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:
where,
= standard Gibbs free energy = +1.67 kJ/mol = +1670 J/mol
R = gas constant = 8.314 J/K.mol
T = temperature =
K = equilibrium constant = ?
Now put all the given values in the above formula, we get:
Now we have to calculate the concentration of fructose-6-phosphate.
The expression of equilibrium constant is:
Therefore, the concentration of fructose-6-phosphate is 0.275 mM
3.34*1000= 3340 j
Now divide joules by grams
3340/10 =334 j/g