Question

Consider the fructose-1,6-bisphosphatase reaction. Calculate the free energy change if the ratio of the concentrations of the products to the concentrations of the reactants is 21.3 and the temperature is 37.0 ° C ? Δ G ° ' for the reaction is − 16.7 kJ/mol .

Answer 1

Answer: The Gibbs free energy of the reaction is -8.82 kJ/mol

Explanation:

The equation used to Gibbs free energy of the reaction follows:

$\Delta G=\Delta G^o+RT\ln K_{eq}$

where,

$\Delta G$ = free energy of the reaction

$\Delta G^o$ = standard Gibbs free energy = -16.7 kJ/mol = -16700 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant = $8.314J/K mol$

T = Temperature = $37^oC=[273+37]K=310K$

$K_{eq}$ = Ratio of concentration of products and reactants = 21.3

Putting values in above equation, we get:

$\Delta G=-16700J/mol+(8.314J/K.mol\times 310K\times \ln (21.3))\\\\\Delta G=-8816.7J/mol=-8.82kJ/mol$

Hence, the Gibbs free energy of the reaction is -8.82 kJ/mol