Question

If the standard free energy change for the conversion of fructopyranose to fructofuranose is 1.7 kJ/mol, what fraction of the total fructose in solution is fructopyranose?

Answer 1

Step-by-step explanation:

We know that relation between Gibb's free energy and temperature is as follows.

`\Delta G = -RT ln K`

We are given that the value of
`\Delta G` is 1.7 kJ/mol.

And,

k =
`([product])/([substrate])`

=
`\frac{\text{[fructofuranose]}}{\text{[fructopyranose]}}`

Since, k = 0.50357 and temperature is equal to 298 K.

Therefore,

`\frac{\text{[fructofuranose]}}{\text{[fructopyranose]}}` = 0.50357

so,
`\frac{\text{[fructofuranose]}}{\text{[fructopyranose]}}` + 1 = 1.50357

=
`\frac{\text{[total fructose solution]}}{\text{[fructopyranose]}}`

=
`(1)/(1.50357)`

= 0.665

Therefore, we can conclude that 0.665 fraction of the total fructose in solution is fructopyranose.