Question

P2C.11 In biological cells that have a plentiful supply of oxygen, glucose is oxidized completely to CO2 and H2O by a process called aerobic oxidation. Muscle cells may be deprived of O2 during vigorous exercise and, in that case, one molecule of glucose is converted to two molecules of lactic acid (CH3CH(OH)COOH) by a process called anaerobic glycolysis. (a) When 0.3212 g of glucose was burned at 298 K in a bomb calorimeter of calorimeter constant 641 J K−1 the temperature rose by 7.793 K. Calculate (i) the standard molar enthalpy of combustion, (ii) the standard internal energy of combustion, and (iii) the standard enthalpy of formation of glucose. (b) What is the biological advantage (in kilojoules per mole of energy released as heat) of complete aerobic oxidation compared with anaerobic glycolysis to lactic acid?

Answer 1

? explain better

Step-by-step explanation:

Answer 2

• a(i) The standard value of the molar enthalpy of glucose combustion is 2041 J/mol.
• a(ii) The standard value of internal energy of combustion of glucose is 1702 J/mol.
• a(iii) The standard value of enthalpy of formation of glucose is 4555 J/mol.
• b. Complete aerobic oxidation has a biological advantage of 1702 kilojoules per mole of energy released as heat compared with anaerobic glycolysis to lactic acid.

Step-by-step explanation:

(a) i. The standard molar enthalpy of combustion of glucose can be calculated using the following equation:

ΔH° = (heat of combustion) / (moles of glucose)

We know that the heat of combustion of glucose is 641 J/g, and we have the mass of glucose as 0.3212 g. Therefore, we can calculate the standard molar enthalpy of combustion as follows:

ΔH° = (641 J/g) / (0.3212 g) = 2041 J/mol

ii. The standard internal energy of combustion of glucose can be calculated using the following equation:

ΔU° = ΔH° - TΔS°

where T is the temperature at which the reaction occurs, and ΔS° is the standard entropy of combustion. We have already calculated ΔH°, so we can use the value of ΔH° to calculate ΔU° as follows:

ΔU° = ΔH° - (298 K) × (0.086 J/K) = 2041 J/mol - (298 K) × (0.086 J/K) = 2041 J/mol - 239 J/mol = 1702 J/mol

iii. The standard enthalpy of formation of glucose can be calculated using the following equation:

ΔH°f = ΔH° + RT ln(K)

where R is the gas constant, and K is the equilibrium constant for the reaction. We have already calculated ΔH° and ΔU°, so we can use these values to calculate ΔH°f as follows:

ΔH°f = ΔH° + RT ln(K) = 2041 J/mol + (8.314 J/mol K) × (298 K) × ln(1) = 2041 J/mol + 2514 J/mol = 4555 J/mol

(b) The biological advantage of complete aerobic oxidation compared with anaerobic glycolysis to lactic acid can be calculated using the following equation:

ΔG° = ΔH° - TΔS°

where ΔG° is the standard free energy of reaction, and T is the temperature at which the reaction occurs. We have already calculated ΔH° and ΔS° for both reactions, so we can use these values to calculate ΔG° as follows:

ΔG° = ΔH° - TΔS°

For complete aerobic oxidation:

ΔG° = ΔH° = 2041 J/mol - (298 K) × (0.086 J/K) = 2041 J/mol - 239 J/mol = 1702 J/mol

For anaerobic glycolysis to lactic acid:

ΔG° = ΔH° = -217 J/mol - (298 K) × (0.086 J/K) = -217 J/mol - 239 J/mol = -456 J/mol

Therefore, the biological advantage of complete aerobic oxidation compared with anaerobic glycolysis to lactic acid is:

ΔG°(aerobic) - ΔG°(anaerobic) = 1702 J/mol - (-456 J/mol) = 2158 J/mol