Consider the following inter-conversion, which occurs in glycolysis: fructose-6-phosphate ↔ glucose-6-phosphate keq’ = 1.97 what is δg°’ for the reaction (keq’ measured at 25°c; and r=8.314 j•mol-1•k-1)? (b) if the concentration of fructose-6-phosphate is adjusted to 1.5 m and that of glucose- 6-phosphate is adjusted to 0.50 m, what is δg? (c) why are δg°’ and δg different?

Explanation:

(a)   The relation between and equilibrium constant is as follows.

As it is given that,

              T = = (25 + 273) K = 298 K

              K = 1.97

              R = 8.314 J/mol K

Hence, putting the given values into the above formula as follows.

                                  =

                                  = -1679.88 J/mol

                                  = 1.679 kJ/mol    (as 1 kJ = 1000 J)

Therefore, the value of is 1.679 kJ/mol.

(b)  Expression for is as follows.

                       =

                        = -4401.8 J/mol

                        = 44.01 kJ/mol

Hence, the value for is 44.01 kJ/mol.

(c)    The value of both and are different because is measured at any concentration whereas is measured at equilibrium concentration.