Question

AG' for the isomerization reaction glucose-1-phosphate (GIP) $ glucose-6-phosphate (G6P) is -7.1 kJ/mol. Calculate the equilibrium ratio of [G1P] to (G6P) at 25°C. Read this carefully to make sure you solve for the correct ratio! SHOW WORK! [G1P][G6P] =

Answer 1

To calculate the equilibrium ratio of [G1P] to [G6P], the Gibbs free energy equation is used with ΔG', universal gas constant R, and the temperature T substituted. The result is that the concentration of G6P is approximately 1.331 times that of G1P at equilibrium and at 25°C.

Step-by-step explanation:

The student is asking about the equilibrium ratio of concentrations of glucose-1-phosphate ([G1P]) to glucose-6-phosphate ([G6P]) at 25°C when the standard free energy change (ΔG') for the isomerization reaction is given as -7.1 kJ/mol. To calculate this ratio, we can use the Gibbs free energy equation for the equilibrium constant (Keq):

ΔG' = -RT ln(Keq)

Where ΔG' is the standard free energy change, R is the universal gas constant (8.314 J/mol K), T is the temperature in Kelvin (25°C + 273.15 = 298.15 K), and Keq is the equilibrium constant which for this reaction is [G6P]/[G1P].

Substituting the values into the equation we get:

-7100 J/mol = -(8.314 J/mol K)(298.15 K) ln([G6P]/[G1P])

Now, we solve for ln([G6P]/[G1P]):

ln([G6P]/[G1P]) = ΔG' / (-R * T)

ln([G6P]/[G1P]) = -7100 J/mol / (-(8.314 J/mol K)(298.15 K))

ln([G6P]/[G1P]) = 0.286

Exponentiating both sides to remove the natural logarithm, we get:

[G6P]/[G1P] = e0.286 = 1.331

Therefore, at equilibrium and at 25°C, the concentration of G6P is approximately 1.331 times that of G1P.

Answer 2

The ratio [G1P]/[G6P] = 5.7 . 10⁻².

Step-by-step explanation:

Let us consider the reaction G1P ⇄ G6P, with ΔG° = -7.1 kJ/mol. According to Hess's Law, we can write the inverse reaction, and Gibbs free energy would have an opposite sign.

G6P ⇄ G1P ΔG° = 7.1 kJ/mol

This is the reaction for which we want to find the equilibrium constant (the equilibrium ratio of [G1P] to [G6P]):

`Kc=([G1P])/([G6P])`

The equilibrium constant and Gibbs free energy are related by the following expression:

`Kc=e^{-\Delta G\si{\textdegree}/R.T } } =e^{-7.1kJ/mol/8.314.10^(-3)kJ/mol.K.298K} } }=5.7.10^(-2)`

where,

R is the ideal gas constant (8.314 . 10⁻3 kJ/mol.K)

T is the absolute temperature (in kelvins)