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 :

(a) The value of for the reaction is -4.7 kJ/mol

(b) The value of for the reaction is 7.55 kJ/mol

(c) The value of for the reaction is -13.7 kJ/mol

Explanation :

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

where,

= standard Gibbs free energy

R = gas constant = 8.314 J/K.mol

T = temperature =

 = equilibrium constant

Now we have to calculate the value of for the following reactions.

(a)    

Thus, the value of for the reaction is -4.7 kJ/mol

(b)    

Thus, the value of for the reaction is 7.55 kJ/mol

(c)    

Thus, the value of for the reaction is -13.7 kJ/mol

is there any options? maybe im just not seeing them but i need the options to

-love shrooms