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

Answer:

a. $-1.68 \frac{kJ}{mol}$

b. $-4.40 \frac{kJ}{mol}$

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

Explanation:

a. The original problem states that:

$K' = 1.97$

For an equilibrium:

$fructose-6-phosphate\rightleftharpoons glucose-6-phosphate$

This means:

$K' = \frac{glucose-6-phosphate}{fructose-6-phosphate}$

Solving for the standard Gibbs free energy change:

$\Delta G^o = -RT ln(K') = -8.314 \frac{J}{K mol}\cdot 298.15 K\cdot ln(1.97) = -1681 \frac{J}{mol} = -1.68 \frac{kJ}{mol}$

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:

$Q = \frac{glucose-6-phosphate}{fructose-6-phosphate}$

We obtain:

$\Delta G = -1681 \frac{J}{mol} + 8.314 \frac{J}{K mol}\cdot 298.15 K\cdot ln(\frac{0.50}{1.5}) = -4404 \frac{J}{mol} = -4.40 \frac{kJ}{mol}$

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.