Solve this:

6 .     W h e n   ( α )   d - g l u cos e   i s   d i s s o l v e d   i n   w a t e r ,   i t   u n d e r g o e s   a   p a r t i a l   c o n v e r s i o n   ( β )   d - g l u cos e   t o   e x h i b i t   m u t a r o t a t i o n .   T h i s   c o n v e r s i o n   s t o p s   w h e n   63 . 6 %   o f   g l u cos e   i s   i n   β - f o r m .   A s s u m i n g   t h a t   e q u d i b r i u m   h a s   b e e n   a t t a i n e d ,   c a l c u l a t e   K c   f o r   m u t a r o t a t i o n .  

Dear Student,

Let us take 100 of alpha glucose is in the reactant and going to the reaction. At equilibrium 63.6% glucose is in bita form. So (100 - 63.6) = 36.4% glucose is unreacted i.e. in the alpha form.
Now we know that 
Kc = [product][reactant] = 63.636.4 = 1.747
So, Kc = 1.747

Regards

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