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

Step-by-step 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* 310K* \ln (21.3))\\\\\Delta G=-8816.7J/mol=-8.82kJ/mol`

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