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

Answer :

(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:

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

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

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

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

$\Delta G^o=-4.7\times 10^{3}J/mol=-4.7kJ/mol$

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

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

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

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

$\Delta G^o=7.55\times 10^{3}J/mol=7.55kJ/mol$

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

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

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

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

$\Delta G^o=-1.37\times 10^{4}J/mol=-13.7kJ/mol$

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