Question

What is ΔG′° for the reaction ( K ′ eq measured at 25 °C)? (b) If the concentration of fructose 6-phosphate is adjusted to 1.5 M and that of glucose 6-phosphate is adjusted to 0.50 M, what is ΔG? (c) Why are ΔG′° and ΔG different?

Answer 1

a.
`-1.68 (kJ)/(mol)`

b.
`-4.40 (kJ)/(mol)`

c. Standard conditions provide 1 M molarity for each component, while non-standard conditions provide specific concentrations

Step-by-step explanation:

a. The original problem states that:

`K' = 1.97`

For an equilibrium:

`fructose-6-phosphate\rightleftharpoons glucose-6-phosphate`

This means:

`K' = (glucose-6-phosphate)/(fructose-6-phosphate)`

Solving for the standard Gibbs free energy change:

`\Delta G^o = -RT ln(K') = -8.314 (J)/(K mol)\cdot 298.15 K\cdot ln(1.97) = -1681 (J)/(mol) = -1.68 (kJ)/(mol)`

b. Now we will solve for the non-standard Gibbs free energy change given by the equation:

`\Delta G = \Delta G^o + RT ln(Q)`

Here:

`Q = (glucose-6-phosphate)/(fructose-6-phosphate)`

We obtain:

`\Delta G = -1681 (J)/(mol) + 8.314 (J)/(K mol)\cdot 298.15 K\cdot ln((0.50)/(1.5)) = -4404 (J)/(mol) = -4.40 (kJ)/(mol)`

c. The standard Gibbs free energy change is measured for 1 M concentration of each molarity, however, the non-standard Gibbs free energy change is measured for the given molarities of 1.5 M and 0.50 M. Due to different conditions, we obtain different values.