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.

Step-by-step 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 * 10^(-5)` M

Glyceraldehyde 3-phosphate =
`3 * 10^(-6)` M

DHAP =
`1.6 * 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 =
`(1.6 * 10^(-5) * 3 * 10^(-6))/(1.4 * 10^(-5))`

=
`3.428 * 10^(-6)`

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

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

=
`23800 + 8.314 * 310.15 * ln(3.428 * 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.