Respuesta :
Explanation:
(a) The relation between [tex]\Delta G^{o}[/tex] and equilibrium constant is as follows.
[tex]\Delta G^{o} = -RT ln K[/tex]
As it is given that,
T = [tex]25^{o}C[/tex] = (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.
[tex]\Delta G^{o} = -RT ln K[/tex]
= [tex]-8.314 J/mol K \times 298 K \times ln (1.97)[/tex]
= -1679.88 J/mol
= 1.679 kJ/mol (as 1 kJ = 1000 J)
Therefore, the value of [tex]\Delta G^{o}[/tex] is 1.679 kJ/mol.
(b) Expression for [tex]\Delta G[/tex] is as follows.
[tex]\Delta G = \Delta G^{o} + RT ln \frac{\text{Glucose-6-phosphate}}{\text{Fructose-6-phosphate}}[/tex]
= [tex]-1679.88 J/mol + 8.314 J/mol K \times 298 K \times \n \frac{0.5}{1.5}[/tex]
= -4401.8 J/mol
= 44.01 kJ/mol
Hence, the value for [tex]\Delta G[/tex] is 44.01 kJ/mol.
(c) The value of both [tex]\Delta G[/tex] and [tex]\Delta G^{o}[/tex] are different because [tex]\Delta G[/tex] is measured at any concentration whereas [tex]\Delta G^{o}[/tex] is measured at equilibrium concentration.
