Answer:
[tex]K_{eq[/tex] = 19
ΔG° of the reaction forming glucose 6-phosphate = -7295.06 J
ΔG° of the reaction under cellular conditions = 10817.46 J
Explanation:
Glucose 1-phosphate ⇄ Glucose 6-phosphate
Given that: at equilibrium, 95% glucose 6-phospate is present, that implies that we 5% for glucose 1-phosphate
So, the equilibrium constant [tex]K_{eq[/tex] can be calculated as:
[tex]K_{eq[/tex] [tex]= \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}[/tex]
[tex]K_{eq[/tex][tex]= \frac{0.95}{0.05}[/tex]
[tex]K_{eq[/tex] = 19
The formula for calculating ΔG° is shown below as:
ΔG° = - RTinK
ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k × 1n(19))
ΔG° = 7295.05957 J
ΔG°≅ - 7295.06 J
b)
Given that; the concentration for glucose 1-phosphate = 1.090 x 10⁻² M
the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M
Equilibrium constant [tex]K_{eq[/tex] can be calculated as:
[tex]K_{eq[/tex] [tex]= \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}[/tex]
[tex]K_{eq}= \frac{1.395*10^{-4}}{1.090*10^{-2}}[/tex]
[tex]K_{eq} =[/tex] 0.01279816514 M
[tex]K_{eq} =[/tex] 0.0127 M
ΔG° = - RTinK
ΔG° = -(8.314*298*In(0.0127)
ΔG° = 10817.45913 J
ΔG° = 10817.46 J
