Determine standard free energy change (ΔG∘), for the reaction NaOH + HCl -> H2O + NaCl when HCl = 1M, NaOH = 5M, H2O = 3M, and NaCl = 2M at 298 K and ΔG is at equilibrium. (R = 8.314 J/Kmol).

Answer 1

concentration of HCl = [HCl]= 1 M

concentration of NaOH = [NaOH]= 5 M

concentration of H2O = [H2O]= 3 M

concentration of NaCl = [NaCl]= 2 M

R = 8.314 J/K

Temperature = 298 K

Given reaction

than, equilibrium constant for given reaction = Kc =  = 2 x 3 / 1 x 5= 1.2

than Gibbs free energy = -R x T x lnK = -8.314 x 298x ln(1.2) = 451.7 J/K mol