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The DG' for the complete oxidation of lactic acid to CO2 and H2O calculated from the information given below is ______ (in cal/mole).
Moles of ATP that could be synthesized in the above process at 40% efficiency is _________ (1 mole of ATP = 7300 calories) (answer in integer)
1. glucose → 2 lactic acid ΔG' = —52.000cal/mole
2. glucose + 6O2 → 6 CO2 + 6H20 ΔG' = —686,000 cal/mole
Moles of ATP that could be synthesized in the above process at 40% efficiency is _________ (1 mole of ATP = 7300 calories) (answer in integer)
1. glucose → 2 lactic acid ΔG' = —52.000cal/mole
2. glucose + 6O2 → 6 CO2 + 6H20 ΔG' = —686,000 cal/mole
Correct answer is '-317000'. Can you explain this answer?
Verified Answer
The DG' for the complete oxidation of lactic acid to CO2and H2O ca...
We want to calculate the ΔG' of following reaction
Lactic acid +3 02,→ 3 CO2 + 3 H20 .....(1)
glucose → 2 lactic acid ΔG,' = -52.000 callmok (Given) ....(3)
2Lactic acid +6 024 6 CO2 + 6 H2O ΔG2' =? Cal/mole ... (4)
(equation (4) is obtained by multiplying equation (1) by 2)
Add equation 3 and 4. we get
Glucose + 6O2 → 6 CO2 + 6 H2O ΔG3' = -686,000 cal/mole (Given)
ΔG3'= +ΔG2'
–686,000 cal/mole = –52,000 cal/mole + ΔG2'
ΔG2'= -634,000 cal/mole
Oxidation of 2 moles of lactic acid gives -634.000 cal/mole. so 1 mole gives -317.000 cal/moles (b) At 40% efficiency = 0.40 x 317000 cal/mole = 127000 cal/mol (Approx)
1ATP gives 7300 cal
No. of ATP required = 127000/7300 = 17.4 ATP
Lactic acid +3 02,→ 3 CO2 + 3 H20 .....(1)
glucose → 2 lactic acid ΔG,' = -52.000 callmok (Given) ....(3)
2Lactic acid +6 024 6 CO2 + 6 H2O ΔG2' =? Cal/mole ... (4)
(equation (4) is obtained by multiplying equation (1) by 2)
Add equation 3 and 4. we get
Glucose + 6O2 → 6 CO2 + 6 H2O ΔG3' = -686,000 cal/mole (Given)
ΔG3'= +ΔG2'
–686,000 cal/mole = –52,000 cal/mole + ΔG2'
ΔG2'= -634,000 cal/mole
Oxidation of 2 moles of lactic acid gives -634.000 cal/mole. so 1 mole gives -317.000 cal/moles (b) At 40% efficiency = 0.40 x 317000 cal/mole = 127000 cal/mol (Approx)
1ATP gives 7300 cal
No. of ATP required = 127000/7300 = 17.4 ATP
This question is part of UPSC exam. View all IIT JAM courses
Most Upvoted Answer
The DG' for the complete oxidation of lactic acid to CO2and H2O ca...
In order to calculate the ΔG for the complete oxidation of lactic acid, we need to know the standard free energy change (ΔG°) for the reaction. However, this information is not given in the question. Without the ΔG° value, we cannot calculate the ΔG for the reaction.
Similarly, without the ΔG value, we cannot calculate the moles of ATP that could be synthesized in the process at 40% efficiency.
Similarly, without the ΔG value, we cannot calculate the moles of ATP that could be synthesized in the process at 40% efficiency.
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The DG' for the complete oxidation of lactic acid to CO2and H2O calculated from the information given below is ______ (in cal/mole).Moles of ATP that could be synthesized in the above process at 40% efficiency is _________ (1 mole of ATP = 7300 calories) (answer in integer)1. glucose →2 lactic acid ΔG' = —52.000cal/mole2. glucose + 6O2 →6 CO2 + 6H20 ΔG' = —686,000 cal/moleCorrect answer is '-317000'. Can you explain this answer?
