Phosphorylation of Glucose in Glycolysis: Calculating Kc at 1298K

To calculate the value of Kc at 1298K for the phosphorylation of glucose in glycolysis, we need to use the relationship between the standard Gibbs free energy change (ΔG°) and the equilibrium constant (Kc) at a given temperature.

1. Background:
The phosphorylation of glucose in glycolysis is an enzymatic reaction that converts glucose into glucose-6-phosphate. The reaction is catalyzed by the enzyme hexokinase and involves the transfer of a phosphate group from ATP to glucose.

The equation for the reaction is as follows:
Glucose + ATP ⇌ Glucose-6-phosphate + ADP

2. Relationship between ΔG° and Kc:
The standard Gibbs free energy change (ΔG°) for a reaction can be related to the equilibrium constant (Kc) through the following equation:

ΔG° = -RT ln(Kc)

Where:
ΔG° is the standard Gibbs free energy change for the reaction,
R is the gas constant (8.314 J/mol·K or 0.008314 kJ/mol·K),
T is the temperature in Kelvin, and
ln represents the natural logarithm.

3. Calculation:
Given:
ΔG° = 13.8 kJ/mol (standard Gibbs free energy change)
T = 1298 K (temperature)

To calculate Kc, we need to convert the units of ΔG° to J/mol and use the appropriate gas constant.

ΔG° = 13.8 kJ/mol × 1000 J/kJ = 13800 J/mol
R = 8.314 J/mol·K

Now, we can plug these values into the equation:

ΔG° = -RT ln(Kc)

13800 J/mol = -(8.314 J/mol·K)(1298 K) ln(Kc)

Simplifying the equation:

ln(Kc) = -13800 J/mol / ((8.314 J/mol·K)(1298 K))

ln(Kc) = -1.403 mol

Taking the exponential of both sides:

Kc = e^(-1.403)

Using a calculator, we find:

Kc ≈ 0.245

4. Conclusion:
The value of Kc at 1298 K for the phosphorylation of glucose in glycolysis is approximately 0.245. This indicates that at equilibrium, the concentration of glucose-6-phosphate is relatively low compared to glucose and ATP, suggesting that the reaction favors the formation of glucose-6-phosphate.

Note: The calculations provided here assume ideal conditions and do not take into account any other factors that may influence the reaction or its equilibrium.

It is 3.81 x 10^-3