Question

Calculation of ΔG′° from an Equilibrium Constant Calculate the standard free-energy change for each of the following metabolically important enzyme-catalyzed reactions, using the equilibrium constants given for the reactions at 25 °C and pH 7.0. ( a ) Glutamate + oxaloacetate aspartate aminotranferase ⇌ aspartate + α -ketoglutarate K ′ eq = 6.8 ( b ) Dihydroxyacetone phosphate triose phosphate isomerase ⇌ glyceraldehyde 3 -phosphate K ′ eq = 0.0475 ( c ) Fructose 6 -phosphate + ATP phosphofructokinase ⇌ fructose 1 , 6 -bisphosphate + ADP K ′ eq = 254

Answer 1

(a) The value of
`\Delta G^o` for the reaction is -4.7 kJ/mol

(b) The value of
`\Delta G^o` for the reaction is 7.55 kJ/mol

(c) The value of
`\Delta G^o` for the reaction is -13.7 kJ/mol

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

`\Delta G^o=-RT* \ln K_(eq)`

where,

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

R = gas constant = 8.314 J/K.mol

T = temperature =
`25^oC=273+25=298K`

`K_(eq)` = equilibrium constant

Now we have to calculate the value of
`\Delta G^o` for the following reactions.

(a)
`\text{Glutamate}+\text{oxaloacetate}\overset{\text{aspartate aminotranferase}}\rightleftharpoons \text{aspartate}+\alpha \text{-ketoglutarate}`
`K_(eq)=6.8`

`\Delta G^o=-RT* \ln K_(eq)`

`\Delta G^o=-(8.314J/K.mol)* (298K)* \ln (6.8)`

`\Delta G^o=-4.7* 10^(3)J/mol=-4.7kJ/mol`

Thus, the value of
`\Delta G^o` for the reaction is -4.7 kJ/mol

(b)
`\text{Dihydroxyacetone phosphate}\overset{\text{triose phosphate isomerase}}\rightleftharpoons \text{glyceraldehyde3-phosphate}`
`K_(eq)=0.0475`

`\Delta G^o=-RT* \ln K_(eq)`

`\Delta G^o=-(8.314J/K.mol)* (298K)* \ln (0.0475)`

`\Delta G^o=7.55* 10^(3)J/mol=7.55kJ/mol`

Thus, the value of
`\Delta G^o` for the reaction is 7.55 kJ/mol

(c)
`\text{Fructose6-phosphate}+\text{ATP}\overset{phosphofructokinase}\rightleftharpoons \text{fructose1,6 -bisphosphate}+\text{ADP}`
`K_(eq)=254`

`\Delta G^o=-RT* \ln K_(eq)`

`\Delta G^o=-(8.314J/K.mol)* (298K)* \ln (254)`

`\Delta G^o=-1.37* 10^(4)J/mol=-13.7kJ/mol`

Thus, the value of
`\Delta G^o` for the reaction is -13.7 kJ/mol