Chemistry, 02.03.2020 17:29, ericchen4399
The value of Δ G ° ' for the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) is + 1.67 kJ/mol . If the concentration of glucose-6-phosphate at equilibrium is 2.95 mM , what is the concentration of fructose-6-phosphate? Assume a temperature of 25.0 ° C .
Answers: 1
![K_c=\frac{[\text{Fructose-6-phosphate}]}{[\text{Glucose-6-phosphate}]}](/tpl/images/0530/2012/7727c.png)
![K_c=\frac{[\text{Fructose-6-phosphate}]}{2.95 mM}](/tpl/images/0530/2012/f71d2.png)
= standard Gibbs free energy = 1.67 kJ/mol = 1670 J/mol (Conversion factor: 1kJ = 1000J)
![25.0^oC=[273+25.0]K=298.0 K](/tpl/images/0530/2012/fee79.png)
![1670 J/mol=-8.314 J/mol K\times 298.0 K\times \ln \frac{[\text{Fructose-6-phosphate}]}{2.95 mM}](/tpl/images/0530/2012/2fb29.png)
![[\text{Fructose-6-phosphate}]=1.503 mM](/tpl/images/0530/2012/8d092.png)
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