-2226 C12H22O11 0 O2 -393 CO2 -286 H2O a) Given the following delta G½ = 30.5 kJ

Question

-2226              C12H22O11
0                               O2
-393                          CO2
-286                          H2O

a) Given the following delta G½ = 30.5 kJ/mol) at standard state conditions by coupling the oxidation of one sucrose molecule into CO2 and H2O? NOTE: ADP + Pi ? ATP ?G½ = 30.5 kJ/mol delta G½? What is the maximum number of ATP molecules that can be made from ADP and Pi ( delta S½ for the oxidation for sucrose is 512 J mol-1 K-1, what is the associated standard free energy change, delta H½ you calculated above and knowing that the delta Hf½ (kJ/mol) Compound -2226 C12H22O11 0 O2 -393 CO2 -286 H2O b) Using the delta H½ for the complete oxidation of sucrose into CO2 and H2O. (Hint: first write a balanced chemical equation.) delta Hf½ values, calculate the

Explanation / Answer

a) the reaction is given by

C12H22O11 + 12 O2 ----> 12 CO2 + 11 H2O

dH rxn = dHfo products - dHfo reactants

dH rxn = 12 x dHfo C02 + 11 x dHfo H20 - 12 x dHfo 02 - dHfo C12H22O11

dHrxn = -12 x 393 - 11 x 286 - 12 x 0 + 2226

dH rxn = -5636 kJ/ mol

so

dHo for complete oxidation is -5636 kJ/ mol

b) we know that

dGo = dH - TdSo

dGo = -5636 x 1000 - 298 x 512

dGo = -5788576 J/ mol

dGo = -5788.576 kJ/mol

no of ATP = total dGo / dGo for one ATP

given dGo for one ATP is 30.5 kJ/mol

no of ATP = 5788.576 / 30.5

no of ATP = 189.79

so the no of ATP that can be made is 189

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