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Vika [28.1K]
3 years ago
11

Need help ASAP please and thank you

Chemistry
1 answer:
Anna35 [415]3 years ago
5 0
The answer is the second choice
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The value of ΔG°′ΔG°′ for the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) is +1.67 kJ/mol+1.67 kJ/mol . If t
elena-14-01-66 [18.8K]

Answer:

The concentration of fructose-6-phosphate is 1.50x10⁻³M

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Gibbs Free energy, ΔG°, is defined as:

ΔG° = -RTln K

<em>Where ΔG° = 1670J/mol</em>

<em>R is gas constant = 8.314J/molK</em>

<em>T is absolute temperature (25°C + 273.15 = 298.15K)</em>

Thus, equilibrium constant, k, is:

1670J/mol = -8.314J/molK* 298.15K ln K

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0.5098 = K

K of the reaction:

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is:

K = 0.5098 = [F6P] / [G6P]

As [G6P] is 2.95x10⁻³M:

0.5098 = [F6P] / [2.95x10⁻³M]

[F6P] = 1.50x10⁻³M

<h3>The concentration of fructose-6-phosphate is 1.50x10⁻³M</h3>
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