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Ответ:
Ответ:
Ответ:
Keq = 19
ΔG° = -7.3kJ/mol
Explanation:
Based on the chemical reaction:
Glucose 1-phosphate ⇄ Glucose 6-phosphate
The equilibrium constant, Keq is defined as:
Keq = [Glucose 6-phosphate] / [Glucose 1-phosphate]
Where [] are equilibrium concentrations of each substance
Replacing:
Keq = [0.19M] / [0.01M]
Keq = 19
Now, standard free energy change, ΔG° is defined as:
ΔG° = -RT ln K
Where R is gas constant 8.314J/molK
T is absolute temperature (25°C + 273.15K = 298.15K)
and K is equilibrium constant = 19
Replacing:
ΔG° = -8.314J/molK*298.15K ln 19
ΔG° = -7299J/mol
ΔG° = -7.3kJ/mol