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?

Explanation:

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

We are given that the value of is 1.7 kJ/mol.

And,  

      k =

         =

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

Therefore,  

  = 0.50357

so, + 1 = 1.50357

          =

           =

          = 0.665

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

answer: the figure is attached herewith.

step-by-step explanation:   we are given to sketch a unit circle and label the terminal points corresponding to

in the attached figure, we have drawn a unit circle, with centre at o(0,0) and radius 1 unit. we have labelled the points a, b, c, d and e corresponding to the angles below

∠aox = 60°,

∠box = 120°,

∠cox = 180°,

∠dox = 240°,

∠eox = 300°.

and the coordinates are a(0.5,1), b(-0.5,1), c(-1,0), d(-0.5,-1) and e(0.5,-1).

thus, the answer is complete.

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