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Feliz [49]
3 years ago
11

When a mixture of glucose 6-phosphate and fructose 6-phosphate is incubated with the enzyme phosphohexose isomerase, the final m

ixture 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.
Chemistry
2 answers:
Anastasy [175]3 years ago
5 0

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

elena-s [515]3 years ago
3 0

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).

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Calculate the freezing point of a solution 1.25 g benzene (C6H6) in 125 g of chloroform (CHCl3).
posledela

Answer:

The freezing point for the solution is -64.09°C

Explanation:

This problem can be solved, by the freezing point depression. This colligative problem shows, that the freezing point of a solution is lower than the freezing point of pure solvent.

ΔT = Kf.  m

ΔT = T° freezing pure solvent - T° freezing solution

Kf = Cryscopic constant, for chloroform is 4.68

T°freezing pure solvent = -63.5°C

m is mol/kg of solvent → molality

Let's determine the moles of benzene

1.25 g / 78 g/mol = 0.0160 mol

Let's convert the mass of solvent to kg

125 g . 1kg / 1000 g = 0.125 kg

m = 0.0160 mol / 0.125 kg → 0.128 m

Let's go to the formula to replace the data

-63.5°C - T° freezing solution = 4.68 °C/m . 0.128 m

T° freezing solution = - (4.68 °C/m . 0.128 m + 63.5°C)

T° freezing solution = - 64.09°C

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<u>Given:</u>

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<u>To determine:</u>

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