=310 K[Pi] = 0.005 M[Glucose]=0.005 M∆G°=-2.303 RT log [Glucose-6-phosphate][Pi] [Glucose]13800=-2.303×8.314×310 log[Glucose-6-phosphate]0.005×0.005-log[Glucose-6-phosphate]0.005×0.005=2.32log0.005×0.005[Glucose-6-phosphate]=2.320.005×0.005[Glucose-6-phosphate]=208.92[Glucose-6-phosphate]=0.005×0.005208.92=1.19×10-7 M or 0.119×10-6 M or 0.119 μMAnswer: The correct answer is option A " 0.00000012 M".