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 21.1 and the temperature is 37.0 ° C ? Δ G ° ' for the reaction is − 16.7 kJ/mol . Δ G = kJ / mol

Answer 1

Answer : The value of
`\Delta G_(rxn)` is -8.84 kJ/mol

Explanation :

The formula used for
`\Delta G_(rxn)` is:

`\Delta G_(rxn)=\Delta G^o+RT\ln Q` ............(1)

where,

`\Delta G_(rxn)` = Gibbs free energy for the reaction = ?

`\Delta G_^o` = standard Gibbs free energy = -16.7 kJ/mol

R = gas constant =
`8.314* 10^(-3)kJ/mole.K`

T = temperature =
`37.0^oC=273+37.0=310K`

Q = reaction quotient =
`(product)/(reactant)` = 21.1

Now put all the given values in the above formula 1, we get:

`\Delta G_(rxn)=(-16.7kJ/mol)+[(8.314* 10^(-3)kJ/mole.K)* (310K)* \ln (21.1)`

`\Delta G_(rxn)=-8.84kJ/mol`

Therefore, the value of
`\Delta G_(rxn)` is -8.84 kJ/mol