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Below are the reduction half reactions for chemolithoautotrophic nitrification, where ammonia is a source of electrons and energy and oxygen is the terminal electron acceptor. • NO₂ +6e-> NH4* (+0.34 volts) • O₂ + 4e -> 2H₂O (+0.82 volts) Using the information given, calculate the AE for this reaction, balance the full reaction to determine the n, the number of electrons transferred when 284 moles of NH4* are oxidized. Finally, use the simplified Nernst Equation AG = -nFAE, where F = 96.5 kJ (mol e x V)-1 to determine the Gibbs Free energy available to do work! • Report your answer in kJ rounded to two decimal places. o Include trailing zeros!! Always report two decimal places even if the answer is a whole number oe.g. 18.00 not 18 Report only the numeric portion of your answer o e.g. 1.01, not 1.01 kj per mole. • Answers should ALWAYS be negative since this is a spontaneous reaction.

Below are the reduction half reactions for chemolithoautotrophic nitrification, where ammonia is a source of electrons and energy and oxygen is the terminal electron acceptor. • NO₂ +6e-> NH4* (+0.34 volts) • O₂ + 4e -> 2H₂O (+0.82 volts) Using the information given, calculate the AE for this reaction, balance the full reaction to determine the n, the number of electrons transferred when 284 moles of NH4* are oxidized. Finally, use the simplified Nernst Equation AG = -nFAE, where F = 96.5 kJ (mol e x V)-1 to determine the Gibbs Free energy available to do work! • Report your answer in kJ rounded to two decimal places. o Include trailing zeros!! Always report two decimal places even if the answer is a whole number oe.g. 18.00 not 18 Report only the numeric portion of your answer o e.g. 1.01, not 1.01 kj per mole. • Answers should ALWAYS be negative since this is a spontaneous reaction.

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Below are the reduction half reactions for chemolithoautotrophic nitrification, where ammonia is a source of electrons and energy and oxygen is the terminal electron acceptor. • NO₂ +6e-> NH4* (+0.34 volts) • O₂ + 4e -> 2H₂O (+0.82 volts) Using the information given, calculate the AE for this reaction, balance the full reaction to determine the n, the number of electrons transferred when 284 moles of NH4* are oxidized. Finally, use the simplified Nernst Equation AG = -nFAE, where F = 96.5 kJ (mol e x V)-1 to determine the Gibbs Free energy available to do work! • Report your answer in kJ rounded to two decimal places. o Include trailing zeros!! Always report two decimal places even if the answer is a whole number oe.g. 18.00 not 18 Report only the numeric portion of your answer o e.g. 1.01, not 1.01 kj per mole. • Answers should ALWAYS be negative since this is a spontaneous reaction.
Transcribed Image Text:Below are the reduction half reactions for chemolithoautotrophic nitrification, where ammonia is a source of electrons and energy and oxygen is the terminal electron acceptor. • NO₂ +6e-> NH4* (+0.34 volts) • O₂ + 4e -> 2H₂O (+0.82 volts) Using the information given, calculate the AE for this reaction, balance the full reaction to determine the n, the number of electrons transferred when 284 moles of NH4* are oxidized. Finally, use the simplified Nernst Equation AG = -nFAE, where F = 96.5 kJ (mol e x V)-1 to determine the Gibbs Free energy available to do work! • Report your answer in kJ rounded to two decimal places. o Include trailing zeros!! Always report two decimal places even if the answer is a whole number oe.g. 18.00 not 18 Report only the numeric portion of your answer o e.g. 1.01, not 1.01 kj per mole. • Answers should ALWAYS be negative since this is a spontaneous reaction.
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