The value of ?G? for the phosphorylation of glucose in glycolysis is 13.8 kJ/mol. Find the value of Kc at 298 K. ?

To determine the equilibrium constant (Kc) for the phosphorylation of glucose in glycolysis, we can use the relationship between the change in Gibbs free energy (ΔG) and the equilibrium constant. The equation that connects these two concepts is:

Understanding the Relationship

The equation is given by:

ΔG = -RT ln(Kc)

Where:

• ΔG is the change in Gibbs free energy (in joules per mole).
• R is the universal gas constant (approximately 8.314 J/(mol·K)).
• T is the temperature in Kelvin (298 K in this case).
• ln(Kc) is the natural logarithm of the equilibrium constant.

Converting Units

First, we need to convert ΔG from kJ/mol to J/mol because the gas constant R is in J/(mol·K). Since 1 kJ = 1000 J, we have:

ΔG = 13.8 kJ/mol × 1000 J/kJ = 13800 J/mol

Substituting Values

Now, we can substitute the values into the equation:

13800 J/mol = - (8.314 J/(mol·K)) × (298 K) × ln(Kc)

Calculating ln(Kc)

Rearranging the equation to solve for ln(Kc), we get:

ln(Kc) = - (13800 J/mol) / (8.314 J/(mol·K) × 298 K)

Calculating the denominator:

8.314 J/(mol·K) × 298 K ≈ 2477.572 J/mol

Now, substituting this back into the equation:

ln(Kc) = - (13800 J/mol) / (2477.572 J/mol) ≈ -5.57

Finding Kc

To find Kc, we need to exponentiate both sides:

Kc = e^(-5.57)

Using a calculator, we find:

Kc ≈ 0.0037

Final Result

Thus, the equilibrium constant (Kc) for the phosphorylation of glucose in glycolysis at 298 K is approximately 0.0037. This low value indicates that at equilibrium, the concentration of reactants is much greater than that of the products, which is consistent with the energetically unfavorable nature of this reaction under standard conditions.