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

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 = $\frac{[product]}{[substrate]}$

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

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

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

           = $\frac{1}{1.50357}$

          = 0.665

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