Question

The value of Δ????°′ΔG°′ for the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) is +1.67 kJ/mol+1.67 kJ/mol . If the concentration of glucose-6-phosphate at equilibrium is 1.85 mM1.85 mM , what is the concentration of fructose-6-phosphate? Assume a temperature of 25.0°C25.0°C .

Answer 1

Answer : The concentration of fructose-6-phosphate is 0.275 mM

Explanation :

First we have to calculate the value of equilibrium constant.

The relation between the equilibrium constant and standard Gibbs free energy is:

`\Delta G^o=-RT* \ln k`

where,

`\Delta G^o` = standard Gibbs free energy = +1.67 kJ/mol = +1670 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature =
`25.0^oC=273+25.0=298K`

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

`+1670J/mol=-(8.314J/K.mol)* (298K)* \ln k`

`k=0.509`

Now we have to calculate the concentration of fructose-6-phosphate.

The expression of equilibrium constant is:

`K=\frac{[\text{fructose-6-phosphate}]}{[\text{glucose-6-phosphate}]}`

`0.509=\frac{[\text{fructose-6-phosphate}]}{1.85mM}`

`\text{fructose-6-phosphate}=0.275mM`

Therefore, the concentration of fructose-6-phosphate is 0.275 mM