Question Description
The DG' for the complete oxidation of lactic acid to CO2and H2O calculated from the information given below is ______ (in cal/mole).Moles of ATP that could be synthesized in the above process at 40% efficiency is _________ (1 mole of ATP = 7300 calories) (answer in integer)1. glucose →2 lactic acid ΔG' = —52.000cal/mole2. glucose + 6O2 →6 CO2 + 6H20 ΔG' = —686,000 cal/moleCorrect answer is '-317000'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about The DG' for the complete oxidation of lactic acid to CO2and H2O calculated from the information given below is ______ (in cal/mole).Moles of ATP that could be synthesized in the above process at 40% efficiency is _________ (1 mole of ATP = 7300 calories) (answer in integer)1. glucose →2 lactic acid ΔG' = —52.000cal/mole2. glucose + 6O2 →6 CO2 + 6H20 ΔG' = —686,000 cal/moleCorrect answer is '-317000'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The DG' for the complete oxidation of lactic acid to CO2and H2O calculated from the information given below is ______ (in cal/mole).Moles of ATP that could be synthesized in the above process at 40% efficiency is _________ (1 mole of ATP = 7300 calories) (answer in integer)1. glucose →2 lactic acid ΔG' = —52.000cal/mole2. glucose + 6O2 →6 CO2 + 6H20 ΔG' = —686,000 cal/moleCorrect answer is '-317000'. Can you explain this answer?.
The DG' for the complete oxidation of lactic acid to CO2and H2O calculated from the information given below is ______ (in cal/mole).Moles of ATP that could be synthesized in the above process at 40% efficiency is _________ (1 mole of ATP = 7300 calories) (answer in integer)1. glucose →2 lactic acid ΔG' = —52.000cal/mole2. glucose + 6O2 →6 CO2 + 6H20 ΔG' = —686,000 cal/moleCorrect answer is '-317000'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about The DG' for the complete oxidation of lactic acid to CO2and H2O calculated from the information given below is ______ (in cal/mole).Moles of ATP that could be synthesized in the above process at 40% efficiency is _________ (1 mole of ATP = 7300 calories) (answer in integer)1. glucose →2 lactic acid ΔG' = —52.000cal/mole2. glucose + 6O2 →6 CO2 + 6H20 ΔG' = —686,000 cal/moleCorrect answer is '-317000'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The DG' for the complete oxidation of lactic acid to CO2and H2O calculated from the information given below is ______ (in cal/mole).Moles of ATP that could be synthesized in the above process at 40% efficiency is _________ (1 mole of ATP = 7300 calories) (answer in integer)1. glucose →2 lactic acid ΔG' = —52.000cal/mole2. glucose + 6O2 →6 CO2 + 6H20 ΔG' = —686,000 cal/moleCorrect answer is '-317000'. Can you explain this answer?.
Solutions for The DG' for the complete oxidation of lactic acid to CO2and H2O calculated from the information given below is ______ (in cal/mole).Moles of ATP that could be synthesized in the above process at 40% efficiency is _________ (1 mole of ATP = 7300 calories) (answer in integer)1. glucose →2 lactic acid ΔG' = —52.000cal/mole2. glucose + 6O2 →6 CO2 + 6H20 ΔG' = —686,000 cal/moleCorrect answer is '-317000'. Can you explain this answer? in English & in Hindi are available as part of our courses for IIT JAM.
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The DG' for the complete oxidation of lactic acid to CO2and H2O calculated from the information given below is ______ (in cal/mole).Moles of ATP that could be synthesized in the above process at 40% efficiency is _________ (1 mole of ATP = 7300 calories) (answer in integer)1. glucose →2 lactic acid ΔG' = —52.000cal/mole2. glucose + 6O2 →6 CO2 + 6H20 ΔG' = —686,000 cal/moleCorrect answer is '-317000'. Can you explain this answer?, a detailed solution for The DG' for the complete oxidation of lactic acid to CO2and H2O calculated from the information given below is ______ (in cal/mole).Moles of ATP that could be synthesized in the above process at 40% efficiency is _________ (1 mole of ATP = 7300 calories) (answer in integer)1. glucose →2 lactic acid ΔG' = —52.000cal/mole2. glucose + 6O2 →6 CO2 + 6H20 ΔG' = —686,000 cal/moleCorrect answer is '-317000'. Can you explain this answer? has been provided alongside types of The DG' for the complete oxidation of lactic acid to CO2and H2O calculated from the information given below is ______ (in cal/mole).Moles of ATP that could be synthesized in the above process at 40% efficiency is _________ (1 mole of ATP = 7300 calories) (answer in integer)1. glucose →2 lactic acid ΔG' = —52.000cal/mole2. glucose + 6O2 →6 CO2 + 6H20 ΔG' = —686,000 cal/moleCorrect answer is '-317000'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The DG' for the complete oxidation of lactic acid to CO2and H2O calculated from the information given below is ______ (in cal/mole).Moles of ATP that could be synthesized in the above process at 40% efficiency is _________ (1 mole of ATP = 7300 calories) (answer in integer)1. glucose →2 lactic acid ΔG' = —52.000cal/mole2. glucose + 6O2 →6 CO2 + 6H20 ΔG' = —686,000 cal/moleCorrect answer is '-317000'. Can you explain this answer? tests, examples and also practice IIT JAM tests.
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