Question

The standard free energy for the reaction A ---> B is -4.00 kJ/mole. If the concentration of A known to be 4.00 mM at equilibrium at 25 oC. What would be equilibrium mM concentration of B. Provide answer to the nearest tenth mM.

Answer 1

The standard free energy for the reaction A ---> B is -4.00 kJ/mole

G⁰    = -4.00KJ/mole = -4000J/mole

T = 25+273 = 298K

G⁰     = -RTlnKc

-4000    = -8.314*298lnKc

-4000    = -8.314*298*2.303logKc

-4000   = -5705.84logKc

logKc   = -4000/-5705.84

logKc    = 0.7

    K      = 10^0.7   = 5.012

K   = [B]/[A]

5.012 = [B]/4

[B]   = 5.012*4    = 20.05mM >>>>answer