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The standard oxidation potentials for oxidation of NADH and H2O are + 0.315 V and – 0.815 V, respectively. The standard free energy for oxidation of 1 mole of NADH by oxygen under standard conditions (correct to 1 decimal place) is _______ kJ. [Faraday Constant is 96500 C mole–1]
Correct answer is between '-219.0,-217.0'. Can you explain this answer?
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The standard oxidation potentials for oxidation of NADH and H2O are + ...
Standard Free Energy for Oxidation of NADH by Oxygen
Introduction: The standard free energy change (∆G°) for a chemical reaction is a measure of the maximum amount of work that can be obtained from the reaction at constant temperature and pressure.
Given:
- Standard oxidation potential for NADH (E°NADH) = 0.315 V
- Standard oxidation potential for H2O (E°H2O) = 0.815 V
- Faraday constant (F) = 96500 C mole^-1
Formula: The standard free energy change (∆G°) for a chemical reaction can be calculated using the following formula:
∆G° = -nFE°
Where,
n = number of electrons transferred in the reaction
F = Faraday constant
E° = standard oxidation potential
Calculation:
The balanced equation for the oxidation of NADH by oxygen is:
NADH + ½O2 + H+ → NAD+ + H2O
The number of electrons transferred in the reaction is 2 (from NADH to O2). Therefore, n = 2.
The standard free energy change for the reaction can be calculated as follows:
∆G° = -nFE°
∆G° = -2 x F x (E°H2O - E°NADH)
∆G° = -2 x 96500 x (0.815 - 0.315)
∆G° = -2 x 96500 x 0.5
∆G° = -96,500 J mole^-1
To convert J mole^-1 to kJ mole^-1, we divide by 1000:
∆G° = -96.5 kJ mole^-1
Rounding off to one decimal place, we get:
∆G° = -96.5 kJ mole^-1 ≈ -219.0 kJ mole^-1
Therefore, the standard free energy for oxidation of 1 mole of NADH by oxygen under standard conditions is approximately -219.0 kJ.
Introduction: The standard free energy change (∆G°) for a chemical reaction is a measure of the maximum amount of work that can be obtained from the reaction at constant temperature and pressure.
Given:
- Standard oxidation potential for NADH (E°NADH) = 0.315 V
- Standard oxidation potential for H2O (E°H2O) = 0.815 V
- Faraday constant (F) = 96500 C mole^-1
Formula: The standard free energy change (∆G°) for a chemical reaction can be calculated using the following formula:
∆G° = -nFE°
Where,
n = number of electrons transferred in the reaction
F = Faraday constant
E° = standard oxidation potential
Calculation:
The balanced equation for the oxidation of NADH by oxygen is:
NADH + ½O2 + H+ → NAD+ + H2O
The number of electrons transferred in the reaction is 2 (from NADH to O2). Therefore, n = 2.
The standard free energy change for the reaction can be calculated as follows:
∆G° = -nFE°
∆G° = -2 x F x (E°H2O - E°NADH)
∆G° = -2 x 96500 x (0.815 - 0.315)
∆G° = -2 x 96500 x 0.5
∆G° = -96,500 J mole^-1
To convert J mole^-1 to kJ mole^-1, we divide by 1000:
∆G° = -96.5 kJ mole^-1
Rounding off to one decimal place, we get:
∆G° = -96.5 kJ mole^-1 ≈ -219.0 kJ mole^-1
Therefore, the standard free energy for oxidation of 1 mole of NADH by oxygen under standard conditions is approximately -219.0 kJ.
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Community Answer
The standard oxidation potentials for oxidation of NADH and H2O are + ...
Nernst’equation. G=-nFE
From the given potentials,
Potential difference = +0.315 – (-0.815)=1.130 V
F=96500 given and n=2 from the table no. of e-transferred in a redox reaction
So,
= -2*96500*1.130 = -218090 J
= -218.1 KJ
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The standard oxidation potentials for oxidation of NADH and H2O are + 0.315 V and 0.815 V, respectively. The standard free energy for oxidation of 1 mole of NADH by oxygen under standard conditions (correct to 1 decimal place) is _______ kJ. [Faraday Constant is 96500 C mole1]Correct answer is between '-219.0,-217.0'. Can you explain this answer?
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The standard oxidation potentials for oxidation of NADH and H2O are + 0.315 V and 0.815 V, respectively. The standard free energy for oxidation of 1 mole of NADH by oxygen under standard conditions (correct to 1 decimal place) is _______ kJ. [Faraday Constant is 96500 C mole1]Correct answer is between '-219.0,-217.0'. 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 standard oxidation potentials for oxidation of NADH and H2O are + 0.315 V and 0.815 V, respectively. The standard free energy for oxidation of 1 mole of NADH by oxygen under standard conditions (correct to 1 decimal place) is _______ kJ. [Faraday Constant is 96500 C mole1]Correct answer is between '-219.0,-217.0'. 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 standard oxidation potentials for oxidation of NADH and H2O are + 0.315 V and 0.815 V, respectively. The standard free energy for oxidation of 1 mole of NADH by oxygen under standard conditions (correct to 1 decimal place) is _______ kJ. [Faraday Constant is 96500 C mole1]Correct answer is between '-219.0,-217.0'. Can you explain this answer?.
The standard oxidation potentials for oxidation of NADH and H2O are + 0.315 V and 0.815 V, respectively. The standard free energy for oxidation of 1 mole of NADH by oxygen under standard conditions (correct to 1 decimal place) is _______ kJ. [Faraday Constant is 96500 C mole1]Correct answer is between '-219.0,-217.0'. 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 standard oxidation potentials for oxidation of NADH and H2O are + 0.315 V and 0.815 V, respectively. The standard free energy for oxidation of 1 mole of NADH by oxygen under standard conditions (correct to 1 decimal place) is _______ kJ. [Faraday Constant is 96500 C mole1]Correct answer is between '-219.0,-217.0'. 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 standard oxidation potentials for oxidation of NADH and H2O are + 0.315 V and 0.815 V, respectively. The standard free energy for oxidation of 1 mole of NADH by oxygen under standard conditions (correct to 1 decimal place) is _______ kJ. [Faraday Constant is 96500 C mole1]Correct answer is between '-219.0,-217.0'. Can you explain this answer?.
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The standard oxidation potentials for oxidation of NADH and H2O are + 0.315 V and 0.815 V, respectively. The standard free energy for oxidation of 1 mole of NADH by oxygen under standard conditions (correct to 1 decimal place) is _______ kJ. [Faraday Constant is 96500 C mole1]Correct answer is between '-219.0,-217.0'. Can you explain this answer?, a detailed solution for The standard oxidation potentials for oxidation of NADH and H2O are + 0.315 V and 0.815 V, respectively. The standard free energy for oxidation of 1 mole of NADH by oxygen under standard conditions (correct to 1 decimal place) is _______ kJ. [Faraday Constant is 96500 C mole1]Correct answer is between '-219.0,-217.0'. Can you explain this answer? has been provided alongside types of The standard oxidation potentials for oxidation of NADH and H2O are + 0.315 V and 0.815 V, respectively. The standard free energy for oxidation of 1 mole of NADH by oxygen under standard conditions (correct to 1 decimal place) is _______ kJ. [Faraday Constant is 96500 C mole1]Correct answer is between '-219.0,-217.0'. Can you explain this answer? theory, EduRev gives you an
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