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If a 0.1 M solution of glucose 1-phosphate is incubated with a catalytic amount of phospho-glucomutase, the glucose 1-phosphate is transformed to glucose 6-phosphate until equilibrium is reached. At equilibrium, the concentration of glucose 1-phosphate is 4.5 x 10–3 M and that of glucose 6-phosphate is 8.6 × 10–2 M. Set up the expressions for the calculation of Keq' and G'° for this reaction (in the direction of glucose 6-phosphate formation). (R = 8.315 J/mol·K; T = 298 K)

Respuesta :

Explanation:

The given data us as follows.

At equilibrium, the concentration of glucose-1-phosphate = [tex]4.5 \times 10^{-3}[/tex] M

At equilibrium, the concentration of glucose-6-phosphate = [tex]8.6 \times 10^{-6}[/tex] M

As,   [tex]K_{eq} = \frac{\text{concentration of glucose-6-phosphate at equilibrium}}{\text{concentration of glucose-1-phosphate at equilibrium}}[/tex]

                  = [tex]\frac{8.6 \times 10^{-6}}{4.5 \times 10^{-3}}[/tex]

                  = 19.11

As it is known that [tex]\Delta G^{o} = -RTln K_{eq}[/tex]

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

           [tex]\Delta G^{o} = -RTln K_{eq}[/tex]

                          = [tex]-8.314 \times 298 \times ln (19.11)[/tex]

                          = 7309.3620 J/mol

or,                       = 7.309 kJ/mol         (as 1 kJ = 1000 J)

Thus, we can conclude that value of [tex]\Delta G^{o}[/tex] is equal to 7.309 kJ/mol.