Question

The standard free-energy change for this reaction in the direction written is 23.8 kJ/mol. The concentrationsof the three intermediates in the hepatocyte of a mammal are: fructose 1,6-bisphosphate,1.4X10-5 M; glyceraldehyde 3-phosphate, 3X10-6 M; and dihydroxyacetone phosphate, 1.6X10-5 M.At body temperature (37C), what is the actual free-energy change for the reaction

Answer 1

Answer: The actual free-energy change for the reaction  -8.64 kJ/mol.

Explanation:

The given reaction is as follows.

  Fructose 1,6-bisphosphate $\rightleftharpoons$ Glyceraldehyde 3-phosphate + DHAP

For the given reaction, $\Delta G^{o}$ is 23.8 kJ/mol.

As we know that,

       $\Delta G = \Delta G^{o} + RT ln Q$

Here,    R = 8.314 J/mol K,       T = $37^{o} C$

                                                    = (37 + 273) K

                                                    = 310.15 K

Fructose 1,6-bisphosphate = $1.4 \times 10^{-5}$ M

Glyceraldehyde 3-phosphate = $3 \times 10^{-6}$ M

DHAP = $1.6 \times 10^{-5}$ M

Expression for reaction quotient of this reaction is as follows.

    Reaction quotient = $\frac{[DHAP][\text{glyceraldehyde 3-phosphate}]}{[/text{Fructose 1,6-bisphosphate}]}$

        Q = $\frac{1.6 \times 10^{-5} \times 3 \times 10^{-6}}{1.4 \times 10^{-5}}$

            = $3.428 \times 10^{-6}$

Now, we will calculate the value of $\Delta G$ as follows.

          $\Delta G = \Delta G^{o} + RT ln Q$

                      = $23800 + 8.314 \times 310.15 \times ln(3.428 \times 10^{-6})$

                      = -8647.73 J/mol

                      = -8.64 kJ/mol

Thus, we can conclude that the actual free-energy change for the reaction  -8.64 kJ/mol.