Thermodynamics of glycolysis PDF

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Summary

Notes on the thermodynamics involved in glycolysis....

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“In biological cells that have a plentiful supply of O 2, glucose is oxidized completely to CO 2 and H2O. Muscle cells may be deprived of O2 during vigorous exercise and, in that case, one molecule of glucose is converted to two moles of lactic acid by the process of glycolysis. Given the thermochemical equations for the combustions of glucose and lactic acid” 6CO2 + 6H20 H° = -2808 kJ mol-1

1)

C6H12O6 + 6O2

2)

CH3CH(OH)COOH + 3O2  3CO2 + 3H2O H° = -1344 kJ mol-1

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Calculate the standard enthalpy for glycolysis: 3)

C6H12O6  2CH3CH(OH)COOH

Is there a biological advantage of complete oxidation of glucose compared to glycolysis? STRATAGY TIME… Well, equations 1 + 2 are combustion equations as oxidation of glucose and lactic acid yield, not surprisingly, carbon dioxide and water. Equation 3 is what matters here, glucose to lactic acid (glycolysis) We need to add (or subtract), etc the thermochemical equations so as to produce ‘the thermochemical equation’ for the reaction required. So lets have a go… We know from equation 3 that ‘1 molecule of glucose in the end produces 2 molecules of lactic acid’. So first step would be glucose is oxidized to carbon dioxide and water. Then carbon dioxide and water will form lactic acid… 1)

C6H12O6 + 6O2



6CO2 + 6H20 H° = -2808 kJ mol-1

(need to reverse equation 2, and by reversing it,negative becomes positive) 2)

3CO2 + 3H2O  CH3CH(OH)COOH + 3O2 H° = +1344 kJ mol-1

BUT we have a problem, things are not balanced, we need 6 carbon dioxides and 6 waters to form 2 lactic acids…so equation above becomes: 6CO2 + 6H2O  2CH3CH(OH)COOH + 6O2 H° = +2688 kJ mol-1 (above value from +1344 kJ mol-1 x 2)

So now we have: 6CO2 + 6H20 H° = -2808 kJ mol-1

1)

C6H12O6 + 6O2

2)

6CO2 + 6H2O  2CH3CH(OH)COOH + 6O2 H° = +2688 kJ mol-1

OVERALL: 3)

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C6H12O6  2CH3CH(OH)COOH

H° = -120 kJ mol-1

(-2808 kJ mol-1 + 2688 kJ mol-1 = -120 kJ mol-1) So, the conversion of glucose to lactic acid during glycolysis is -120 kJ mol-1 (less than 5% of the enthalpy of combustion (equation 1), so full oxidation of glucose from a metabolic point of view is much more useful than glycolysis because with oxidation of glucose more ‘energy’ is available for work !!!...