One glucose molecule, C 6 H 12 O 6 (s), is converted to two lactic acid molecules,
Question:
One glucose molecule, C6H12O6(s), is converted to two lactic acid molecules, CH3CH(OH)COOH(s) during glycolysis. Given the combustion reactions of glucose and lactic acid, determine the standard enthalpy for glycolysis.
Transcribed Image Text:
C6H12O6(s) + 602(g) 6 CO2(g) + 6H₂O(1) A,H° ΔΗ° CH₂CH(OH)COOH(s) + 3O₂(g) 3CO2(g) +3H,O(1) AH° -2808 kJ mol-1 -1344 kJ mol-1
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To determine the standard enthalpy for glycolysis using the given information we can use Hesss Law H...View the full answer
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Related Book For
General Chemistry Principles And Modern Applications
ISBN: 9780132931281
11th Edition
Authors: Ralph Petrucci, Jeffry Madura, F. Herring, Carey Bissonnette
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