The oxidation of solid glucose (C6H12O6) to carbon dioxide gas(CO2) and liquid w

Question

The oxidation of solid glucose (C6H12O6) to carbon dioxide gas(CO2) and liquid water
(H2O) at 298.15 K and 1 bar pressure is accompanied by an enthalpychange of -2808
kJ/mol.

C6H12O6(s) + 6O2 (g)->6CO2 (g) + 6H2O(l)
a) What is the standard Gibbs energy change of this reaction at298.15K?
b) What is the change in the TOTAL entropy of the system +surroundings after this
reaction takes place?
c) If the pressure is increased slightly, does this make thereaction more or less
spontaneous? Why? hide problem

Explanation / Answer

           C6H12O6(s) + 6O2(g).......>6CO2 (g) + 6H2O(l) a.   G0rxn ={6G0(CO2(g))+6G0(H2O(l))}-{1G0(C6H12O6(s))+6G0(6O2(g))}                    = {6(-394.4)+6(-237.13)kJ}-{1(-910.4)+6(230.1)}kJ                     =-3789.18kJ -470.2kJ                    = -4259.38kJ b.   H0rxn = {6H0(CO2(g))+6H0(H2O(l))}-{1H0(C6H12O6(s))+6H0(6O2(g))}                    = {6(-393.5)+6(-285.83)}-{1(-1273.02)+6(0)}kJ       H0rxn  =-2802.96kJ          S0surr = -H0sys / T                       = -(-2802.96kJ) / 298.15K                       = 9.4011kJ/K          S0surr = 9401.1J/K         S0rxn = {6Sf0(CO2(g))+6Sf0((H2O(l))} -{1Sf0((C6H12O6(s))+6Sf0(O2(g))}                      ={6(213.6)+6(69.91)}J/K-{1(212.1)+6(205.0)}J/K          S0rxn =258.96J/K        S0rxn = S0sys =258.96J/K             S0univ =S0sys + S0surr                      = 258.96J/K + 9401.1J/K                      =9660.06J/K                     

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