Question

Under anaerobic conditions, glucose is broken down in muscle tissues to form lactic acid according to the reaction, C6H12O6(s) → 2CH3CHOCOOH(s). Thermodynamic data for glucose and lactic acid at 298 K are given below. ΔHfº (kJ/mol) Cp.m (J/mol∙K) Smº (J/mol∙K) Glucose −1273.1 219.3 209.2 Lactic Acid −673.6 127.6 192.1 (a) Calculate ΔS for the system, surrounding and universe at T = 325 K. Assume the heat capacities are constant between 298 K and 325 K. (b) Is this reaction spontaneous or reversible at 325 K? Explain. (c) At what temperature, will this reaction establish the equilibrium?.

Answer 1

   C6H12O6(s) ----> 2CH3CHOCOOH(s)

DH0rxn(at 298) = (2*DH0f,CH3CHOCOOH) - (1*DH0f,C6H12O6)

       = (2*-1273.1)-(1*-673.6)

       = -1872.6 kj/mol

DHrxn(at 325k) = DH0rxn(at 298k)+ DCpreaction*(T2-T1)

DCp reaction = Cp products - Cp reactant

         = (2*127.6)-(219.3)

         = 35.9 j/mol.k

T2 = 325 k , T1 = 298 k

DHrxn(at 325k) = -1872.6 + 35.9*10^-3*(325-298)

               = -1871.63 Kj/mol

DSsurroundings = -DHrxn/T

               = -(-1871.63)/325

               = 5.76 kj/mol.k

               = 5760 j/mol.k

DS0rxn(at 298) = (2*S0f,CH3CHOCOOH) - (1*S0f,C6H12O6)

       = (2*209.2)-(1*192.1)

       = 226.3 joule/k.mol

DSrxn(at 325k) = DS0rxn(at 298k)+ DCprxn*ln(T2/T1)

DCp rxn = Cp products - Cp reactant

         = (2*127.6)-(219.3)

         = 35.9 j/k

T2 = 325 k , T1 = 298 k

DSrxn(at 325k) = 226.3 + 35.9ln(325/298)

DSrxn(at 325k) = 229.4 j/k.mol

DSsys at 325 k = 229.4 j/k.mol

DSuniv at 325 k = DSsys + DS surroundings

       = 229.4 + 5760

DSuniv at 325 k = 5989.4 j/mol.k

b) DG = DH - TDS

at 325 k , DH = -1871.63 kj/mol

DS = 229.6 j/mol.k

DG = (-1871.63)-(325*229.6*10^-3)

   = -1946.25 kj/mol

as DG = -ve, The reaction is spontaneous.

c) For the reaction to be at equilibrium, DG = 0

DG = DH - TDS

DH = -1872.6 kj/mol , DS = 226.3 j/mol.k

0 = (-1872.6)-(T*226.3*10^-3)

T = -8274 k

It does not exist.so that, reaction dot reaches equilibrium