K'eq and delta G'o question plz help and explain thnx much?

Question

if a .1 M solution of glucose-1-phosphate is converted to glucose-6-phosphate and at equilibrium, the concentrations of the reaction components are

glucose-1-phosphate<-->glucose-6-phosp...

4.5 x 10^-3 M 9.6x10^-2 M

calculate K'eq and delta G for this reaction at 25 degreCelsiusius? thnx much

Answer 1

The concentrations at equilibrium should be

( 9.55 x 10^ -2 ) M for G-6-P  and  ( 4.5 x 10^ -3 ) M for G-1-P

But we will follow all values that you give .  

Keq = [ Product Concentration ] / [ Reactant Concentration ]

= ( 9.6 x 10^ -2 ) / 4.5 x 10^ -3 )

= 21.33

At Equilibrium  :  

Delta G (zero)  =  - RT Ln ( Keq )  

= - ( 8.314 ) x ( 298.1 ) Ln ( 21.33 )

Unit of R  is Joule per Degree-Mol

Unit of T  is Degree

Unit of Delta G(zero) is Joule per mol  

Sorry, I do not have a Scientific Calculator to complete the answer.