Question

The value of ΔG° for the phosphorylation of glucose in glycolysis is 13.8 kJ/mol.What is the value of the equilibrium constant for the reaction at 25.0°C?

Answer 1

Step-by-step explanation

Given:

ΔG° = 13.8 kJ/mol = (13.8 x 1000) J/mol = 13800 J/mol

Temperature, T = 25.0°C. = 25.0°C + 273 = 298.0 K

What to find:

the value of the equilibrium constant, K for the reaction at 25.0°C.

Step-by-step solution:

Both K and ΔG° can be used to predict the ratio of products to reactants at equilibrium for a given reaction.

ΔG° is related to K by the equation ΔG°= −RTlnK.

R is the molar gas constant, ( R = 8.3144598 J/K/mol)

If ΔG° < 0, then K > 1, and products are favored over reactants at equilibrium.

The next step is to substitute the values of ΔG° and T into the equation to get K.

ΔG°= −RTlnK

13800 J/mol = -(8.3144598 J/K/mol x 298 K x lnK)

13800 J/mol = -(2477.70902 J/mol x lnK)

Divide both sides by 2477.70902 J/mol

`\begin{gathered} \frac{13800\text{ J/mol}}{2477.70902\text{ J/mol}}=-\frac{2477.70902\text{ J/mol }*\ln K}{2477.70902\text{ J/mol}} \\ 5.5697=-\ln K \\ \ln K=-5.5697 \\ K=\ln ^(-1)(-5.5697) \\ \end{gathered}`