Question

When a mixture of glucose 6-phosphate and fructose 6-phosphate is incubated with the enzyme phosphohexose isomerase, the final mixture contains twice as much glucose 6-phosphate as fructose 6-phosphate. Which one of the following statements is most nearly correct, when applied to the reaction below (R = 8.315 J/mol·K and T = 298 K)?
Glucose 6-phosphate : fructose 6-phosphate
A.) ΔG'° is +1.7 kJ/mol.
B.) ΔG'° is –1.7 kJ/mol.
C.) ΔG'° is incalculably large and negative.
D.) ΔG'° is incalculably large and positive.
E.) ΔG'° is zero.

Answer 1

Answer:

The correct option is B

Explanation:

The reaction is given as

             $Glucose\ 6-phosphate ---->fractose \ 6-phosphate$

The equilibrium constant for this reaction is mathematically represented as

               $K_c = \frac{[concentration \ of \ product ]}{concentration \ of \ reactant } = \frac{[fructose \ 6 -phosphate]}{[Glucose \ 6 - phosphate]}$

From the question we are told that

    [Glucose 6-phosphate] = 2 × [fructose 6-phosphate]

So  

      $K_c = \frac{[fructose \ 6 -phosphate]}{[2 (fructose \ 6 -phosphate)]} = 0.5$

Generally change in free energy $\Delta G$ is mathematically represented as

          $\Delta G = - RTlnK_c$

Given that T = 298 K

                 R = 8.315 J/mol·K

        $\Delta G = -8.314 * 298 * ln (0.5)$

               $= -1717.32 J/mol$

               $= -1.7 \ kJ/mol$

Answer 2

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).