Answer:
The correct option is B
Explanation:
The reaction is given as
[tex]Glucose\ 6-phosphate ---->fractose \ 6-phosphate[/tex]
The equilibrium constant for this reaction is mathematically represented as
[tex]K_c = \frac{[concentration \ of \ product ]}{concentration \ of \ reactant } = \frac{[fructose \ 6 -phosphate]}{[Glucose \ 6 - phosphate]}[/tex]
From the question we are told that
[Glucose 6-phosphate] = 2 × [fructose 6-phosphate]
So
[tex]K_c = \frac{[fructose \ 6 -phosphate]}{[2 (fructose \ 6 -phosphate)]} = 0.5[/tex]
Generally change in free energy [tex]\Delta G[/tex] is mathematically represented as
[tex]\Delta G = - RTlnK_c[/tex]
Given that T = 298 K
R = 8.315 J/mol·K
[tex]\Delta G = -8.314 * 298 * ln (0.5)[/tex]
[tex]= -1717.32 J/mol[/tex]
[tex]= -1.7 \ kJ/mol[/tex]
Answer:
The correct answer is A: (ΔG'° is +1.7 kJ/mol).
Explanation:
Step 1: Data given
The final mixture contains twice as much glucose 6-phosphate as fructose 6-phosphate.
R = 8.315 J/mol*K
T = 298 K
Step 2:
Keq = [fructose 6-phosphate]/[glucose 6-phosphate] = 1/2
ΔG'° = -RT ln (Keq)
ΔG'° = -RT ln (1/2)
⇒with R = 8.315 J/mol
⇒with T = the temperature = 298 K
ΔG'° = -8.315 * 298 * ln (1/2)
ΔG'° = -8.315 * 298 * -0*693
ΔG'° = 1717 J/mol
ΔG'° = 1.7 * 10³ J/mol = 1.7 kJ/mol
The correct answer is A: (ΔG'° is +1.7 kJ/mol).
