Question

The difference between AG and AG' (Basic Applied) In glycolysis the following reaction occurs: Fructose 6-phosphate Glucose 6-phosphate K'eq=1.97 a) What is the AGʻº for the reaction if K'eq is measured at 25°C? (2 points) b) If the concentration of fructose 6-phosphate is 1.5M and that of glucose 6-phosphate is adjusted to 0.50M what is AG? (3 points) c) Why are AG and AG' different? (1 point)

Answer 1

Equilibrium : fructose-6-phosphate a g lucose-6-phosphate Equilibrium constant, Kg =1.97 Temperature, T = 25°C +273 = 298 K Gas constant, R= 8.314 JK-m01-1 Standard free energy change of the reaction, AG = - RT in Ka =-(8.314 JK 'mo1-) (298 K)In (1.97 ) =-1680 I, 1kJ mol 10% ] = -1.68 kJ/mol ---------------------------------------------- Equation: fructose-6-phosphate A glucose-6-phosphate ;4Gº = -1.68 kJ/mol Free energy change, AG = Gº +RTinQ AG = AGⓇ +RT in (glucos-6-phosphate] fructose-6-phosphate) = ( - 1.68 kJ/mol ) +(8.314x108 kJX+molº)x(298 K)in(95) =(-1.68 kJ/mol )+(8.314x 109 kJX+mo1™)x(298 K)(-1.1) = ( -1.68 kJim ol )+(-2.72 kJ/mol) = -4.40 kJ/mol - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Conditions for AG" → The pressure is 101.3 kPa (1 atm) → The temperature is 298 K → The reactant and product are at their equilibrium concentrations AG is measured at the standard consitions where as AG is measured at the given conditions