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 21.9, and the temperature is 37.0 �C? ?G�\' for the reaction is �16.7 kJ/mol. The constant R = 8.3145 J/(mol�K)

Answer 1

Answer:

The free energy change for the reaction at 37.0°C is -8.741 kJ.

Explanation:

The free energy of the reaction is given by :

$\Delta G=\Delta G^o+RT\ln K$

where,

$\Delta G^o$ = standard Gibbs free energy

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

T = temperature in Kelvins

K = equilibrium constant

We have :

$\Delta G^o=-16.7 kJ/mol=-16,700 J/mol$

1 kJ = 1000 J

T = 37.0 C = 37 +273.15 K = 310.15 K

Ratio of concentrations of the products to the concentrations of the reactants =K = 21.9

$\Delta G=-16,700 J/mol+8.314J/K mol\times 310.15 K \ln[21.9]$

$=-8,741.22 J = -8.741 kJ$

The free energy change for the reaction at 37.0°C is -8.741 kJ.