Answer:
a. [tex]-1.68 \frac{kJ}{mol}[/tex]
b. [tex]-4.40 \frac{kJ}{mol}[/tex]
c. Standard conditions provide 1 M molarity for each component, while non-standard conditions provide specific concentrations
Explanation:
a. The original problem states that:
[tex]K' = 1.97[/tex]
For an equilibrium:
[tex]fructose-6-phosphate\rightleftharpoons glucose-6-phosphate[/tex]
This means:
[tex]K' = \frac{glucose-6-phosphate}{fructose-6-phosphate}[/tex]
Solving for the standard Gibbs free energy change:
[tex]\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}[/tex]
b. Now we will solve for the non-standard Gibbs free energy change given by the equation:
[tex]\Delta G = \Delta G^o + RT ln(Q)[/tex]
Here:
[tex]Q = \frac{glucose-6-phosphate}{fructose-6-phosphate}[/tex]
We obtain:
[tex]\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}[/tex]
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.
